CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 07

Question. The centre of a circle is C(2, k). If A(2, 1) and B(5, 2) are two points on its circumference, then the value of k is   
(A) 6
(B) 2
(C) –6
(D) –2

Answer: A

Question. The ratio in which the line joining (1, 3) an (2, 7) is divided by 3x + y = 9 is   
(A) 3 : 4
(B) 2 : 4
(C) 1 : 2
(D) 3 : 1

Answer: A

Question. The distance between the points (2k + 4, 5k) and (2k, –3 + 5k) in units is   
(A) 1
(B) 2
(C) 4
(D) 5

Answer: D

Question. The distance between the points (3k + 1, –3k) and (3k – 2, –4 –3k) (in units) is   
(A) 3k
(B) 5k
(C) 5
(D) 3

Answer: C

Question. The point which divides the line joining the points A(1, 2) and B(–1, 1) internally in the ratio 1 : 2 is _________.   
(A) (-1/5 ,5/3)
(B) (1/3 , 5/3)
(C) (–1, 5)
(D) (1, 5)

Answer: B

Question. The ratio in which the line joining (a + b, b + a) and (a – b, b – a) is divided by the point (a, b) is ___________.   
(A) b : a internally
(B) 1 : 1 internally
(C) a : b externally
(D) 2 : 1 externally

Answer: B

Question. If the line (3x -8y+5) +a(5x-3y +10) = 0 is parallel to X-axis, then a is   
(A) -(8/3)
(B) -(3/5)
(C) –2
(D) 1/2

Answer: B

Question. Find the area of a triangle formed by the lines 4x-y-8 = 0, 2x+y-10 = 0 and y = 0 (in sq units). 
(A) 5
(B) 6
(C) 4
(D) 

Answer: B

Question. Find the length of the longest side of the triangle formed by the line 3x + 4y =12with the coordinate axes.   
(A) 9
(B) 16
(C) 5
(D) 7

Answer: C

Question. Find the area of the triangle formed by the line 3x - 4y +12 = 0with the coordinate axes.   
(A) 6 units2
(B) 12 units2
(C) 1 units2
(D) 36 units2

Answer: A

Question. Find the equation of a line which divides the line segment joining the points (1, 1) and (2, 3) in the ratio 2 : 3 perpendicularly,   
(A) 5x - 5y + 2 = 0
(B) 5x + 5y + 2 = 0
(C) x + 2y - 5 = 0
(D) x + 2y + 7 = 0

Answer: C

Question. If A(-2,3) and B(2,3) are two vertices of ΔABCand G(0, 0) is its centroid, then the coordinates of C are 
(A) (0,-6)
(B) (-4,0)
(C) (4,0)
(D) (0,6)

Answer: A

Question. Let ΔABCbe a right angled triangle in which A(0, 2) and B(2, 0). Then the coordinates of C can be   
(A) (0, 0)
(B) (2, 2)
(C) either (A) or (B)
(D) none of these

Answer: C

Question. If ΔABCis a right angled triangle in which A(3, 0) and B(0, 5), then the coordinates of C can be   
(A) (5, 3)
(B) (3, 5)
(C) (0, 0)
(D) both (B) and (C)

Answer: D

Question. A triangle is formed by the lines x - y - 8,X-axis and Y-axis. Find its centroid.   
(A) (8/3 , 8/3 )
(B) (8, 8)
(C) (4, 4)
(D) (0, 0)

Answer: A

VERY SHORT ANSWER TYPE QUESTIONS

Question. If the distance between the points (x, 0) and (5, 8) is 10 units, find the value(s) of x.

Answer: –1 or 11

Question. What is the distance of the point A(3, –4) from y-axis?

Answer: 3 units

Question. In what ratio is the line joining the points P(7, 7) and Q(–4, 4) is divided by (0, –1)?

Answer: 4 : 7

Question. What are the coordinates of the centroid of triangle formed by points A(–2, 4), B(7, –3) and C(1, 5).

Answer: (2, 2)

Question. Find the coordinates of points which divides line joining (–4, 0) and (0, 6) in the ratio 1 : 3. 

Answer: (-3 , 3/2)

Question. Point A(3, –4) lies on circle of radius 5 cm with centre (0, 0). Write the coordinates of the other end of the diameter whose one end is A.

Answer: (–3, 4)

Question. Find the third vertex of a triangle if two of its vertices are (3, –6) and (–5, 2) and its centroid is at the point (2, 0).

Answer: (8, 4)

Question. What is the distance between the points (1, –2) and (–3, 2)?

Answer: 4√2 units

Question. What is the area of triangle formed by the points (–2, 0), (4, 0) and (2, 3).

Answer: 9 sq. units

Question. Find the mid-point of the line segment joining the points P(–3, 4) and Q (9, –6).

Answer: (3, –1)

Question. C is point on the perpendicular bisector of AB. What is the relation between A, B, C?

Answer: AC = BC

Question. What is the ordinate of any point on x-axis?

Answer: 0

Question. Find the coordinates of fourth vertex of the rectangle formed by the points (0, 0), (3, 0) and (0, 5).

Answer: (3, 5)

Question. What is the distance of the point (–6, 8) from origin?

Answer: 10 units

Question. What is the area of ΔABC, if points A, B and C are collinear?

Answer: zero

Question. In what ratio does the line segment joining the points (6, 4) and (1, –7) is divided internally by the axis of x?
Solution. 4 : 7

Question. Find the coordinates of the point of trisection of the line segment AB whose end points are A (2, 1) and B (5, –8).
Solution. (3, –2) and (4, –5)

Question. Find the length of median AD of a triangle whose vertices are A(–1, 3), B(1, –1) and C(5, 1).
Solution. 5 units

Question. Find the area of the quadrilateral, the coordinates of whose vertices are (1, 2), (6, 2), (5, 3) and (3, 4).
Solution. 11/2 sq.units

Question. What point on x-axis is equidistant from the points (–3, 4) and (7, 6)?
Solution. (3, 0)

Question. The coordinates of the centroid of a triangle are (1, 3) and the two vertices are (8, 5) and (–7, 6). Find the third vertex of the triangle.
Solution. (2, –2)

Question. If (3, 2), (4, 4) and (1, 3) are the mid-points of the sides of a triangle, find the coordinates of the vertices of the triangle.
Solution. (0, 1), (6, 3) and (2, 5)

Question. Three vertices of a parallelogram, taken in order are (3, 1), (2, 2) and (–2, 1) respectively. Find the coordinates of fourth vertex.
Solution. (–1, 0)

Question. Find the ratio in which the y-axis divides the segment joining (-3,6) and(12,-3).
Solution. 1/4

Question. Find the value of x for which the distance between the points P(4,-5) and Q ( 12 , x ) i s10 units.
Solution. 1, -11

Question. If the points A (4,3) and B(x,5) are on the circle with Centre O(2,3) then find the value of x.
Solution. 2

Question. What is the distance between the point A(c,0) and B(0,-c)?
Solution. √2 c

Question. For what value of p, are the points (-3,9),(2,p) and(4,-5) collinear?
Solution. -1

Question. Find the ratio in which the point P (x,2) divides the line-segments joining the points A(12,5)and B(4,- 3).Also ,find the value of x.
Solution. 3:5, x=9

Question. If the points A (-2, 1),B (a, b) and C(4,-1) are collinear and a-b=1.Find the value of a and b.
Solution. a=1, b=0

Question. In what ratio does the point (-4, 6) divides the line segment joining the points A (-6, 10)
Solution. 2/7

Question. Show that the points (3,2),(0,5),(-3,2) and (0,-1) are the vertices of a square.
Solution. Proof

Question. Point P divides the line segment joining the points A (2,1) and B(5,-8) such that AP:
AB=1:3 If P lies on the line2x-y+k=0, then find the value of k.
Solution. K=-8

Question. If the distance between the points (4,p) &(1,0) is 5, then find the value of p.
Solution. ±4

Question. If the point A(1,2), B(0,0) and C(a ,b) are collinear, then find the relation between a and b.
Solution. 2a=b

Question. Find the ratio in which the Y-axis divides the line segment joining the points (5,-6) and (-1,-4). Also find the coordinates of the point of division.
Solution. 5:1, (0,-13/3)

Question. Find the distance between the points P (7,5) and Q(2,5).
Solution. 5

Question. If P (α/3 ,4) is the midpoint of the line segment joining the points Q(-6,5) and R( -2,3),then find the value of α.
Solution. -12

Question. By distance formula, show that the points (1,-1), (5,2) and (9,5) are collinear.
Solution. Proof

Question. Find the relation between x and y if the points(2,1),(x , y)and(7,5) are collinear
Solution. 4x - 5y - 3=0

Question. Find the ratio in which the line2x+3y=10 divides the line segment joining the points (1,2) and (2,3).
Solution. 2:3

Question. Find the coordinates of the point on y-axis which is nearest to the point (- 2, 5). 
Solution. The point on y-axis that is nearest to the point(-2,5) is (0,5).

Question. If 18, a ,b ,- 3 are in A.P., then find a + b. 
Solution. Since 18, a, b, and - 3 are in A.P., Then
a - 18 = - 3 - b
or, a + b = - 3 + 18
or, a + b = 15

PRACTICE EXERCISE

Question. A point P is at a distance of √13 units from the point (5, 4). Find the coordinates of P, if its ordinate is thrice of its abscissa.
Solution. (2,6) or (7/5 , 21/5)

Question. The area of a triangle is 5 square units. Two of its vertices are (2, 1) and (3, –2). The third vertex lies on y = x + 3. Find the third vertex.
Solution. (7/2 , 13/2) or ( -3/2 , 3/2)

Question. Show that the points A(2, –1), B(3, 4), C(–2, 3) and D(–3, –2) forms a rhombus but not a square. Find the area of the rhombus also.
Solution. 24 sq. units.

Question. Find the value of p if the distance between the points (3, p) and (4, 1) is 10 units.
Solution. p = 4 or –2

Question. Find the coordinates of the point which divides the line segment joining the points (4, –7) and (–5, 6) internally in the ratio 7 : 2.
Solution. (-3, -28/9)

Question. If the coordinates of two points A and B are (3, 4) and (5, –2) respectively. Find the coordinates of any point P, if PA = PB and area of ΔPAB = 10 square units.
Solution. (7, 2) or (1, 0)

Question. Find the coordinate of point P which divides the join of A(6, 5) and B(9, 2) in the ratio 1 : 2.
Solution. P(7, 4)

Question. If the coordinates of the mid-points of the sides of a triangle are (4, –3), (4, 5) and (–2, 3). Find the coordinates of its centroid.
Solution. (2 , 5/3)

Question. Find the area of the triangle whose vertices are :
(i) A (1, –1), B(–4, 6) and C(–3, –5) (ii) P (4, 2), Q(4, 5) and R (–2, 2)
(iii) A (1, 2), B(–2, 3) and C(–3, –4) (iv) P (5, 2), Q(4, 7) and R (7, –4)
Solution. (i) 24 sq. units (ii) 9 sq. units (iii) 11 sq. units. (iv) 2 sq. units

Question. A and B are the points (1, 2) and (2, 3). Find the coordinates of point C on the line segment AB such that 3AC = 4 BC.
Solution. C (11/7 ,18/7)

Question. Find the points of trisection of the line segment joining the points :
(i) (3, –2) and (–3, –4) (ii) (1, –2) and (–3, 4)
Solution. (i) (1 , -8/3 ) , (-1 , -10/3) (ii) (-1/3 , 0 ) , (-5/3 , 2)

Question. ABCD is a square with the opposite angular points A(3, 4) and C(1, –1). Find the coordinates of B and D.
Solution. (9/2 , 1/2) and (-1 /2 , 5/2)

Question. Find the coordinates of the circumcentre of a triangle whose vertices are A(5, 1), B(11, 1) and C(11, 9).
Solution. (8, 5)

Question. Three consecutive vertices of a parallelogram are (–2, –1), (1, 0) and (4, 3). Find the coordinate of the fourth vertex.
Solution. (1, 2)

Question. Find the value of y if the distance between the points (2, –3) and (10, y) be 10 units.
Solution. y = 3 or –9

Question. Find the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).
Solution. 2 : 1

Question. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, –1), (1, 3) and (x, 8) respectively.
Solution. x = – 3 or 5

Question. Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7).
Solution. 3 : 4

Question. In what ratio the point (–3, k) divides the line segment joining the points (–5, –4) and (–2, 3). Hence, find the value of k.
Solution. 2 : 1; k = 2/3

Question. Find the ratio in which the line segment joining (–2, –3) and (5, 6) is divided by (i) x-axis (ii) y-axis. Also, find the coordinates of the point of division in each case.
Solution. (i) 1 : 2, (1/3 , 0) (ii) 2 : 5; (0,-3/7)

Question. If the coordinates of the mid-points of the sides of a triangle are (1, 2), (0, –1) and (2, –1). Find the coordinates of its vertices.
Solution. (1, –4), (3, 2) and (–1, 2)

Question. Find the point on x-axis which is equidistant from the points (–4, 6) and (5, 9).
Solution.(3, 0)

Question. Find the point on the y-axis which is equidistant from the points (3, 2) and (–5, –2).
Solution. (0, –2)

Question. The coordinates of the middle points D, E, F of the sides BC, CA and AB respectively of a ΔABC are (–3, 2), (5, –7) and (11, 7) respectively, find the coordinates of the vertices A, B and C.
Solution. (19, –2), (3, 16), (–9, –12)

Question. The line segment joining the points (–6, 8) and (8, –6) is divided into four equal parts. Find the coordinates of the point of section.
Solution. (-5/2 , 9/2);(1, 1) ;(9/2 , -5/2)

Question. For what value of x, the distance between P(x, 7) and Q(–2, 3) is 4 5 units.
Solution. x = 6 or –10

Question. Find the ratio in which the line-segment joining the points (6, 4) and (1, –7) is divided internally by x-axis.
Solution. 4 : 7

Question. The line joining the points (2, 1) and (5, –8) is trisected at the points P and Q. If P lies on the line 2x – y + k = 0, find the value of k. 
Solution. k = – 8 or – 13

Question. The centre of a circle is (3k + 1, 2k – 1). If the circle passes through the point (–1, –3) and the length of its diameter be 20 units, find the value of k.
Solution. k = 2, - 46/13

Question. If the points (10, 5), (8, 4) and (6, 6) are the mid-points of the sides of a triangle, find its vertices.
Solution. (8, 7), (12, 3), (4, 5)

Question. Find the coordinates of the points on the x-axis which are at a distance of 5 units from the point (5, 4).
Solution. (8, 0) and (2, 0)

Question. Find the coordinates of the points on the y-axis which are at a distance of 13 units from the point (12, 9).
Solution. (0, 14) and (0, 4)

Question. Two vertices of a triangle are (1, 2) and (3, 5) and its centroid is at the origin. Find the coordinates of the third vertex.
Solution. (–4, –7)

Question. A (3, 2) and B(–2, 1) are two vertices of ΔABC whose centroid G has the coordinates (5/3 , -1/3) Find the coordinates of the third vertex C of the triangle.
Solution. (4, –4)

Question. If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3) and (3, 4). Find its centroid.
Solution. (2 , 2/3)

Question. Find the ratio in which the line segment joining the points (7, 3) and (–4, 5) is divided internally by y-axis.
Solution. 7 : 4

Question. Find the centre of a circle, the end points of whose one diameter are (–3, –1) and (5, 8).
Solution. (1 , 7/2)

Question. Find the lengths of medians of a ΔABC having the vertices A(5, 1), B(1, 5) and C(–3, –1).
Solution. AD = √37 , BE = 5, CF = 2√13

Question. The line segment joining the points (3, –4) and (1, 2) is trisected at the ponts P and Q. If the coordinates of P and Q are (p, –2) and (5/3 , q) respectively, find the values of p and q.
Solution. p = 7/3 , q = 0

Question. (i) For what value of k, the points (k, –1), (5, 7) and (8, 11) are collinear?
(ii) For what value of k are the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) lie on a straight line?
Solution. (i) k = – 1 (ii) k = - 1 or 1/2

Question. The vertices of some triangles are given below alongwith their areas. Find the value of a.
Vertices Area
(i) (2, 3), (6, –2), (–2, a) 6
(ii) (3, 8), (4, a), (5, –2) 8
Solution. (i) a = 5 or 11 (ii) a = – 5 or 11

Question. Find the distance between the points:
(i) A (–3, –2) and B (–6, –7)
(ii) P(a, 0) and Q(0, b)
(iii) A (–m, –n) and B (m, n)
(iv) R( 3 √1, 1) and S(0, 3)
(v) M(3√ 3, 3√ 3) and N(0, 0)
(vi) A (a sin α, – b cos α) and B (–a cos α, b sin α)
Solution.
(i) √34 units
(ii) √a2 + b2 units
(iii) √2 m2 + n2 units
(iv) 2√ 2 units
(v) 2 √6 units
(vi) √a2 + b2 (sin α + cosα)

Question. An equilateral triangle has two vertices at the points (0, 0) and (3, 3) . Find the coordinates of the third vertex.
Solution. (0, 2 √3) or (3, √ 3)

Question. Find the centroid of a triangle whose vertices are :
(i) (–2, 3), (2, –1), (4, 0) (ii) (4, –8), (–9, 7), (18, 13)
Solution. (i)(4/3 , 2/3) (ii) (13/3 , 4)

Question. Find the coordinates of the points equidistant from three given points A(5, 1), B(–3, –7) and C(7, –1).
Solution. (2, –4)

Question. Find the coordinates of a point whose distance from (3, 5) is 5 units and that from (0, 1) is 10 units.
Solution. (6, 9)

Question. A circle passes through the points A(3, 1), B(1, –3) and C(6, –8). Find the coordinates of the centre of the circle.
Solution. (6, –3)

Question. The coordinates of a vertex of a triangle are (2, 5) and the coordinates of the mid-points of the sides passing through this vertex are (8, 0) and (9, 3). Find the coordinates of the remaining vertices.
Solution. (14, –5) and (16, 1)

Question. Find the coordinates of the point of intersection of medians of ΔABC whose vertices are A(–7, 5), B(– 1, –3) and C(5, 7).
Solution. (–1, 3)

Question. Find the area of the quadrilateral, the coordinates of whose vertices are :
(i) (–3, 2), (5, 4), (7, –6) and (–5, –4) (ii) (–4, –2), (–3, –5), (3, –2) and (2, 3)
(iii) (–5, 7), (–4, –5), (–1, –6) and (4, 5) (iv) (6, 9), (7,4), (4, 2) and (3, 7)
Solution. (i) 80 sq. units (ii) 28 sq. units (iii) 72 sq. units (iv) 17 sq. units

Question. If the area of the quadrilateral whose angular points, taken in order, are (1, 2), (–5, 6), (7, –4), (p, –2) be zero, find the value of p.
Solution. p = 3

MCQ

Question 1. Distance between the points (5, −3) and (8, 1) is
(a) 5 units
(b) 6 units
(c) 25 units
(d) none of these
Answer: (a) 5 units
In simple words: Using the distance formula, we find the change in the horizontal coordinates is 3 and the vertical change is 4. The hypotenuse of this right triangle is 5.

Exam Tip: Remember the basic distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Always identify and list your coordinates carefully before substituting them.

Question 2. If the distance between (4, 0) and (0, x) is 5 units, then x is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
In simple words: The distance formula gives the relation \( 16 + x^2 = 25 \). Solving this equation tells us that x must be equal to 3.

Exam Tip: Square both sides of the distance equation immediately to remove the radical sign and simplify the quadratic equation.

Question 3. Three points are collinear if they lie on a
(a) line
(b) plane
(c) both (a) & (b)
(d) none of these
Answer: (a) line
In simple words: By definition, collinear points are points that are situated along the exact same straight line.

Exam Tip: Make sure you know the difference between coplanar (points lying in the same plane) and collinear (points lying on the same line).

Question 4. If the points (x, y), (2, 3) & (−3, 4) are collinear then
(a) x + y = 17
(b) x − y = 17
(c) x − 5y = 17
(d) x + 5y = 17
Answer: (d) x + 5y = 17
In simple words: For points to be collinear, the area of the triangle formed by them must be zero. Simplifying this relation leads to the equation \( x + 5y = 17 \).

Exam Tip: Setting the area of the triangle formula equal to zero is a standard and robust method to verify collinearity.

Question 5. P is a point on x-axis at a distance 3 units from y-axis to its right the coordinates of P are
(a) (3, 0)
(b) (0, 3)
(c) (3, 3)
(d) (−3, 3)
Answer: (a) (3, 0)
In simple words: Since the point lies on the x-axis, its y-coordinate is 0. Being 3 units to the right of the y-axis makes the x-coordinate positive 3.

Exam Tip: Any point on the x-axis always has a y-coordinate of 0. Similarly, any point on the y-axis always has an x-coordinate of 0.

Question 6. The distance of point A(4, −3) from the origin is
(a) 1 unit
(b) 7 units
(c) 5 units
(d) 3 units
Answer: (c) 5 units
In simple words: The distance of any point \( (x, y) \) from the origin is \( \sqrt{x^2 + y^2} \). Here, \( \sqrt{16 + 9} = 5 \).

Exam Tip: Origin distance problems are common and simple. Use the direct formula \( \sqrt{x^2 + y^2} \) to save time.

Question 7. The co-ordinates of 2 points are (6, 0) & (0, 8). The co-ordinates of the midpoints are
(a) (3, 4)
(b) (6, 8)
(c) (0, 0)
(d) (4, 3)
Answer: (a) (3, 4)
In simple words: The midpoint coordinates are found by taking the average of the x-coordinates and the average of the y-coordinates.

Exam Tip: Remember the midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Question 8. What point on the x-axis is equilistant from the point A(7, 6) & B(−3, 4) ?
(a) (0, 4)
(b) (−4, 0)
(c) (3, 0)
(d) (0, 3)
Answer: (c) (3, 0)
In simple words: A point on the x-axis has coordinates \( (x, 0) \). Setting up the equal distance relation from both points shows that \( x = 3 \).

Exam Tip: Always make sure to write the coordinates of a point on the x-axis as \( (x, 0) \) right away to ensure your equations remain simple.

Question 9. X-axis divides the join of A(2, −3) & B(5, 6) in ratio
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3
Answer: (a) 1 : 2
In simple words: The ratio in which the x-axis divides a line segment is given by the formula \( -y_1 : y_2 \). This simplifies to \( 3 : 6 \), which is \( 1 : 2 \).

Exam Tip: The shortcut formula \( -y_1 : y_2 \) is highly efficient for division by the x-axis. For division by the y-axis, use \( -x_1 : x_2 \).

Question 10. If the distance of P(x, y) from A(5, 1) and B(−1, 5) is same the which of the following is true
(a) 3x = 4y
(b) x = 2y
(c) 3x = 2y
(d) x = 3y
Answer: (c) 3x = 2y
In simple words: Equating the squared distances from P to both points gives us the relation \( 12x = 8y \), which simplifies down to \( 3x = 2y \).

Exam Tip: Expand the binomial squared terms carefully and cancel out common quadratic variables from both sides of the equation.

Question 11. If P(−1, 1) is the middle point of the line segment joining Q(−3, b) and R(1, b+4), then b is
(a) 1
(b) −1
(c) 2
(d) 0
Answer: (b) −1
In simple words: The y-coordinate of the midpoint is the average of the y-coordinates. This gives \( \frac{2b+4}{2} = 1 \), which leads to \( b = -1 \).

Exam Tip: Solve the linear equation systematically. Clear fractions by multiplying both sides by 2 first.

Question 12. The 3rd vertex of an equilateral triangle whose other & vertices are (1, 1) and (−1, −1) is
(a) \( (\sqrt{3}, -\sqrt{3}) \)
(b) both a & b
(c) \( (-\sqrt{3}, \sqrt{3}) \)
(d) none of these
Answer: (b) both a & b
In simple words: An equilateral triangle can be constructed on either side of the base. Both points \( (\sqrt{3}, -\sqrt{3}) \) and \( (-\sqrt{3}, \sqrt{3}) \) are equidistant from the base vertices and form valid equilateral triangles.

Exam Tip: In symmetric geometry problems, always consider that there may be two symmetrical solutions on opposite sides of a line segment.

Question 13. The centroid of the triangle having vertices (7, 5), (5, 7) and (−3, 3) is
(a) (3, −5)
(b) (−3, 5)
(c) (−3, −5)
(d) (3, 5)
Answer: (d) (3, 5)
In simple words: The centroid is found by averaging the coordinates of the three vertices, giving \( \left(\frac{7+5-3}{3}, \frac{5+7+3}{3}\right) = (3, 5) \).

Exam Tip: The centroid of a triangle is simply the average of its vertices: \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).

Question 14. 2 vertices of ∆ABC are A(−1, 4) & B(5, 2) and its centroid is G(0, −3). The coordinate of C are
(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)
Answer: (c) (−4, −15)
In simple words: Rearranging the centroid formulas allows us to solve for the missing vertex's coordinates, which are \( -4 \) and \( -15 \).

Exam Tip: Rearrange the centroid formula: \( x_3 = 3x_G - x_1 - x_2 \) and \( y_3 = 3y_G - y_1 - y_2 \) to find the third vertex quickly.

Question 15. If the vertices of a triangle be (x1, y1), (x2, y2) and (x3, y3) then the coordinates of its centroid are
(a) \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
(b) \( \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_2}{2} \right) \)
(c) \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
(d) None of these
Answer: (c) \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
In simple words: The coordinates of the centroid are found by taking the arithmetic mean of the coordinates of all three vertices.

Exam Tip: Make sure you do not confuse the midpoint formula (dividing by 2) with the centroid formula (dividing by 3).

Question 16. The coordinate of 3rd vertex of a triangle having other two vertices (−3, 1) (0, −2) & coordinate of the centroid (0, 0) is
(a) (1, 3)
(b) (3, 1)
(c) (−1, 3)
(d) (−1, −3)
Answer: (b) (3, 1)
In simple words: The sum of the x-coordinates must equal 0, so \( -3 + 0 + x = 0 \implies x = 3 \). Similarly, for the y-coordinates, \( 1 - 2 + y = 0 \implies y = 1 \).

Exam Tip: When the centroid is at the origin \( (0, 0) \), the sum of the coordinates of all three vertices must be zero.

Question 17. The centre of circle is (−1, 3) and one end of a diameter has coordinate (2, 5). The co-ordinate of other ends are :
(a) (−4, 1)
(b) (1, −4)
(c) (4, −1)
(d) None of these
Answer: (a) (−4, 1)
In simple words: The center of the circle is the midpoint of the diameter. Solving the midpoint equations gives the coordinates of the other end as \( (-4, 1) \).

Exam Tip: A diameter is bisected by the center of the circle. This makes midpoint formulas perfect for solving this type of problem.

Question 18. If 2 vertices of a llgm are (3, 2) and (−1, 0) & the diagonals intersect at (2, −5), then other 2 vertices are
(a) (1, −10), (5, −12)
(b) (1, −12), (5, −10)
(c) (2, −10), (5, −12)
(d) (1, −10), (2, −12)
Answer: (b) (1, −12), (5, −10)
In simple words: Since the diagonals of a parallelogram bisect each other, their intersection point is the midpoint of both diagonals. We use this to find the opposite vertices.

Exam Tip: The intersection point of the diagonals of a parallelogram is the midpoint of both diagonals. Apply this midpoint property to solve opposite pairs separately.

Question 19. If the co-ordinate of middle point of the line segment joining the points (2, 1) and (1, −3) be (α, β) then which of the following is true ?
(a) α + β − 1 = 0
(b) 6α + β = 8
(c) α + 6β = 8
(d) α + β − 8 = 0
Answer: (b) 6α + β = 8
In simple words: We calculate the midpoint to be \( \alpha = 1.5 \) and \( \beta = -1 \). Checking the choices shows that only \( 6(1.5) - 1 = 8 \) is true.

Exam Tip: First calculate the numerical values of \( \alpha \) and \( \beta \) and then substitute them into the given options to find which equation holds true.

Question 20. 3 consecutive vertices of a llgm are (1, −2), (3, 6) & (5, 10). The co-ordinate of 4th vertices are
(a) (−3, 2)
(b) (2, −3)
(c) (3, 2)
(d) (−2, −3)
Answer: (c) (3, 2)
In simple words: Let the 4th vertex be \( (x, y) \). Since the midpoints of the diagonals AC and BD are identical, we solve to find that \( x = 3 \) and \( y = 2 \).

Exam Tip: Let the consecutive vertices be A, B, C, and D. Apply the midpoint property of diagonals: \( M_{AC} = M_{BD} \).

Question 21. The vertices of llgm are (3, −2), (4, 0), (6, −3) & (5, −5). The diagonals intersect at M. The coordinate of M are :
(a) \( \left( \frac{9}{2}, \frac{-5}{2} \right) \)
(b) \( \left( \frac{7}{2}, \frac{-5}{2} \right) \)
(c) \( \left( \frac{7}{2}, \frac{-3}{2} \right) \)
(d) None of these
Answer: (a) \( \left( \frac{9}{2}, \frac{-5}{2} \right) \)
In simple words: The intersection point of the diagonals is the midpoint of the diagonal joining \( (3, -2) \) and \( (6, -3) \), which is \( \left(\frac{9}{2}, -\frac{5}{2}\right) \).

Exam Tip: Select any pair of opposite vertices (e.g., first and third, or second and fourth) and find their midpoint to get the coordinates of the diagonal intersection.

Question 22. The perimeter of the ∆ formed by points (0, 0), (1, 0) & (0, 1) is
(a) \( \pm 2\sqrt{1} \)
(b) \( \sqrt{2} + 1 \)
(c) 3
(d) \( 2 + \sqrt{2} \)
Answer: (d) \( 2 + \sqrt{2} \)
In simple words: The lengths of the sides are 1, 1, and \( \sqrt{2} \). Adding these sides together gives the perimeter as \( 2 + \sqrt{2} \).

Exam Tip: Plotting the points can help you see that they form a right-angled isosceles triangle with legs of length 1 and a hypotenuse of length \( \sqrt{2} \).

Question 23. The condition that the point (x, y) may lie on the line joining (3, 4) and (−5, −6) is
(a) 5x + 4y + 1 = 0
(b) 5x − 4y + 1 = 0
(c) 5x − 4y − 1 = 0
(d) 5x + 4y − 1 = 0
Answer: (b) 5x − 4y + 1 = 0
In simple words: By finding the slope of the line as \( \frac{5}{4} \) and writing the equation, we obtain \( 5x - 4y + 1 = 0 \).

Exam Tip: Use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) first, then apply the point-slope form \( y - y_1 = m(x - x_1) \) to get the linear equation.

Question 24. The points (−4, 0), (4, 0), (0, 3) are vertices of
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer: (b) isosceles triangle
In simple words: Calculating the side lengths shows that two sides are equal (5 units each) and the base is 8 units, making it an isosceles triangle.

Exam Tip: Calculate all three side lengths using the distance formula to determine the classification of any triangle.

Question 25. The point which divides the line segment joining the points (7, −6), (3, 4) in ratio 1 : 2 internally
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Answer: (d) IV quadrant
In simple words: The section formula yields the coordinates of the point as \( \left(\frac{17}{3}, -\frac{8}{3}\right) \). Since x is positive and y is negative, it lies in the IV quadrant.

Exam Tip: Use the section formula: \( \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \) to find the coordinates, and check their signs to identify the quadrant.

Question 26. The perpendicular bisector of the line segment joining A(1, 5) and B(4, 6) cuts the y-axis at
(a) (0, 13)
(b) (0, −13)
(c) (0, 12)
(d) (13, 0)
Answer: (a) (0, 13)
In simple words: The equation of the perpendicular bisector is \( 3x + y = 13 \). Putting \( x = 0 \) gives the y-intercept at \( (0, 13) \).

Exam Tip: Remember that a line cuts the y-axis where \( x = 0 \). Find the equation of the perpendicular bisector first, then substitute \( x = 0 \).

Question 27. The co-ordinate of the point which is equidistant from the 3 vertices of the ∆AOB as shown in figure is
(a) (x, y)
(b) (y, x)
(c) \( \left( \frac{x}{2}, \frac{y}{2} \right) \)
(d) \( \left( \frac{y}{2}, \frac{x}{2} \right) \)
Answer: (c) \( \left( \frac{x}{2}, \frac{y}{2} \right) \)
In simple words: For a right-angled triangle, the point equidistant from all three vertices is the midpoint of the hypotenuse, which is \( \left(\frac{x}{2}, \frac{y}{2}\right) \).

Exam Tip: The circumcentre of any right-angled triangle always lies at the midpoint of its hypotenuse.

Question 28. A line intersects the y-axis and x-axis at the point P & Q. If (2, −5) is the midpoint of PQ, then coordinates of P & Q are
(a) (0, −5) & (2, 0)
(b) (0, 10) & (−4, 0)
(c) (0, 4) & (−10, 0)
(d) (0, −10) & (4, 0)
Answer: (d) (0, −10) & (4, 0)
In simple words: Since P is on the y-axis and Q is on the x-axis, we write them as \( (0, y_P) \) and \( (x_Q, 0) \). Their midpoint is \( \left(\frac{x_Q}{2}, \frac{y_P}{2}\right) = (2, -5) \), which yields \( (0, -10) \) and \( (4, 0) \).

Exam Tip: Assume general coordinates \( (0, y) \) for the y-intercept and \( (x, 0) \) for the x-intercept to keep your midpoint calculations clear.

Question 29. The area of ∆ with vertices (a, b + c), (b, c + a) & (c, a + b) is
(a) \( (a + b + c)^2 \)
(b) 0
(c) a + b + c
(d) abc
Answer: (b) 0
In simple words: Substituting these coordinates into the area formula of a triangle simplifies to exactly 0, meaning these three points are collinear.

Exam Tip: When evaluating a complex area formula, expand and group identical positive and negative terms to see if they cancel out to zero.

Question 30. If the distance between the points (4, p) and (1, 0) is 5, then P is
(a) 4 only
(b) ±4
(c) −4 only
(d) 0
Answer: (b) ±4
In simple words: The distance formula gives \( 9 + p^2 = 25 \), which leads to \( p^2 = 16 \). This gives two valid solutions: \( p = \pm 4 \).

Exam Tip: Always consider both the positive and negative roots when taking a square root during distance equation solving.

Question 31. If the points A(1, 2), 0(0, 0) and C(a, b) are co-linear then
(a) a = b
(b) a = 2b
(c) 2a = b
(d) a = −b
Answer: (c) 2a = b
In simple words: Since the points are collinear, the slope from the origin to A must be the same as the slope from the origin to C, giving us \( \frac{2}{1} = \frac{b}{a} \implies 2a = b \).

Exam Tip: When one of the collinear points is the origin \( (0, 0) \), the collinearity condition simplifies directly to equating the ratios of coordinates: \( \frac{y_1}{x_1} = \frac{y_2}{x_2} \).

Question 32. The area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order is
(a) 24 sq. units
(b) 12 sq. units
(c) 35 sq units
(d) 55 sq units
Answer: (a) 24 sq. units
In simple words: We find the lengths of both diagonals to be \( 4\sqrt{2} \) and \( 6\sqrt{2} \). The area is half their product, which equals 24.

Exam Tip: The area of a rhombus is given by \( \frac{1}{2} \times d_1 \times d_2 \). Calculate the lengths of both diagonals using the distance formula.

Co-ordinate Geometry Practice Sheet

Question 33. The area ∆ whose vertices are A(5, 2), B(4, 7), C(7, −4) is
(a) 5 sq units
(b) 2 sq. units
(c) 4 sq units
(d) 0 sq. units
Answer: (b) 2 sq. units
In simple words: Using the coordinate area formula, the absolute value of the expansion is 2. Thus, the area of the triangle is 2 square units.

Exam Tip: Always remember that area cannot be negative. If your formula yields a negative value, take its absolute value to get the correct answer.

Question 34. 2 vertices of ∆ABC are A(1, 4) & B(5, 2) & its centroid G(0, −3). The coordinates of C is
(a) (4, 6)
(b) (6, 4)
(c) (−15, −4)
(d) (−4, −15)
Answer: (d) (−4, −15)
In simple words: Since the x-coordinates sum to 0, we find \( -1 + 5 + x_C = 0 \implies x_C = -4 \). For y-coordinates, \( 4 + 2 + y_C = -9 \implies y_C = -15 \).

Exam Tip: Rearrange the centroid formula to calculate the missing vertex directly: \( x_3 = 3x_G - x_1 - x_2 \).

Question 35. If the points (a, 0), (0, b) and (1, 1) are collinear then \( \frac{1}{a} + \frac{1}{b} = ? \)
(a) −1
(b) 0
(c) 1
(d) 2
Answer: (c) 1
In simple words: Equating the area of the triangle formed by these three points to zero gives \( ab = a + b \). Dividing both sides by \( ab \) yields \( \frac{1}{a} + \frac{1}{b} = 1 \).

Exam Tip: This is a standard coordinate algebraic identity. Remember the simplified step where you divide the equation by the product variable \( ab \).

Short Answers (2 marks)

Question 1. If A(0, 2) is equidistant from B(3, a) and C(a, 5). Find a
Answer: Since A is equidistant from B and C, we have \( AB = AC \implies AB^2 = AC^2 \).
Using the distance formula:
\( (3-0)^2 + (a-2)^2 = (a-0)^2 + (5-2)^2 \)
\( \implies 9 + a^2 - 4a + 4 = a^2 + 9 \)
\( \implies 13 - 4a = 9 \)
\( \implies 4a = 4 \)
\( \implies a = 1 \).
Therefore, the value of a is 1.
In simple words: Setting the squared distance from A to B equal to the squared distance from A to C allows us to cancel the quadratic terms and solve for a directly to get 1.

Exam Tip: Squaring both sides of the distance equation helps to eliminate the radical sign immediately and simplifies the quadratic term cancellations.

Question 2. Point A(a, 6) and B(2, 8) are equidistant from point C(1, 1). Find a.
Answer: Since A and B are equidistant from C, we have \( AC^2 = BC^2 \).
Applying the distance formula:
\( (1-a)^2 + (1-6)^2 = (1-2)^2 + (1-8)^2 \)
\( \implies (1-a)^2 + 25 = 1 + 49 \)
\( \implies (1-a)^2 = 25 \)
\( \implies 1 - a = \pm 5 \)

Case 1: \( 1 - a = 5 \implies a = -4 \)
Case 2: \( 1 - a = -5 \implies a = 6 \)
Thus, the values of a are \( -4 \) and \( 6 \).
In simple words: Setting the distance relations equal tells us \( (1-a)^2 = 25 \). Taking the square root gives two possibilities, so a can be either \( -4 \) or \( 6 \).

Exam Tip: Always remember that taking the square root of a positive real number yields both a positive and a negative solution.

Question 3. Find the value of x such that PQ = QR, where coordinates of P, Q & R are (6, −1), (1, 3) and (x, 8) respectively.
Answer: We are given \( PQ = QR \implies PQ^2 = QR^2 \).
Using the distance formula:
\( (1-6)^2 + (3 - (-1))^2 = (x-1)^2 + (8-3)^2 \)
\( \implies (-5)^2 + 4^2 = (x-1)^2 + 5^2 \)
\( \implies 25 + 16 = (x-1)^2 + 25 \)
\( \implies (x-1)^2 = 16 \)
\( \implies x-1 = \pm 4 \)

Case 1: \( x - 1 = 4 \implies x = 5 \)
Case 2: \( x - 1 = -4 \implies x = -3 \)
Thus, the values of x are \( 5 \) and \( -3 \).
In simple words: Setting the squared distances equal yields the quadratic relation \( (x-1)^2 = 16 \). Taking the square root reveals that x can be either \( 5 \) or \( -3 \).

Exam Tip: Subtracting \( 25 \) from both sides of the equation early on saves significant computational effort.

Question 4. (i) If A(3, 2) and B(−4, −5) are equidistant from P(x, y) then show that x + y + 2 = 0.
(ii) If the distances of P(x, y) from the points A(3, 6) & B(−3, 4) are equal, prove 3x + y = 5.
Answer:

(i) Since P(x, y) is equidistant from A and B:
\( PA^2 = PB^2 \)
\( \implies (x-3)^2 + (y-2)^2 = (x+4)^2 + (y+5)^2 \)
\( \implies x^2 - 6x + 9 + y^2 - 4y + 4 = x^2 + 8x + 16 + y^2 + 10y + 25 \)
\( \implies -6x - 4y + 13 = 8x + 10y + 41 \)
\( \implies 14x + 14y + 28 = 0 \)
Dividing by 14:
\( x + y + 2 = 0 \). (Hence proved)

(ii) Since P(x, y) is equidistant from A(3, 6) and B(-3, 4):
\( PA^2 = PB^2 \)
\( \implies (x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2 \)
\( \implies x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \)
\( \implies -6x - 12y + 45 = 6x - 8y + 25 \)
\( \implies 12x + 4y = 20 \)
Dividing both sides by 4:
\( 3x + y = 5 \). (Hence proved)
In simple words: For both parts, we set the squared distance from the variable point to the two known points equal. Expanding and simplifying these equations yields the required linear relations.

Exam Tip: Be meticulous with algebraic signs when expanding squares like \( (x+4)^2 \) and \( (y+5)^2 \) to avoid any numerical errors.

Question 5. If the point (x, y) is equidistant from the points {(a + b), (b − a)} and (a − b, a + b). Prove that bx = ay.
Answer: Let \( P(x, y) \), \( A(a+b, b-a) \), and \( B(a-b, a+b) \).
Since P is equidistant from A and B, \( PA^2 = PB^2 \):
\( [x - (a+b)]^2 + [y - (b-a)]^2 = [x - (a-b)]^2 + [y - (a+b)]^2 \)
\( \implies x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(b-a) + (b-a)^2 = x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2 \)

Cancelling \( x^2, y^2, (a+b)^2 \), and \( (a-b)^2 \) from both sides:
\( -2x(a+b) - 2y(b-a) = -2x(a-b) - 2y(a+b) \)
Divide both sides by -2:
\( x(a+b) + y(b-a) = x(a-b) + y(a+b) \)
\( \implies ax + bx + by - ay = ax - bx + ay + by \)
\( \implies bx - ay = -bx + ay \)
\( \implies 2bx = 2ay \)
\( \implies bx = ay \). (Hence proved)
In simple words: Setting the squared distances equal and cancelling out common squared algebraic terms simplifies down to \( 2bx = 2ay \), which proves that \( bx = ay \).

Exam Tip: Group terms like \( (a+b)^2 \) and keep them together rather than expanding them fully, as they cancel out cleanly on both sides.

Question 6. Find the point on the x-axis which is equidistant from the points (7, 6) and (−3, 4)
Answer: Any point on the x-axis can be written as \( P(x, 0) \).
Since P is equidistant from \( A(7, 6) \) and \( B(-3, 4) \), we have \( PA^2 = PB^2 \):
\( (x-7)^2 + (0-6)^2 = (x - (-3))^2 + (0-4)^2 \)
\( \implies x^2 - 14x + 49 + 36 = x^2 + 6x + 9 + 16 \)
\( \implies -14x + 85 = 6x + 25 \)
\( \implies 20x = 60 \)
\( \implies x = 3 \).
Therefore, the required point is \( (3, 0) \).
In simple words: A point on the x-axis has a y-coordinate of 0. Equating its squared distance from both points lets us solve for the x-coordinate, which is 3.

Exam Tip: Remember that any point lying on the x-axis has coordinates of the form \( (x, 0) \), which instantly reduces one variable in your equations.

Question 7. A point is equidistant from A(−6, 4) and B(2, −8). Find its coordinates if the abscissa and ordinate are equal.
Answer: Since the abscissa (x-coordinate) and ordinate (y-coordinate) are equal, let the coordinates of the point be \( P(x, x) \).
Since P is equidistant from A and B, \( PA^2 = PB^2 \):
\( (x - (-6))^2 + (x - 4)^2 = (x - 2)^2 + (x - (-8))^2 \)
\( \implies (x+6)^2 + (x-4)^2 = (x-2)^2 + (x+8)^2 \)
\( \implies x^2 + 12x + 36 + x^2 - 8x + 16 = x^2 - 4x + 4 + x^2 + 16x + 64 \)
\( \implies 4x + 52 = 12x + 68 \)
\( \implies 8x = -16 \)
\( \implies x = -2 \).
Thus, the coordinates of the point are \( (-2, -2) \).
In simple words: We assume the point has matching coordinates \( (x, x) \). Equating its squared distance from both endpoints yields a simple linear equation, which gives the point as \( (-2, -2) \).

Exam Tip: "Abscissa and ordinate are equal" simply means \( x = y \). Setting both coordinates to \( x \) keeps the distance calculations straightforward.

Question 8. The distance between A(4, 2) and B(1, y) is 5. Find the value of y.
Answer: We are given the distance \( AB = 5 \implies AB^2 = 25 \).
Using the distance formula:
\( (1-4)^2 + (y-2)^2 = 25 \)
\( \implies (-3)^2 + (y-2)^2 = 25 \)
\( \implies 9 + (y-2)^2 = 25 \)
\( \implies (y-2)^2 = 16 \)
\( \implies y-2 = \pm 4 \)

Case 1: \( y - 2 = 4 \implies y = 6 \)
Case 2: \( y - 2 = -4 \implies y = -2 \)
Thus, the value of y can be \( 6 \) or \( -2 \).
In simple words: Using our distance formula gives \( (y-2)^2 = 16 \). Taking the square root gives two possible values for y, which are \( 6 \) and \( -2 \).

Exam Tip: Avoid expanding \( (y-2)^2 \) into a quadratic equation if you can solve it directly by taking square roots on both sides.

Question 9. A point P is at a distance of \( \sqrt{10} \) from the point (2, 3). Find the coordinates of the point P if its y coordinate is thrice of the x co-ordinate.
Answer: Let the coordinates of P be \( (x, 3x) \).
We are given that the distance to \( (2, 3) \) is \( \sqrt{10} \), so the squared distance is 10:
\( (x-2)^2 + (3x-3)^2 = 10 \)
\( \implies x^2 - 4x + 4 + 9x^2 - 18x + 9 = 10 \)
\( \implies 10x^2 - 22x + 13 = 10 \)
\( \implies 10x^2 - 22x + 3 = 0 \).
Using the quadratic formula to solve for x:
\( x = \frac{22 \pm \sqrt{(-22)^2 - 4(10)(3)}}{20} \)
\( \implies x = \frac{22 \pm \sqrt{484 - 120}}{20} \)
\( \implies x = \frac{22 \pm \sqrt{364}}{20} = \frac{22 \pm 2\sqrt{91}}{20} = \frac{11 \pm \sqrt{91}}{10} \).
The y-coordinate is \( 3x \), so:
\( y = 3\left(\frac{11 \pm \sqrt{91}}{10}\right) = \frac{33 \pm 3\sqrt{91}}{10} \).
Thus, the coordinates are \( \left(\frac{11 \pm \sqrt{91}}{10}, \frac{33 \pm 3\sqrt{91}}{10}\right) \).
In simple words: We write the point as \( (x, 3x) \). Solving the distance relation leads to a quadratic equation, which we solve to get the coordinates.

Exam Tip: Meticulously apply the quadratic formula when the quadratic expression cannot be factored with simple integers.

Question 10. Find the ordinates of the points whose abscissa is 2 & which are at a distance of \( 3\sqrt{5} \) units from the point (5, 1).
Answer: Since the abscissa is 2, the coordinates of the points are \( (2, y) \).
The distance to \( (5, 1) \) is \( 3\sqrt{5} \) units, so the squared distance is 45:
\( (5-2)^2 + (1-y)^2 = 45 \)
\( \implies 3^2 + (1-y)^2 = 45 \)
\( \implies 9 + (1-y)^2 = 45 \)
\( \implies (1-y)^2 = 36 \)
\( \implies 1 - y = \pm 6 \)

Case 1: \( 1 - y = 6 \implies y = -5 \)
Case 2: \( 1 - y = -6 \implies y = 7 \)
Therefore, the required ordinates are \( -5 \) and \( 7 \).
In simple words: Placing the known coordinate in our distance formula gives \( (1-y)^2 = 36 \). Solving this simple equation gives the two ordinates as \( -5 \) and \( 7 \).

Exam Tip: Remember that "ordinate" refers to the y-coordinate of a point, while "abscissa" refers to the x-coordinate.

Question 11. A is a point on the y-axis whose ordinate is 5 & B is a point whose coordinates are (−3, 1). Calculate AB.
Answer: Since A lies on the y-axis and has an ordinate of 5, its coordinates are \( A(0, 5) \).
The coordinates of B are \( B(-3, 1) \).
Using the distance formula:
\( AB = \sqrt{(-3-0)^2 + (1-5)^2} \)
\( \implies AB = \sqrt{(-3)^2 + (-4)^2} \)
\( \implies AB = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
Thus, the distance AB is 5 units.
In simple words: The coordinates of A are \( (0, 5) \). Applying our standard distance formula to A and B gives us exactly 5 units.

Exam Tip: Writing the coordinates of point A correctly as \( (0, 5) \) is the key first step to solving this problem.

Question 12. Distance between A(x, y) and B(−4, 7) is \( \sqrt{41} \) . Find x, y is it’s a’s ordinate is thrice of its abscissa.
Answer: Let the coordinates of A be \( (x, 3x) \).
Since the distance to B is \( \sqrt{41} \), the squared distance is 41:
\( (-4 - x)^2 + (7 - 3x)^2 = 41 \)
\( \implies (x+4)^2 + (3x-7)^2 = 41 \)
\( \implies x^2 + 8x + 16 + 9x^2 - 42x + 49 = 41 \)
\( \implies 10x^2 - 34x + 65 = 41 \)
\( \implies 10x^2 - 34x + 24 = 0 \)
Divide the entire equation by 2:
\( 5x^2 - 17x + 12 = 0 \)
Splitting the middle term:
\( 5x^2 - 5x - 12x + 12 = 0 \)
\( \implies 5x(x - 1) - 12(x - 1) = 0 \)
\( \implies (5x - 12)(x - 1) = 0 \)
\( \implies x = 1 \) or \( x = \frac{12}{5} \).

Case 1: If \( x = 1 \), then \( y = 3 \). Point is \( (1, 3) \).
Case 2: If \( x = \frac{12}{5} \), then \( y = \frac{36}{5} \). Point is \( \left(\frac{12}{5}, \frac{36}{5}\right) \).
In simple words: We assume point A has coordinates \( (x, 3x) \). Solving the distance relation leads to a quadratic equation, which gives two possible points: \( (1, 3) \) or \( \left(\frac{12}{5}, \frac{36}{5}\right) \).

Exam Tip: Be sure to compute and state both the x and y values for both cases to get full marks.

Question 13. Using the distance formula show that the points (−1, −1), (2, 3) and (8, 11) are collinear.
Answer: Let the points be \( A(-1, -1) \), \( B(2, 3) \), and \( C(8, 11) \).
Let's calculate the distances between each pair:
\( AB = \sqrt{(2 - (-1))^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \)
\( BC = \sqrt{(8-2)^2 + (11-3)^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \)
\( AC = \sqrt{(8 - (-1))^2 + (11 - (-1))^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \)

We can observe that:
\( AB + BC = 5 + 10 = 15 = AC \).
Since the sum of two smaller distances is exactly equal to the largest distance, the points are collinear.
In simple words: We find the distances between the points are 5, 10, and 15. Since \( 5 + 10 = 15 \), the points must lie in a perfectly straight line.

Exam Tip: To prove collinearity using the distance formula, you must always show that the sum of the two shorter segments equals the longest segment.

Question 14. Find the centroid of the triangle whose vertices are (7, −8), (−9, 7), (8, 3).
Answer: The coordinates of the centroid \( G(x, y) \) are found by averaging the coordinates of the three vertices:
\( x = \frac{7 + (-9) + 8}{3} = \frac{6}{3} = 2 \)
\( y = \frac{-8 + 7 + 3}{3} = \frac{2}{3} \).
Therefore, the coordinates of the centroid are \( \left(2, \frac{2}{3}\right) \).
In simple words: We take the average of the x-coordinates and the average of the y-coordinates to get the centroid as \( \left(2, \frac{2}{3}\right) \).

Exam Tip:Centroid calculations are very straightforward. Keep fractional results in their simplest form.

Question 15. For what value of x are the points (7, x), (−5, 2) and (3, 6) collinear.
Answer: For the three points to be collinear, the area of the triangle formed by them must be 0:
\( \frac{1}{2} | 7(2 - 6) + (-5)(6 - x) + 3(x - 2) | = 0 \)
\( \implies | 7(-4) - 30 + 5x + 3x - 6 | = 0 \)
\( \implies | -28 - 36 + 8x | = 0 \)
\( \implies | 8x - 64 | = 0 \)
\( \implies 8x = 64 \)
\( \implies x = 8 \).
Thus, the value of x is 8.
In simple words: Setting our triangle area formula to zero and expanding the expression yields \( 8x - 64 = 0 \), showing that x must be 8.

Exam Tip: The factor of \( \frac{1}{2} \) in front of the area formula can be dropped immediately when setting the area to zero.

3 Mark Question :

Question 1. Find the area of a rhombus is its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order.
Answer: Let the vertices of the rhombus taken in order be \( A(3, 0) \), \( B(4, 5) \), \( C(-1, 4) \), and \( D(-2, -1) \).
The diagonals are AC and BD. Let's calculate their lengths:
\( d_1 = AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2} \)
\( d_2 = BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2} \)

The area of a rhombus is half the product of its diagonals:
\( \text{Area} = \frac{1}{2} \cdot d_1 \cdot d_2 \)
\( = \frac{1}{2} \cdot 4\sqrt{2} \cdot 6\sqrt{2} \)
\( = \frac{1}{2} \cdot 24 \cdot 2 = 24 \) sq. units.
Thus, the area of the rhombus is 24 square units.
In simple words: We find the lengths of both diagonals to be \( 4\sqrt{2} \) and \( 6\sqrt{2} \). Multiplying them and dividing by 2 gives us 24 square units.

Exam Tip: Remember the area formula for a rhombus: \( \text{Area} = \frac{1}{2} \times d_1 \times d_2 \). Be careful to pair opposite vertices for the diagonals, not adjacent ones.

Question 2. Two vertices of an isosceles triangle are (−2, 5) and (4, −1). Find the third vertex if the length of equal sides is \( 3\sqrt{2} \) units.
Answer: Let the vertices of the isosceles triangle be \( A(-2, 5) \), \( B(4, -1) \), and \( C(x, y) \).
Since the two equal sides are AC and BC, we have \( AC^2 = BC^2 = (3\sqrt{2})^2 = 18 \).
Setting \( AC^2 = BC^2 \):
\( (x+2)^2 + (y-5)^2 = (x-4)^2 + (y+1)^2 \)
\( \implies x^2 + 4x + 4 + y^2 - 10y + 25 = x^2 - 8x + 16 + y^2 + 2y + 1 \)
\( \implies 4x - 10y + 29 = -8x + 2y + 17 \)
\( \implies 12x - 12y + 12 = 0 \)
\( \implies x - y + 1 = 0 \)
\( \implies y = x + 1 \) (Equation 1)

Now, substitute \( y = x + 1 \) into the equation \( AC^2 = 18 \):
\( (x+2)^2 + (x+1-5)^2 = 18 \)
\( \implies (x+2)^2 + (x-4)^2 = 18 \)
\( \implies x^2 + 4x + 4 + x^2 - 8x + 16 = 18 \)
\( \implies 2x^2 - 4x + 20 = 18 \)
\( \implies 2x^2 - 4x + 2 = 0 \)
\( \implies x^2 - 2x + 1 = 0 \)
\( \implies (x - 1)^2 = 0 \)
\( \implies x = 1 \).
Using Equation 1 to find y:
\( y = 1 + 1 = 2 \).
Thus, the coordinates of the third vertex are \( C(1, 2) \).
In simple words: Setting the squared distances from the third vertex to both endpoints equal tells us \( y = x+1 \). Substituting this relation back into the distance formula gives \( x=1 \) and \( y=2 \), so the point is \( (1, 2) \).

Exam Tip: Equate the squared side lengths to create a linear relationship between \( x \) and \( y \) before attempting to solve the quadratic equation.

Question 3. Prove that A(8, −10), B(7, −3) and C(0, −4) are the vertices of a right angled triangle.
Answer: Let us calculate the squared side lengths of the triangle:
\( AB^2 = (7-8)^2 + (-3 - (-10))^2 = (-1)^2 + 7^2 = 1 + 49 = 50 \)
\( BC^2 = (0-7)^2 + (-4 - (-3))^2 = (-7)^2 + (-1)^2 = 49 + 1 = 50 \)
\( AC^2 = (0-8)^2 + (-4 - (-10))^2 = (-8)^2 + 6^2 = 64 + 36 = 100 \)

We can observe that:
\( AB^2 + BC^2 = 50 + 50 = 100 = AC^2 \).
Since the sum of the squares of two sides equals the square of the longest side, by the converse of Pythagoras' theorem, \( \Delta ABC \) is a right-angled triangle.
In simple words: The squared side lengths are 50, 50, and 100. Since \( 50 + 50 = 100 \), this satisfies Pythagoras' theorem, proving it is a right triangle.

Exam Tip: Always show that the relation \( a^2 + b^2 = c^2 \) holds true to verify a right-angled triangle, and mention "Converse of Pythagoras' Theorem".

Question 4. Show that the points (7, 10), (−2, 5) and (3, −4) are the vertices of an isosceles right triangle.
Answer: Let the vertices be \( A(7, 10) \), \( B(-2, 5) \), and \( C(3, -4) \).
Let's calculate the squared side lengths:
\( AB^2 = (-2-7)^2 + (5-10)^2 = (-9)^2 + (-5)^2 = 81 + 25 = 106 \)
\( BC^2 = (3 - (-2))^2 + (-4-5)^2 = 5^2 + (-9)^2 = 25 + 81 = 106 \)
\( AC^2 = (3-7)^2 + (-4-10)^2 = (-4)^2 + (-14)^2 = 16 + 196 = 212 \)

We can observe two properties:
1. \( AB^2 = BC^2 = 106 \implies AB = BC \), so the triangle is isosceles.
2. \( AB^2 + BC^2 = 106 + 106 = 212 = AC^2 \), which satisfies Pythagoras' theorem, so the triangle is right-angled.
Therefore, the points form an isosceles right triangle.
In simple words: The squared side lengths are 106, 106, and 212. Since two sides are equal and \( 106 + 106 = 212 \), it is both isosceles and right-angled.

Exam Tip: For an "isosceles right triangle", you must explicitly prove both conditions: two equal sides, and the Pythagorean relation.

Question 5. Show that the points (0, 0), (3a, \( \sqrt{3} \)a ) and (3a, − \( \sqrt{3} \)a ) are the vertices of an equilateral triangle.
Answer: Let the vertices be \( A(0, 0) \), \( B(3a, \sqrt{3}a) \), and \( C(3a, -\sqrt{3}a) \).
Let's calculate the squared side lengths:
\( AB^2 = (3a - 0)^2 + (\sqrt{3}a - 0)^2 = 9a^2 + 3a^2 = 12a^2 \)
\( BC^2 = (3a - 3a)^2 + (-\sqrt{3}a - \sqrt{3}a)^2 = 0 + (-2\sqrt{3}a)^2 = 12a^2 \)
\( AC^2 = (3a - 0)^2 + (-\sqrt{3}a - 0)^2 = 9a^2 + 3a^2 = 12a^2 \)

Since \( AB^2 = BC^2 = AC^2 = 12a^2 \implies AB = BC = AC = \sqrt{12}a \), all three sides are equal.
Therefore, the points form an equilateral triangle.
In simple words: Calculating the distance between each pair of vertices shows that all three side lengths are exactly equal to \( \sqrt{12}a \), proving it is an equilateral triangle.

Exam Tip: Be careful with variables in the coordinates. The variable \( a \) will carry through your calculation but will verify that the sides are equal.

Question 6. Derive a relationship between a & b where P(13, 8), Q(a, b) and R(6, 0) are the vertices of right triangle with ∠R = 90o.
Answer: Since the triangle is right-angled at R, by Pythagoras' theorem, we have \( PR^2 + RQ^2 = PQ^2 \).
Using the distance formula:
\( PR^2 = (6-13)^2 + (0-8)^2 = (-7)^2 + (-8)^2 = 49 + 64 = 113 \)
\( RQ^2 = (a-6)^2 + (b-0)^2 = a^2 - 12 a + 36 + b^2 \)
\( PQ^2 = (a-13)^2 + (b-8)^2 = a^2 - 26a + 169 + b^2 - 16b + 64 \)

Substituting these into the relation:
\( 113 + (a^2 - 12a + 36 + b^2) = a^2 - 26a + 169 + b^2 - 16b + 64 \)
\( \implies a^2 + b^2 - 12a + 149 = a^2 + b^2 - 26a - 16b + 233 \)
Subtracting \( a^2 + b^2 \) from both sides:
\( -12a + 149 = -26a - 16b + 233 \)
\( \implies 26a - 12a + 16b = 233 - 149 \)
\( \implies 14a + 16b = 84 \)
Dividing by 2:
\( 7a + 8b = 42 \).
Thus, the required relationship is \( 7a + 8b = 42 \).
In simple words: Since the angle at R is 90 degrees, we apply Pythagoras' theorem. Substituting the coordinate distances and simplifying the algebraic terms yields the relation \( 7a + 8b = 42 \).

Exam Tip: Be sure to write down the theorem relation \( PR^2 + RQ^2 = PQ^2 \) corresponding to \( \angle R = 90^\circ \) before writing out the coordinates.

Question 7. Show that A(3, 5), B(−1, 3), C(0, −1) & D(4, 1) form a parallelogram ABCD.
Answer: For a quadrilateral ABCD to be a parallelogram, its diagonals AC and BD must bisect each other, meaning they share the same midpoint.
Let's calculate the midpoints of both diagonals:
Midpoint of diagonal AC: \( \left(\frac{3+0}{2}, \frac{5-1}{2}\right) = (1.5, 2) \)
Midpoint of diagonal BD: \( \left(\frac{-1+4}{2}, \frac{3+1}{2}\right) = (1.5, 2) \)

Since the midpoint of diagonal AC is identical to the midpoint of diagonal BD, the diagonals bisect each other.
Therefore, ABCD is a parallelogram.
In simple words: The exact midpoint of both diagonal AC and diagonal BD is \( (1.5, 2) \). Since they share the same midpoint, the diagonals bisect each other, proving ABCD is a parallelogram.

Exam Tip: Showing that the midpoints of the diagonals are identical is the quickest and cleanest way to prove a quadrilateral is a parallelogram.

Question 8. Show that the points A(2, −2), B(14, 10), C(11, 13) & D(−1, 1) are the vertices of a rectangle.
Answer: A rectangle is a parallelogram with equal diagonals.
First, let's verify if ABCD is a parallelogram by checking midpoints:
Midpoint of AC: \( \left(\frac{2+11}{2}, \frac{-2+13}{2}\right) = (6.5, 5.5) \)
Midpoint of BD: \( \left(\frac{14-1}{2}, \frac{10+1}{2}\right) = (6.5, 5.5) \)
Since the midpoints are identical, ABCD is a parallelogram.

Now, let's check the diagonal lengths:
\( AC^2 = (11-2)^2 + (13 - (-2))^2 = 9^2 + 15^2 = 81 + 225 = 306 \)
\( BD^2 = (-1-14)^2 + (1-10)^2 = (-15)^2 + (-9)^2 = 225 + 81 = 306 \)
Since \( AC^2 = BD^2 \implies AC = BD \), the diagonals are equal.
Therefore, the vertices form a rectangle.
In simple words: First we show the diagonals bisect each other because they have the same midpoint, \( (6.5, 5.5) \). Then we show the diagonals are of equal length (\( \sqrt{306} \) units), proving it is a rectangle.

Exam Tip: Proving a rectangle requires two distinct steps: first prove it is a parallelogram, then prove the diagonals are of equal length.

Question 9. The 3 vertices of a rhombus taken in order are (3, 4), (−2, 3) and (−3, −2). What are the co-ordinates of the 4th vertex ?
Answer: Let the vertices be \( A(3, 4) \), \( B(-2, 3) \), \( C(-3, -2) \), and let the 4th vertex be \( D(x, y) \).
Since a rhombus is a parallelogram, its diagonals AC and BD bisect each other, meaning they share the same midpoint:
\( M_{AC} = M_{BD} \)
\( \implies \left(\frac{3-3}{2}, \frac{4-2}{2}\right) = \left(\frac{-2+x}{2}, \frac{3+y}{2}\right) \)
\( \implies (0, 1) = \left(\frac{-2+x}{2}, \frac{3+y}{2}\right) \)

Equating coordinates:
1. \( \frac{-2+x}{2} = 0 \implies -2+x = 0 \implies x = 2 \)
2. \( \frac{3+y}{2} = 1 \implies 3+y = 2 \implies y = -1 \)
Thus, the coordinates of the 4th vertex are \( D(2, -1) \).
In simple words: The diagonals share the same midpoint, which is \( (0, 1) \). Setting up the midpoint equation for diagonal BD lets us find that the 4th vertex is \( (2, -1) \).

Exam Tip: Using the diagonal midpoint bisection property is the most direct way to solve for the missing vertex of any parallelogram, rhombus, rectangle, or square.

Question 10. (i) Show that the points A(3, 5), B(1, 1), C(5, 3) and D(7, 7) all the vertices of a rhombus.
(ii) Show that A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus.
Answer:

(i) For A(3, 5), B(1, 1), C(5, 3), D(7, 7):
First check the diagonal midpoints:
Midpoint of AC: \( \left(\frac{3+5}{2}, \frac{5+3}{2}\right) = (4, 4) \).
Midpoint of BD: \( \left(\frac{1+7}{2}, \frac{1+7}{2}\right) = (4, 4) \).
Since the midpoints are identical, ABCD is a parallelogram.
Now calculate adjacent side lengths:
\( AB^2 = (1-3)^2 + (1-5)^2 = (-2)^2 + (-4)^2 = 4 + 16 = 20 \)
\( BC^2 = (5-1)^2 + (3-1)^2 = 4^2 + 2^2 = 16 + 4 = 20 \)
Since \( AB = BC = \sqrt{20} \), the adjacent sides are equal. Thus, ABCD is a rhombus.

(ii) For A(-3, 2), B(-5, -5), C(2, -3), D(4, 4):
First check the diagonal midpoints:
Midpoint of AC: \( \left(\frac{-3+2}{2}, \frac{2-3}{2}\right) = (-0.5, -0.5) \).
Midpoint of BD: \( \left(\frac{-5+4}{2}, \frac{-5+4}{2}\right) = (-0.5, -0.5) \).
Since the midpoints are identical, ABCD is a parallelogram.
Now check adjacent side lengths:
\( AB^2 = (-5 - (-3))^2 + (-5-2)^2 = (-2)^2 + (-7)^2 = 4 + 49 = 53 \)
\( BC^2 = (2 - (-5))^2 + (-3 - (-5))^2 = 7^2 + 2^2 = 49 + 4 = 53 \).
Since \( AB = BC = \sqrt{53} \), the adjacent sides are equal. Thus, ABCD is a rhombus.
In simple words: For both parts, we prove the shape is a parallelogram by showing the diagonals bisect each other. Then, we show adjacent sides have equal lengths, which confirms the shape is a rhombus.

Exam Tip: A rhombus has all sides equal. Proving that it is a parallelogram with two adjacent sides equal is quicker than calculating all four side lengths.

Question 11. Find the coordinates of the circumcentre of a triangle whose vertices are (3, 7), (0, 6) & (−1, 5). Find its circumradius.
Answer: Let \( A(3, 7) \), \( B(0, 6) \), and \( C(-1, 5) \) be the vertices. Let the circumcentre be \( P(x, y) \).
Since P is equidistant from A, B, and C:
\( PA^2 = PB^2 \)
\( \implies (x-3)^2 + (y-7)^2 = x^2 + (y-6)^2 \)
\( \implies x^2 - 6x + 9 + y^2 - 14y + 49 = x^2 + y^2 - 12y + 36 \)
\( \implies -6x - 2y + 58 = 36 \)
\( \implies 6x + 2y = 22 \)
\( \implies 3x + y = 11 \) (Equation 1)

Also, \( PB^2 = PC^2 \):
\( x^2 + (y-6)^2 = (x+1)^2 + (y-5)^2 \)
\( \implies x^2 + y^2 - 12y + 36 = x^2 + 2x + 1 + y^2 - 10y + 25 \)
\( \implies -12y + 36 = 2x - 10y + 26 \)
\( \implies 2x + 2y = 10 \)
\( \implies x + y = 5 \) (Equation 2)

Subtracting Equation 2 from Equation 1:
\( 2x = 6 \implies x = 3 \)
Substitute \( x = 3 \) in Equation 2:
\( 3 + y = 5 \implies y = 2 \).
The circumcentre is \( P(3, 2) \).
The circumradius \( R \) is the distance PB:
\( R = \sqrt{(3-0)^2 + (2-6)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 \) units.
Thus, the circumcentre is \( (3, 2) \) and the circumradius is 5 units.
In simple words: The circumcentre is equidistant from all three corners. Setting these distance equations equal lets us solve for the center point, which is \( (3, 2) \). The distance from this point to any corner is the radius, which is 5 units.

Exam Tip: Equate \( PA^2 = PB^2 \) and \( PB^2 = PC^2 \) separately to generate two simple linear equations in \( x \) and \( y \), and solve them simultaneously.

Question 12. Find the radius of the circle whose centre is (0, 0) & which passes through (−6, 8)
Answer: The radius \( R \) of the circle is the distance between the centre \( (0, 0) \) and the point \( (-6, 8) \) on the circumference:
\( R = \sqrt{(-6 - 0)^2 + (8 - 0)^2} \)
\( \implies R = \sqrt{(-6)^2 + 8^2} \)
\( \implies R = \sqrt{36 + 64} = \sqrt{100} = 10 \) units.
Thus, the radius of the circle is 10 units.
In simple words: Since the circle goes through \( (-6, 8) \) from the origin, the radius is the distance between these points, which calculates to exactly 10 units.

Exam Tip: Origin distance problems are simple. Just use the formula \( R = \sqrt{x^2 + y^2} \) to calculate the radius in one step.

Question 13. If (7, 1), (x, 9) and (−1, y) are 3 concyclic points whose centre is (3, 4). Find x & y.
Answer: Let \( C(3, 4) \) be the centre of the circle, and let the points be \( A(7, 1) \box \), \( B(x, 9) \), and \( D(-1, y) \).
Since the points are concyclic (lying on the same circle), they are equidistant from the centre:
\( CA^2 = CB^2 = CD^2 = R^2 \)

First, let's find \( R^2 \) using point A:
\( R^2 = CA^2 = (7-3)^2 + (1-4)^2 = 4^2 + (-3)^2 = 16 + 9 = 25 \)

Now, use point B to find x:
\( CB^2 = 25 \)
\( \implies (x-3)^2 + (9-4)^2 = 25 \)
\( \implies (x-3)^2 + 5^2 = 25 \)
\( \implies (x-3)^2 = 0 \)
\( \implies x = 3 \)

Now, use point D to find y:
\( CD^2 = 25 \)
\( \implies (-1-3)^2 + (y-4)^2 = 25 \)
\( \implies (-4)^2 + (y-4)^2 = 25 \)
\( \implies 16 + (y-4)^2 = 25 \)
\( \implies (y-4)^2 = 9 \)
\( \implies y-4 = \pm 3 \)
If \( y - 4 = 3 \implies y = 7 \).
If \( y - 4 = -3 \implies y = 1 \).
Thus, the values are \( x = 3 \) and \( y = 7 \) or \( 1 \).
In simple words: The distance from the center to any point on the circle is 5. Solving this distance relation for the other points gives \( x = 3 \), and y can be either \( 7 \) or \( 1 \).

Exam Tip: First calculate the radius squared \( R^2 \) using the completely known point before setting up equations for the other variables.

Question 14. 3 consecutive vertices of a llgm ABCD are A(1, 2), B(1, 0) and C(4, 0). Find the 4th vertex D.
Answer: Let the 4th vertex be \( D(x, y) \).
Since ABCD is a parallelogram, its diagonals AC and BD bisect each other, sharing the same midpoint:
\( M_{AC} = M_{BD} \)
\( \implies \left(\frac{1+4}{2}, \frac{2+0}{2}\right) = \left(\frac{1+x}{2}, \frac{0+y}{2}\right) \)
\( \implies \left(\frac{5}{2}, 1\right) = \left(\frac{1+x}{2}, \frac{y}{2}\right) \)

Equating coordinates:
1. \( \frac{1+x}{2} = \frac{5}{2} \implies 1+x = 5 \implies x = 4 \)
2. \( \frac{y}{2} = 1 \implies y = 2 \)
Therefore, the coordinates of the 4th vertex are \( D(4, 2) \).
In simple words: The diagonals must have the same midpoint, which is \( (2.5, 1) \). Setting up the midpoint formula for diagonal BD tells us that the 4th vertex is \( (4, 2) \).

Exam Tip: Midpoint equivalence of diagonals is the most reliable way to find the fourth vertex of a parallelogram.

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Question 15. If (1, 2), (4, y), (x, 6) and (3, 5) ar the vertices of a parallelogram taken in order. Find x & y.
Answer: Let the consecutive vertices of the parallelogram be \( A(1, 2) \), \( B(4, y) \box \), \( C(x, 6) \), and \( D(3, 5) \).
The diagonals AC and BD bisect each other, so their midpoints are identical:
\( M_{AC} = M_{BD} \)
\( \implies \left(\frac{1+x}{2}, \frac{2+6}{2}\right) = \left(\frac{4+3}{2}, \frac{y+5}{2}\right) \)
\( \implies \left(\frac{1+x}{2}, 4\right) = \left(\frac{7}{2}, \frac{y+5}{2}\right) \)

Equating coordinates:
1. \( \frac{1+x}{2} = \frac{7}{2} \implies 1+x = 7 \implies x = 6 \)
2. \( \frac{y+5}{2} = 4 \implies y+5 = 8 \implies y = 3 \)
Thus, the values are \( x = 6 \) and \( y = 3 \).
In simple words: The diagonals share the same midpoint. Equating their coordinates gives \( x = 6 \) and \( y = 3 \).

Exam Tip: Be sure to keep the vertices in the stated order when checking diagonal pairs (AC and BD).

Question 16. P(5, a), Q(−4, 3) & R(b, −2) are the vertices of a ∆PQR, whose centroid is the origin. Find a & b.
Answer: Since the centroid \( G(x_G, y_G) \) is at the origin \( (0, 0) \):
1. For the x-coordinate:
\( x_G = \frac{5 - 4 + b}{3} = 0 \)
\( \implies 1 + b = 0 \)
\( \implies b = -1 \)

2. For the y-coordinate:
\( y_G = \frac{a + 3 - 2}{3} = 0 \)
\( \implies a + 1 = 0 \)
\( \implies a = -1 \).
Thus, the values are \( a = -1 \) and \( b = -1 \).
In simple words: When the centroid is at \( (0, 0) \), the sum of the coordinates of all three points must be zero. This gives \( a = -1 \) and \( b = -1 \).

Exam Tip: When the centroid is at the origin, the sum of all x-coordinates is zero, and the sum of all y-coordinates is zero.

Question 17. Find the value of P for which the points A(−5, 1), B(1, p) and C(4, −2) are collinear.
Answer: For the points to be collinear, the area of the triangle must be 0:
\( \frac{1}{2} | -5(p - (-2)) + 1(-2 - 1) + 4(1 - p) | = 0 \)
\( \implies | -5(p+2) - 3 + 4 - 4p | = 0 \)
\( \implies | -5p - 10 + 1 - 4p | = 0 \)
\( \implies | -9p - 9 | = 0 \)
\( \implies -9p - 9 = 0 \)
\( \implies -9p = 9 \)
\( \implies p = -1 \).
Thus, the value of p is -1.
In simple words: We set our area formula to zero. Simplifying the algebraic expression leads to the relation \( -9p - 9 = 0 \), showing that p must be -1.

Exam Tip: Ensure that you expand the terms inside the absolute value brackets carefully to avoid missing any negative signs.

Question 18. Find a relation between x & y if the points (x, y), (1, 2) and (7, 0) are collinear.
Answer: For collinear points, the area of the triangle is 0:
\( \frac{1}{2} | x(2 - 0) + 1(0 - y) + 7(y - 2) | = 0 \)
\( \implies | 2x - y + 7y - 14 | = 0 \)
\( \implies | 2x + 6y - 14 | = 0 \)
\( \implies 2x + 6y - 14 = 0 \)
Dividing by 2:
\( x + 3y - 7 = 0 \).
Thus, the relation is \( x + 3y - 7 = 0 \).
In simple words: Setting our collinear relation to zero and expanding gives \( 2x + 6y - 14 = 0 \). Dividing by 2 yields the final relation \( x + 3y - 7 = 0 \).

Exam Tip: Simplify the final equation by dividing by any common factors to present the relation in its simplest form.

Question 19. Find the values of k for which the points A(−5, 1), B(1, k) and C(4, −2) are collinear. Find the ratio in which B divides AC.
Answer: From the collinearity condition, we find \( k = -1 \) (as solved in Question 17).
Let B(1, -1) divide AC in the ratio \( \lambda : 1 \) internally.
Using the section formula for the x-coordinate:
\( x_B = \frac{\lambda(4) + 1(-5)}{\lambda + 1} = 1 \)
\( \implies 4\lambda - 5 = \lambda + 1 \)
\( \implies 3\lambda = 6 \)
\( \implies \lambda = 2 \).
Thus, \( k = -1 \), and B divides AC in the ratio 2 : 1 internally.
In simple words: The collinear condition gives \( k = -1 \). Applying the section formula shows that point B divides the segment in the ratio of 2 to 1 internally.

Exam Tip: Using the x-coordinate of the dividing point is usually simpler than using the y-coordinate when solving for the division ratio.

Question 20. If the points A(x, y), B(5, 5) and C(10, 7) are collinear show that 5y − 2x = 15.
Answer: Since the points are collinear, the area of the triangle is 0:
\( \frac{1}{2} | x(5 - 7) + 5(7 - y) + 10(y - 5) | = 0 \)
\( \implies | -2x + 35 - 5y + 10 y - 50 | = 0 \)
\( \implies | -2x + 5y - 15 | = 0 \)
\( \implies 5y - 2x - 15 = 0 \)
\( \implies 5y - 2x = 15 \). (Hence proved)
In simple words: Setting the area of the triangle to zero and expanding the expression yields \( -2x + 5y - 15 = 0 \), which directly proves \( 5y - 2x = 15 \).

Exam Tip: Be sure to keep the order of terms same as shown in the prove statement to make verification simple.

Question 21. (i) Find the coordinates of x, y such that the point P(x, y) lies on the line segment joining A(1, 4) & B(−3, 16)
(ii) If P(x, y) lies on the line segment joining the points (1, −3) and (−4, 2), prove that x + y + 2 = 0.
Answer:

(i) Any point P(x, y) lying on the line joining A and B satisfies the equation of the line AB:
Slope of AB, \( m = \frac{16 - 4}{-3 - 1} = \frac{12}{-4} = -3 \).
Equation of the line:
\( y - 4 = -3(x - 1) \)
\( \implies y - 4 = -3x + 3 \)
\( \implies 3x + y - 7 = 0 \).
Thus, the coordinates \( (x, y) \) must satisfy the relationship \( 3x + y = 7 \).

(ii) Since P(x, y) lies on the segment joining (1, -3) and (-4, 2), the points are collinear:
\( x(-3 - 2) + 1(2 - y) + (-4)(y - (-3)) = 0 \)
\( \implies -5x + 2 - y - 4y - 12 = 0 \)
\( \implies -5x - 5y - 10 = 0 \)
Divide the entire equation by -5:
\( x + y + 2 = 0 \). (Hence proved)
In simple words: For the first part, the relation between the coordinates is \( 3x + y = 7 \). For the second part, setting the area formula of collinear points to zero directly yields \( x + y + 2 = 0 \).

Exam Tip: Proving that a point lies on a line segment is identical to proving that the three points are collinear.

4 Mark Questions

Question 1. Find the co-ordinates of the points of trisection of the line segment joing the points (2, 3) & (6, 5)
Answer: Let \( A(2, 3) \) and \( B(6, 5) \). The points of trisection divide AB in the ratio 1 : 2 and 2 : 1 internally.

1. For the first point \( P \), dividing AB in the ratio 1 : 2:
\( x_P = \frac{1(6) + 2(2)}{1 + 2} = \frac{10}{3} \)
\( y_P = \frac{1(5) + 2(3)}{1 + 2} = \frac{11}{3} \)
So, \( P = \left(\frac{10}{3}, \frac{11}{3}\right) \).

2. For the second point \( Q \), dividing AB in the ratio 2 : 1:
\( x_Q = \frac{2(6) + 1(2)}{2 + 1} = \frac{14}{3} \)
\( y_Q = \frac{2(5) + 1(3)}{2 + 1} = \frac{13}{3} \)
So, \( Q = \left(\frac{14}{3}, \frac{13}{3}\right) \).
Therefore, the coordinates of the points of trisection are \( \left(\frac{10}{3}, \frac{11}{3}\right) \) and \( \left(\frac{14}{3}, \frac{13}{3}\right) \).
In simple words: Trisection means dividing into three equal parts. Applying the section formula with ratios 1:2 and 2:1 gives the two points of trisection.

Exam Tip: Do not approximate fractions like \( \frac{10}{3} \). Leave them in their exact fractional form for full marks.

Question 2. P is a point on the line segment joining A(4, 3) & B(−2, 6) such that 5AP = 2BP. Find coordinates of P.
Answer: We are given:
\( 5AP = 2BP \implies \frac{AP}{BP} = \frac{2}{5} \).
This means point P divides the line segment AB in the ratio 2 : 5 internally.
Using the section formula:
\( x = \frac{2(-2) + 5(4)}{2 + 5} = \frac{-4 + 20}{7} = \frac{16}{7} \)
\( y = \frac{2(6) + 5(3)}{2 + 5} = \frac{12 + 15}{7} = \frac{27}{7} \).
Therefore, the coordinates of P are \( \left(\frac{16}{7}, \frac{27}{7}\right) \).
In simple words: The given relation means P divides the segment AB in a 2 to 5 ratio. Applying the section formula yields the coordinates of P as \( \left(\frac{16}{7}, \frac{27}{7}\right) \).

Exam Tip: Be sure to write the ratio \( AP:BP \) clearly as your first step before applying the section formula.

Question 3. In what ratio does the point (3, a) divide the join of (1, 7) & (6, −3) ? Also find a.
Answer: Let the point \( P(3, a) \) divide the segment joining \( (1, 7) \) and \( (6, -3) \) in the ratio \( \lambda : 1 \) internally.
Using the section formula for the x-coordinate:
\( x_P = \frac{\lambda(6) + 1(1)}{\lambda + 1} = 3 \)
\( \implies 6\lambda + 1 = 3\lambda + 3 \)
\( \implies 3\lambda = 2 \)
\( \implies \lambda = \frac{2}{3} \).
Thus, the ratio is 2 : 3 internally.

Now, find the value of a using the y-coordinate of the section formula:
\( a = \frac{\frac{2}{3}(-3) + 1(7)}{2/3 + 1} = \frac{-2 + 7}{5/3} = \frac{5}{5/3} = 3 \).
Therefore, the ratio is 2 : 3 and \( a = 3 \).
In simple words: The x-coordinate tells us the division ratio is 2:3. Substituting this ratio into the y-coordinate section formula yields \( a = 3 \).

Exam Tip: Solving for the ratio using the fully known coordinate first is the correct sequence of steps.

Question 4. Determine the ratio in which the straight line x − y = 0 divides the segment joining A(3, −1) and B(8, 9)
Answer: Let the line divide the segment AB at point P in the ratio \( k : 1 \) internally.
The coordinates of the point of division P are:
\( P = \left( \frac{8k + 3}{k + 1}, \frac{9k - 1}{k + 1} \right) \)

Since P lies on the line \( x - y = 0 \):
\( \frac{8k + 3}{k + 1} - \frac{9k - 1}{k + 1} = 0 \)
\( \implies 8k + 3 - (9k - 1) = 0 \)
\( \implies -k + 4 = 0 \)
\( \implies k = 4 \).
Therefore, the ratio in which the line divides the segment is 4 : 1 internally.
In simple words: We assume the division ratio is k:1. Setting the coordinates of the dividing point into the line equation \( x - y = 0 \) yields \( k = 4 \), meaning the ratio is 4:1.

Exam Tip: For any line dividing a segment, find the general coordinates using the ratio \( k:1 \) and substitute them into the line equation.

Question 5. Determine the ratio in which the line 3x + 4y − 9 = 0 divides the line segment joining (1, 3) and (2, 7)
Answer: Let the line divide the line segment joining \( A(1, 3) \) and \( B(2, 7) \) at point P in the ratio \( k : 1 \) internally.
The coordinates of the dividing point P are:
\( P = \left( \frac{2k + 1}{k + 1}, \frac{7k + 3}{k + 1} \right) \)

Since P lies on the line \( 3x + 4y - 9 = 0 \):
\( 3\left(\frac{2k+1}{k+1}\right) + 4\left(\frac{7k+3}{k+1}\right) - 9 = 0 \)
\( \implies 3(2k + 1) + 4(7k + 3) - 9(k + 1) = 0 \)
\( \implies 6k + 3 + 28k + 12 - 9k - 9 = 0 \)
\( \implies 25k + 6 = 0 \)
\( \implies k = -\frac{6}{25} \).
The negative sign indicates that the division is external. Thus, the line divides the segment externally in the ratio 6 : 25.
In simple words: The ratio is found to be \( -6/25 \). The negative value shows that the line divides the segment externally in a 6 to 25 ratio.

Exam Tip: A negative value for the ratio \( k \) indicates external division. Always mention this detail in your final statement.

Question 6. Find the ratio in which the line joining the points (2, −6) and (8, 4) is divides by the x-axis. Find the co-ordinates of the point of division.
Answer: Let the x-axis divide the segment joining \( A(2, -6) \) and \( B(8, 4) \) in the ratio \( \lambda : 1 \) internally at point \( P(x, 0) \).
Since P lies on the x-axis, its y-coordinate is 0:
\( y_P = \frac{\lambda(4) + 1(-6)}{\lambda + 1} = 0 \)
\( \implies 4\lambda - 6 = 0 \)
\( \implies \lambda = \frac{3}{2} \).
So, the ratio is 3 : 2 internally.

Now, find the x-coordinate of P:
\( x = \frac{\frac{3}{2}(8) + 1(2)}{3/2 + 1} = \frac{12 + 2}{5/2} = \frac{14}{2.5} = 5.6 \) (or \( \frac{28}{5} \)).
Thus, the point of division is \( (5.6, 0) \) (or \( \left(\frac{28}{5}, 0\right) \)).
In simple words: The y-coordinate being 0 tells us that the division ratio is 3:2. Substituting this ratio into the x-coordinate formula gives the point as \( (5.6, 0) \).

Exam Tip: Since the x-axis divides the segment, set the y-coordinate of the dividing point to 0 to find the ratio first.

Question 7. The midpoint of the line joining A(2, p) and B(q. 4) is (3, 5). Calculate the values of p, q.
Answer: Using the midpoint formula for the segment AB:
\( \text{Midpoint} = \left( \frac{2 + q}{2}, \frac{p + 4}{2} \right) = (3, 5) \)

Equating coordinates:
1. \( \frac{2+q}{2} = 3 \implies 2+q = 6 \implies q = 4 \)
2. \( \frac{p+4}{2} = 5 \implies p+4 = 10 \implies p = 6 \).
Thus, the values are \( p = 6 \) and \( q = 4 \).
In simple words: We take the midpoint average of the coordinates. Equating these to \( (3, 5) \) yields \( q = 4 \) and \( p = 6 \).

Exam Tip: Midpoint problems are highly scoring. Meticulously pair up coordinates and solve each equation separately.

Question 8. The coordinates of the point of the line joining points (3p, 4) & (−2, 2q) are (5, p). Find p & q.
Answer: Assuming the given coordinates represent the midpoint of the line segment:
\( \text{Midpoint} = \left( \frac{3p - 2}{2}, \frac{4 + 2q}{2} \right) = (5, p) \)

Equating coordinates:
1. \( \frac{3p-2}{2} = 5 \implies 3p-2 = 10 \implies 3p = 12 \implies p = 4 \)
2. \( \frac{4+2q}{2} = p \)
Substitute \( p = 4 \):
\( \frac{4+2q}{2} = 4 \implies 4+2q = 8 \implies 2q = 4 \implies q = 2 \).
Thus, the values are \( p = 4 \) and \( q = 2 \).
In simple words: The midpoint equation for the x-coordinate gives \( p = 4 \). Substituting this value of p into the y-coordinate midpoint equation yields \( q = 2 \).

Exam Tip: Solve the equation containing only one variable first, then substitute its value to solve the remaining equation.

Question 9. In the figure p(2, 3) is the midpoint of the line segment AB. Write the co-ordinates of AB.
Answer: Let \( A(x, 0) \) be the point on the x-axis, and let \( B(0, y) \) be the point on the y-axis.
Since \( P(2, 3) \) is the midpoint of AB:
\( \left( \frac{x+0}{2}, \frac{0+y}{2} \right) = (2, 3) \)
\( \implies \frac{x}{2} = 2 \implies x = 4 \)
\( \implies \frac{y}{2} = 3 \implies y = 6 \).
Therefore, the coordinates of the points are \( A(4, 0) \) and \( B(0, 6) \).
In simple words: Since A lies on the x-axis and B lies on the y-axis, we use their midpoint \( (2, 3) \) to find \( A = (4, 0) \) and \( B = (0, 6) \).

Exam Tip: Ensure that you identify the points of intercept on the axes correctly before formulating the midpoint equations.

Question 10. A line segment meets x-axis at A & y-axis at B. If the coordinates of the midpoint of AB are (3, 4), find the co-ordinates of A & B and length of AB.
Answer: Let the coordinates of A on the x-axis be \( (x, 0) \) and B on the y-axis be \( (0, y) \).
Since the midpoint of AB is \( (3, 4) \):
\( \left( \frac{x+0}{2}, \frac{0+y}{2} \right) = (3, 4) \)
\( \implies \frac{x}{2} = 3 \implies x = 6 \)
\( \implies \frac{y}{2} = 4 \implies y = 8 \).
So, \( A = (6, 0) \) and \( B = (0, 8) \).

Now, calculate the length of AB:
\( AB = \sqrt{(0-6)^2 + (8-0)^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units.
Thus, the coordinates are \( A(6, 0) \), \( B(0, 8) \), and the length of AB is 10 units.
In simple words: The midpoint coordinates tell us that the intercepts are \( A(6, 0) \) and \( B(0, 8) \). Applying the distance formula between these points gives the length of AB as 10 units.

Exam Tip: Break down the problem: solve for coordinates first, then apply the distance formula to find the segment length.

Question 11. A & B are points (−2, −5) & (4, −3) respectively. Find the coordinates of point C on AB produced such that AC = 2BC.
Answer: We are given that C lies on AB produced such that \( AC = 2BC \).
This implies that B is the midpoint of the line segment AC, since:
\( AC = AB + BC = 2BC \implies AB = BC \).

Let the coordinates of C be \( (x, y) \). Since \( B(4, -3) \) is the midpoint of AC:
1. For the x-coordinate:
\( \frac{-2+x}{2} = 4 \implies -2+x = 8 \implies x = 10 \)

2. For the y-coordinate:
\( \frac{-5+y}{2} = -3 \implies -5+y = -6 \implies y = -1 \).
Therefore, the coordinates of point C are \( (10, -1) \).
In simple words: The relation \( AC = 2BC \) means B is the midpoint of AC. Using our midpoint formula with A and B lets us solve for C as \( (10, -1) \).

Exam Tip: Draw a simple line sketch to visualize the positions of points A, B, and C before writing down the midpoint equations.

Question 12. The line segment joining the points (−3, 9) and (6, 3) is trisected. Find the coordinates of point of trisection.
Answer: Let \( A(-3, 9) \) and \( B(6, 3) \) be the given endpoints. The points of trisection P and Q divide the segment AB internally in the ratio 1 : 2 and 2 : 1 respectively.

1. For point P, dividing AB in the ratio 1 : 2:
\( x_P = \frac{1(6) + 2(-3)}{1 + 2} = \frac{6 - 6}{3} = 0 \)
\( y_P = \frac{1(3) + 2(9)}{1 + 2} = \frac{3 + 18}{3} = 7 \)
So, \( P = (0, 7) \).

2. For point Q, dividing AB in the ratio 2 : 1:
\( x_Q = \frac{2(6) + 1(-3)}{2 + 1} = \frac{12 - 3}{3} = 3 \)
\( y_Q = \frac{2(3) + 1(9)}{2 + 1} = \frac{6 + 9}{3} = 5 \)
So, \( Q = (3, 5) \).
Thus, the coordinates of the points of trisection are \( (0, 7) \) and \( (3, 5) \).
In simple words: The two trisection points divide the line segment into three equal parts. Using the section formula with ratios 1:2 and 2:1 gives the points as \( (0, 7) \) and \( (3, 5) \).

Exam Tip: Be consistent with your endpoints \( A \) and \( B \) when applying different ratios like \( 1:2 \) and \( 2:1 \).

Question 13. The linesegment joining the points (3, −4) & (1, 2) is trisected at P & Q. If coordinates of P & Q are (P, −2) and \( \left( \frac{5}{3}, q \right) \). Find p & q.
Answer: Let the endpoints of the line segment be \( A(3, -4) \) and \( B(1, 2) \).
The points of trisection P and Q divide AB in the ratio 1 : 2 and 2 : 1 internally.

1. Point P, dividing AB in the ratio 1 : 2:
\( x_P = \frac{1(1) + 2(3)}{1 + 2} = \frac{7}{3} \)
\( y_P = \frac{1(2) + 2(-4)}{1 + 2} = \frac{2 - 8}{3} = -2 \)
Comparing this with the given coordinates of P\( (p, -2) \), we find \( p = \frac{7}{3} \).

2. Point Q, dividing AB in the ratio 2 : 1:
\( x_Q = \frac{2(1) + 1(3)}{2 + 1} = \frac{5}{3} \)
\( y_Q = \frac{2(2) + 1(-4)}{2 + 1} = \frac{4 - 4}{3} = 0 \)
Comparing this with the given coordinates of Q\( \left(\frac{5}{3}, q\right) \), we find \( q = 0 \).
Therefore, \( p = \frac{7}{3} \) and \( q = 0 \).
In simple words: Calculating the two trisection points gives \( \left(\frac{7}{3}, -2\right) \) and \( \left(\frac{5}{3}, 0\right) \). Comparing these to the given templates directly yields \( p = \frac{7}{3} \) and \( q = 0 \).

Exam Tip: Verifying that the computed y-coordinate for P (\( -2 \)) and x-coordinate for Q (\( \frac{5}{3} \)) match the given coordinates is a great way to confirm your calculations are correct.

Question 14. The line segment joining the point (3, 2) and (6, 8) is divided into 4 equal parts. Find the coordinates of points of section.
Answer: Let the endpoints be \( A(3, 2) \) and \( B(6, 8) \). Let the three dividing points be P, Q, and R.
1. Q is the midpoint of the entire segment AB:
\( Q = \left( \frac{3+6}{2}, \frac{2+8}{2} \right) = \left(4.5, 5\right) \)

2. P is the midpoint of the segment AQ:
\( P = \left( \frac{3+4.5}{2}, \frac{2+5}{2} \right) = \left(3.75, 3.5\right) \)

3. R is the midpoint of the segment QB:
\( R = \left( \frac{4.5+6}{2}, \frac{5+8}{2} \right) = \left(5.25, 6.5\right) \).
Thus, the coordinates of the points of section are \( (3.75, 3.5) \), \( (4.5, 5) \), and \( (5.25, 6.5) \).
In simple words: We find the exact middle of the segment first to get Q. Finding the midpoints of both halves gives us the other two points, P and R.

Exam Tip: Finding the midpoints of nested segments is much faster and simpler than using the section formula with ratios \( 1:3 \), \( 2:2 \), and \( 3:1 \).

Question 15. Line segment PQ is divided into 5 equal parts at A, B, C, D. If Ais (−3, −7) & C(1, 1). Find P, B, D, Q.
Answer: The points in order are P, A, B, C, D, Q.
Since B lies exactly between A and C, it is the midpoint of AC:
\( B = \left( \frac{-3+1}{2}, \frac{-7+1}{2} \right) = (-1, -3) \).

Let \( \vec{d} \) be the constant difference vector between any two consecutive points:
\( \vec{d} = B - A = (-1 - (-3), -3 - (-7)) = (2, 4) \).

We can find the other points by adding or subtracting the difference vector:
\( P = A - \vec{d} = (-3-2, -7-4) = (-5, -11) \)
\( D = C + \vec{d} = (1+2, 1+4) = (3, 5) \)
\( Q = D + \vec{d} = (3+2, 5+4) = (5, 9) \).
Therefore, the points are \( P(-5, -11) \), \( B(-1, -3) \), \( D(3, 5) \), and \( Q(5, 9) \).
In simple words: Finding the midpoint of A and C gives B as \( (-1, -3) \). Since the points are equally spaced, the step size is \( (2, 4) \), which we use to find the remaining points.

Exam Tip: Using the concept of a constant difference vector (or AP-like steps) is an extremely elegant way to solve multi-point segment division problems.

Question 16. Show that the line segment joining (2, 5), (−8, 3) and (−3, 1), (−3, 7) bisects each other.
Answer: If two line segments bisect each other, they must share the exact same midpoint.
1. Midpoint of the segment joining \( (2, 5) \) and \( (-8, 3) \):
\( M_1 = \left( \frac{2-8}{2}, \frac{5+3}{2} \right) = (-3, 4) \)

2. Midpoint of the segment joining \( (-3, 1) \) and \( (-3, 7) \):
\( M_2 = \left( \frac{-3-3}{2}, \frac{1+7}{2} \right) = (-3, 4) \)

Since \( M_1 = M_2 = (-3, 4) \), the two line segments bisect each other.
In simple words: We calculate the midpoint for both line segments. Since both midpoints are exactly the same point \( (-3, 4) \), the lines bisect each other.

Exam Tip: Proving bisection is equivalent to showing that the midpoint coordinates of both segments are identical.

Question 17. A(3, 3), B(6, −1) and D(−5, 9) are vertices of llgm ABCD. Find the coordinates of vertex C.
Answer: Let the coordinates of vertex C be \( (x, y) \).
Since ABCD is a parallelogram, its diagonals AC and BD bisect each other, sharing the same midpoint:
\( M_{AC} = M_{BD} \)
\( \implies \left( \frac{3+x}{2}, \frac{3+y}{2} \right) = \left( \frac{6-5}{2}, \frac{-1+9}{2} \right) \)
\( \implies \left( \frac{3+x}{2}, \frac{3+y}{2} \right) = \left( \frac{1}{2}, 4 \right) \)

Equating coordinates:
1. \( \frac{3+x}{2} = \frac{1}{2} \implies 3+x = 1 \implies x = -2 \)
2. \( \frac{3+y}{2} = 4 \implies 3+y = 8 \implies y = 5 \).
Therefore, the coordinates of vertex C are \( (-2, 5) \).
In simple words: The diagonals share the same midpoint, which is \( (0.5, 4) \). Setting up the midpoint equation for diagonal AC lets us find that vertex C is \( (-2, 5) \).

Exam Tip: Always make sure to pair opposite vertices (AC and BD) correctly, as pairing adjacent ones will give incorrect coordinates.

Question 18. If A(4, 4), B(0, 0) & C(6, 2) are the vertices of ∆ABC, find the length of median through A.
Answer: The median through vertex A meets side BC at its midpoint, M.
First, find the coordinates of midpoint M:
\( M = \left( \frac{0+6}{2}, \frac{0+2}{2} \right) = (3, 1) \).

Now, calculate the length of the median AM using the distance formula:
\( AM = \sqrt{(3-4)^2 + (1-4)^2} \)
\( \implies AM = \sqrt{(-1)^2 + (-3)^2} \)
\( \implies AM = \sqrt{1 + 9} = \sqrt{10} \) units.
Thus, the length of the median through A is \( \sqrt{10} \) units.
In simple words: We find the midpoint of side BC is \( (3, 1) \). The distance from vertex A to this midpoint is the length of the median, which is \( \sqrt{10} \).

Exam Tip: Break this problem into two clear parts: first find the midpoint of the opposite side, then apply the distance formula to find the length.

Question 19. Find the 3rd vertex of a triangle if its 2 vertices are (5, 3) and (7, −3) & midpoint of 1 side is (2, 2)
Answer: Let the vertices be \( A(5, 3) \), \( B(7, -3) \), and the third vertex be \( C(x, y) \).
Assuming the given midpoint \( (2, 2) \) is the midpoint of side AC:
\( \left( \frac{5+x}{2}, \frac{3+y}{2} \right) = (2, 2) \)

Equating coordinates:
1. \( \frac{5+x}{2} = 2 \implies 5+x = 4 \implies x = -1 \)
2. \( \frac{3+y}{2} = 2 \implies 3+y = 4 \implies y = 1 \).
Therefore, the coordinates of the 3rd vertex are \( (-1, 1) \).
In simple words: The midpoint of AC is given as \( (2, 2) \). Using our midpoint formula with vertex A lets us solve for the third vertex C as \( (-1, 1) \).

Exam Tip: Clearly state which side's midpoint you are using in your solution to make your steps easy for the examiner to follow.

Question 20. Find the value of P for which the area formed by the triangle with vertices A(P, 2p), B(−2, 6) and C(3, 1) is 10 sq. units.
Answer: The area of the triangle is given as 10 square units:
\( \text{Area} = \frac{1}{2} | p(6 - 1) + (-2)(1 - 2p) + 3(2p - 6) | = 10 \)
\( \implies | 5p - 2 + 4p + 6p - 18 | = 20 \)
\( \implies | 15p - 20 | = 20 \)
This yields two cases:

Case 1: \( 15p - 20 = 20 \implies 15p = 40 \implies p = \frac{8}{3} \).
Case 2: \( 15p - 20 = -20 \implies 15p = 0 \implies p = 0 \).
Therefore, the value of p can be \( \frac{8}{3} \) or \( 0 \).
In simple words: Substituting the coordinates into the triangle area formula gives the absolute relation \( |15p - 20| = 20 \). Solving this gives two possible values for p: \( \frac{8}{3} \) or \( 0 \).

Exam Tip: Remember to use absolute values when setting up area equations, as this accounts for both positive and negative cases of the expression.

Question 21. Find the area of the triangle formed by the joining of the midpoints of the sides of triangle whose vertices are (0, −1), (2, 1) and (0, 3). Find the ratio of this area to the area of given triangle.
Answer: Let the vertices of the given triangle be \( A(0, -1) \), \( B(2, 1) \), and \( C(0, 3) \).
Let's calculate the area of this given triangle:
\( \text{Area}_{\text{given}} = \frac{1}{2} | 0(1-3) + 2(3 - (-1)) + 0(-1-1) | \)
\( \implies \text{Area}_{\text{given}} = \frac{1}{2} | 2(4) | = 4 \) sq. units.

The area of the triangle formed by joining the midpoints of the sides is always exactly \( \frac{1}{4} \) of the area of the given triangle:
\( \text{Area}_{\text{midpoint}} = \frac{1}{4} \cdot \text{Area}_{\text{given}} = \frac{1}{4} \cdot 4 = 1 \) sq. unit.

The ratio of this area to the area of the given triangle is:
\( \text{Ratio} = 1 : 4 \).
In simple words: The area of the main triangle is 4 square units. Since the midpoint triangle is always one-fourth of the main triangle's area, its area is 1 square unit, and the ratio is 1:4.

Exam Tip: You can state the midpoint area theorem directly to find the area and ratio without needing to compute the actual midpoint coordinates.

Page 3

Question 22. Find the area of quadrilateral ABCD formed by the points A(−2, −2), B(5, 1), C(2, 4) and D(−1, 5)
Answer: We can find the area of the quadrilateral by dividing it into two triangles, ABC and ACD, and summing their areas:
1. Area of \( \Delta ABC \):
\( \text{Area}_{ABC} = \frac{1}{2} | -2(1-4) + 5(4 - (-2)) + 2(-2-1) | \)
\( = \frac{1}{2} | -2(-3) + 5(6) + 2(-3) | \)
\( = \frac{1}{2} | 6 + 30 - 6 | = 15 \) sq. units.

2. Area of \( \Delta ACD \):
\( \text{Area}_{ACD} = \frac{1}{2} | -2(4-5) + 2(5 - (-2)) + (-1)(-2-4) | \)
\( = \frac{1}{2} | -2(-1) + 2(7) + (-1)(-6) | \)
\( = \frac{1}{2} | 2 + 14 + 6 | = 11 \) sq. units.

Sum of areas:
\( \text{Area}_{ABCD} = 15 + 11 = 26 \) sq. units.
Thus, the area of the quadrilateral is 26 square units.
In simple words: We split the quadrilateral into two triangles. The area of the first triangle is 15 and the second is 11, giving a total area of 26 square units.

Exam Tip: Be sure to keep the order of the vertices consistent when splitting the quadrilateral into two triangles.

Question 23. Find the area of the quadrilateral ABCD formed by the points A(2, 3), B(−4, −2), C(−3, −5) and D(3, −2) taken in order
Answer: Let us divide the quadrilateral into two triangles, ABC and ACD, and sum their areas:
1. Area of \( \Delta ABC \):
\( \text{Area}_{ABC} = \frac{1}{2} | 2(-2 - (-5)) + (-4)(-5-3) + (-3)(3 - (-2)) | \)
\( = \frac{1}{2} | 2(3) + (-4)(-8) + (-3)(5) | \)
\( = \frac{1}{2} | 6 + 32 - 15 | = \frac{23}{2} = 11.5 \) sq. units.

2. Area of \( \Delta ACD \):
\( \text{Area}_{ACD} = \frac{1}{2} | 2(-5 - (-2)) + (-3)(-2-3) + 3(3 - (-5)) | \)
\( = \frac{1}{2} | 2(-3) + (-3)(-5) + 3(8) | \)
\( = \frac{1}{2} | -6 + 15 + 24 | = \frac{33}{2} = 16.5 \) sq. units.

Sum of areas:
\( \text{Area}_{ABCD} = 11.5 + 16.5 = 28 \) sq. units.
Thus, the area of the quadrilateral is 28 square units.
In simple words: Splitting the quadrilateral into two triangles gives areas of 11.5 and 16.5. Adding them together gives the total area of the quadrilateral as 28 square units.

Exam Tip: Double check your arithmetic when evaluating double negative signs during triangle area calculations.

Question 24. Two vertices of a ∆are (8, −6) and (−4, 6). The area of the triangle is 120 sq. units. Find the 3rd vertex, if it lies on x − 2y = 6.
Answer: Let the coordinates of the third vertex be \( C(x, y) \). Since it lies on the line \( x - 2y = 6 \), we have:
\( x = 2y + 6 \) (Equation 1)

The area of the triangle is given as 120 square units:
\( \text{Area} = \frac{1}{2} | 8(6 - y) + (-4)(y - (-6)) + x(-6 - 6) | = 120 \)
\( \implies | 48 - 8y - 4y - 24 - 12x | = 240 \)
\( \implies | 24 - 12y - 12x | = 240 \)
Divide the expression inside the modulus by 12:
\( \implies | 2 - y - x | = 20 \)

Substituting \( x = 2y + 6 \) from Equation 1:
\( | 2 - y - (2y + 6) | = 20 \)
\( \implies | -3y - 4 | = 20 \)
This gives two possible cases:

Case 1: \( -3y - 4 = 20 \implies -3y = 24 \implies y = -8 \).
Substitute \( y = -8 \) in Equation 1:
\( x = 2(-8) + 6 = -10 \). Vertex is \( (-10, -8) \).

Case 2: \( -3y - 4 = -20 \implies -3y = -16 \implies y = \frac{16}{3} \).
Substitute \( y = \frac{16}{3} \) in Equation 1:
\( x = 2\left(\frac{16}{3}\right) + 6 = \frac{50}{3} \). Vertex is \( \left(\frac{50}{3}, \frac{16}{3}\right) \).
Thus, the coordinates of the third vertex are \( (-10, -8) \) or \( \left(\frac{50}{3}, \frac{16}{3}\right) \).
In simple words: The third root lies on \( x = 2y + 6 \). Solving the triangle area equation gives two possible coordinates for the third vertex: \( (-10, -8) \) or \( \left(\frac{50}{3}, \frac{16}{3}\right) \).

Exam Tip: Substituting the linear equation into the area relation early on reduces the equation to a single variable, making it much easier to solve.

Value Based Questions.

Question 1. Aadya and Nitya planted some trees in a square garden as shown in the fig.1 both arguing that they have planted them in a straight line. Find out who is correct? Justify your decision. (N stands for Nitya and A stands of Aadya)
Answer: To determine who is correct, we must check if the coordinates of the trees planted by Aadya and Nitya are collinear.

1. For Aadya's trees, the coordinates are \( A_1(2, 1) \), \( A_2(4, 3) \), and \( A_3(6, 5) \).
Slope of \( A_1 A_2 = \frac{3-1}{4-2} = \frac{2}{2} = 1 \)
Slope of \( A_2 A_3 = \frac{5-3}{6-4} = \frac{2}{2} = 1 \)
Since the slopes are equal, the points are collinear. Thus, Aadya's trees are in a straight line.

2. For Nitya's trees, the coordinates are \( N_1(2, 3) \), \( N_2(3, 4) \), and \( N_3(4, 6) \).
Slope of \( N_1 N_2 = \frac{4-3}{3-2} = \frac{1}{1} = 1 \)
Slope of \( N_2 N_3 = \frac{6-4}{4-3} = \frac{2}{1} = 2 \)
Since the slopes are not equal, the points are not collinear. Thus, Nitya's trees are not in a straight line.
Therefore, Aadya is correct.
In simple words: Calculating the slopes between the coordinates of the trees shows that only Aadya's trees have equal slopes (1), meaning they lie in a straight line. Aadya is correct.

Exam Tip: Showing that the slopes between consecutive points are equal is a very elegant and quick way to prove collinearity.

Question 2. The students of class X of a school undertake to work for the campaign “Say no to plastic” in a city. They took the map of the city and form co-ordinate plane on it to divide their areas. Group A book the region covered between the co-ordinates (1, 1), (-3, 2), (-2,-2) and (1, -3) taken in order. Find the area of the region covered by group 4.
a) What are the harmful effects of using plastic?
b) How can you contribute in spreading awareness for such campaign?
Answer: Let the coordinates of the region be \( P(1, 1) \), \( Q(-3, 2) \box \), \( R(-2, -2) \), and \( S(1, -3) \).
We can find the total area by splitting the quadrilateral into two triangles, PQR and PRS:

1. Area of \( \Delta PQR \):
\( \text{Area}_{PQR} = \frac{1}{2} | 1(2 - (-2)) + (-3)(-2 - 1) + (-2)(1 - 2) | \)
\( = \frac{1}{2} | 4 + 9 + 2 | = 7.5 \) sq. units.

2. Area of \( \Delta PRS \):
\( \text{Area}_{PRS} = \frac{1}{2} | 1(-2 - (-3)) + (-2)(-3 - 1) + 1(1 - (-2)) | \)
\( = \frac{1}{2} | 1 + 8 + 3 | = 6 \) sq. units.

Total Area = \( 7.5 + 6 = 13.5 \) sq. units.

a) Harmful effects of using plastic:
- Plastic is non-biodegradable and causes long-term soil and water pollution.
- It harms marine and terrestrial wildlife when ingested.
- Burning plastic releases toxic gases, contributing to air pollution.

b) Spreading awareness:
- Encouraging the use of reusable cloth or jute bags instead of single-use plastic.
- Educating friends and family about the importance of recycling and the 3 Rs.
In simple words: The total area of the region is 13.5 square units. Plastics are harmful because they pollute the soil and water and do not rot. We can spread awareness by using cloth bags and educating others.

Exam Tip: For value-based questions, write clear and concise points for the subjective questions to secure full marks easily.