CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 05

Question. Three given points will be collinear, if the area of the triangle formed by these points is 
(a) 0 sq. units
(b) 1 sq. units
(c) -1 sq. units
(d) 2 sq. units
Answer: (a) 0 sq. units
Explanation: Three points that lie on a same straight line are colinear, and the area of the triangle formed by these collinear points is zero. Hence Three given points will be collinear, if the area of the triangle formed by these points is 0 sq. units.

 

Question. The area of the triangle with vertices \( (a, b+c) \), \( (b, c+a) \) and \( (c, a+b) \) is 
(a) \( a + b + c \)
(b) \( a^2 + b^2 + c^2 \)
(c) 0
(d) \( (a + b + c)^2 \)
Answer: (c) 0
Explanation: Given: Vertices of a triangle ABC, \( A(a, b+c) \), \( B(b, c+a) \) and \( C(c, a+b) \)
\( \therefore \text{ar}(\triangle ABC) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
\( = \frac{1}{2} |a(c+a - (a+b)) + b(a+b - (b+c)) + c(b+c - (c+a))| \)
\( = \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)| \)
\( = \frac{1}{2} |ac - ab + ab - bc + bc - ac| \)
\( = \frac{1}{2} \times 0 = 0 \text{ sq. units} \)
Also, therefore, the three given points are collinear.

 

Question. The points A(4, – 1), B(6, 0), C(7, 2) and D(5, 1) are the vertices of a 
(a) Square
(b) Parallelogram
(c) Rhombus
(d) Rectangle
Answer: (c) Rhombus
Explanation: Given: The points A(4, – 1), B(6, 0), C(7, 2) and D(5, 1).
\( \therefore AB = \sqrt{(6 - 4)^2 + (0 + 1)^2} = \sqrt{4 + 1} = \sqrt{5} \text{ units} \)
\( BC = \sqrt{(7 - 6)^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5} \text{ units} \)
\( CD = \sqrt{(5 - 7)^2 + (1 - 2)^2} = \sqrt{4 + 1} = \sqrt{5} \text{ units} \)
\( AD = \sqrt{(5 - 4)^2 + (1 + 1)^2} = \sqrt{1 + 4} = \sqrt{5} \text{ units} \)
Therefore all 4 sides AB, BC, CD and DA are equal
and diagonal \( AC = \sqrt{(7 - 4)^2 + (2 + 1)^2} = \sqrt{9 + 9} = 3\sqrt{2} \text{ units} \)
and \( BD = \sqrt{(5 - 6)^2 + (1 - 0)^2} = \sqrt{1 + 1} = \sqrt{2} \text{ units} \)
Therefore diagonals AC and BD are not equal
Since, all sides are equal and both diagonals are not equal. Therefore, the given quadrilateral is a rhombus.

 

Question. If the co – ordinates of a point are ( – 5, 11), then its abscissa is 
(a) -5
(b) 11
(c) 5
(d) -11
Answer: (a) -5
Explanation: Since \( x \)–coordinate of a point is called abscissa.
Therefore, abscissa is \( -5 \).

 

Question. The vertices of a quadrilateral are (1, 7), (4, 2), ( – 1, – 1) and ( – 4, 4). The quadrilateral is a 
(a) rectangle
(b) parallelogram
(c) square
(d) Rhombus
Answer: (c) square
Explanation: Let A (1, 7), B (4, 2), C (– 1, – 1) and D (– 4, 4) are the vertices of a quadrilateral ABCD.
\( \therefore AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} \)
\( = \sqrt{9 + 25} = \sqrt{34} \text{ units} \)
and \( BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} \)
\( = \sqrt{25 + 9} = \sqrt{34} \text{ units} \)
and \( CD = \sqrt{(-4 + 1)^2 + (4 + 1)^2} \)
\( = \sqrt{9 + 25} = \sqrt{34} \text{ units} \)
and \( AD = \sqrt{(-4 - 1)^2 + (4 - 7)^2} = \sqrt{25 + 9} = \sqrt{34} \text{ units} \)
\( AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} \)
\( = \sqrt{4 + 64} = 2\sqrt{17} \text{ units} \)
\( BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} \)
\( = \sqrt{64 + 4} = 2\sqrt{17} \text{ units} \)
Since, all sides are equal and both diagonals are also equal. Therefore, the given quadrilateral is a square.

 

Question. If the points (0, 0), (1, 2) and (x, y) are collinear,then find x. 
Answer: The points are collinear, then area of triangle = 0
\( \therefore \frac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] = 0 \)
or, \( \frac{1}{2} [0(2 - y) + 1(y - 0) + x(0 - 2)] = 0 \)
or, \( \frac{1}{2} [y - 2x] = 0 \)
or, \( 2x - y = 0 \)
\( \therefore x = \frac{y}{2} \)

 

Question. Find the point on the X-axis which is equidistant from the points (-1,0) and (5,0) 
Answer: Let A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).

\( \implies (x + 1)^2 = (x - 5)^2 \)

\( \implies x^2 + 2x + 1 = x^2 - 10x + 25 \)

\( \implies 2x + 1 = -10x + 25 \)

\( \implies 2x + 10x = 25 - 1 \)

\( \implies 12x = 24 \)

\( \implies x = 24/12 \)

\( \implies x = 2 \)
Therefore , Required point is (2,0).

 

Question. Find the distance of the point \( (\alpha, \beta) \) from y-axis. 
Answer: Distance of the point \( (\alpha, \beta) \) from y-axis is the positive value of its x-coordinate.
\( \therefore \text{Distance} = |\alpha| \)

 

Question. Find the area of the triangle with vertices (0, 0) (6, 0) and (0, 5). 
Answer: We have to find the area of the triangle with vertices (0,0) (6,0) and (0,5).
Area of triangle
\( = \frac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] \)
\( = \frac{1}{2} [0(0 - 5) + 6(5 - 0) + 0(0 - 0)] \)
\( = \frac{1}{2} [6 \times 5] \)
= 15 sq. units.

 

Question. If the centre and radius of circle is (3, 4) and 7 units respectively, then what is the position of the point A(5,8) with respect to circle? 
Answer: Distance of the point, from the centre
\( a = \sqrt{(5 - 3)^2 + (8 - 4)^2} \)
\( = \sqrt{4 + 16} = \sqrt{20} \)
\( = 2\sqrt{5} \)
Since, \( 2\sqrt{5} \) is less than radius=7
\( \therefore \) The point lies inside the circle.

 

Question. Find the distance between the points P (-4, 7) and Q(2, -5). 
Answer: The given points are P (-4, 7) and Q(2, -5).
Then, \( x_1 = -4, y_1 = 7 \) and \( x_2 = 2, y_2 = -5 \).
\( \therefore PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{[(2 - (-4)]^2 + (-5 - 7)^2} = \sqrt{6^2 + (-12)^2} \)
\( = \sqrt{36 + 144} = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \text{ units.} \)

 

Question. Prove that the coordinates of the centroid of a triangle ABC, with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \) and \( C(x_3, y_3) \) are given by \( (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}) \). (2)
Answer: Let the coordinates of vertices of \( \triangle ABC \) be \( A(x_1, y_1) \), \( B(x_2, y_2) \) and \( C(x_3, y_3) \) respectively. Let D be the midpoint of BC. Using section formula, coordinates of D will be \( (\frac{x_3+x_2}{2}, \frac{y_3+y_2}{2}) \). Now since centroid G will divide the line joining A and D in the ratio of 2 : 1, therefore again using section formula, coordinates of G will be
\( \left[ \frac{\left( \frac{x_3+x_2}{2} \right) \cdot 2 + x_1 \cdot 1}{2+1}, \frac{\left( \frac{y_3+y_2}{2} \right) \cdot 2 + y_1 \cdot 1}{2+1} \right] \)
\( = \left( \frac{x_3+x_2+x_1}{3}, \frac{y_3+y_2+y_1}{3} \right) \)

 

Question. Find the area of the rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. 
Answer: Let A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)
\( AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} = 4\sqrt{2} \)
\( BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{36 + 36} = 6\sqrt{2} \)
Area of rhombus \( = \frac{1}{2} d_1 \times d_2 \)
\( = \frac{1}{2} AC \times BD \)
\( = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24 \text{ Sq. unit.} \)

 

Question. Find the coordinates of the centre of the circle passing through the points (0, 0), (-2, 1) and (-3, 2). Also, find its radius. 
Answer: Let P (x, y) be the centre of the circle passing through the points O (0, 0), A (-2, 1) and B (-3, 2). Then, OP = AP = BP
Now, OP = AP

\( \implies OP^2 = AP^2 \)

\( \implies x^2 + y^2 = (x + 2)^2 + (y - 1)^2 \)

\( \implies x^2 + y^2 = x^2 + y^2 + 4x - 2y + 5 \)

\( \implies 4x - 2y + 5 = 0 \) ...(i)
and, OP = BP

\( \implies OP^2 = BP^2 \)

\( \implies x^2 + y^2 = (x + 3)^2 + (y - 2)^2 \)

\( \implies x^2 + y^2 = x^2 + y^2 + 6x - 4y + 13 \)

\( \implies 6x - 4y + 13 = 0 \) .........(ii)
On solving equations (i) and (ii), we get \( x = \frac{3}{2} \) and \( y = \frac{11}{2} \)
Thus, the coordinates of the centre are \( (\frac{3}{2}, \frac{11}{2}) \)
Therefore, Radius = OP \( = \sqrt{x^2 + y^2} = \sqrt{\frac{9}{4} + \frac{121}{4}} = \frac{1}{2}\sqrt{130} \text{ sq. units.} \)

 

Question. If the points P (-3, 9), Q (a, b) and R (4, -5) are collinear and a + b = 1, find the values of a and b. 
Answer: It is given that the points P (-3, 9), Q (a, b) and R(4,-5) are collinear.
\( \therefore \text{Area of } \triangle PQR = 0 \)

\( \implies |\{ -3b - 5a + 36 \} - \{ 15 + 4b + 9a \}| = 0 \)

\( \implies |( -14a - 7b + 21 )| = 0 \)

\( \implies 14a + 7b - 21 = 0 \)

\( \implies 2a + b - 3 = 0 \) .....(i)
It is given that a + b = 1 ... (ii)
Solving (i) and (ii), we obtain a = 2 and b = - 1.

 

Question. The centre of a circle is (2a, a -7). Find the values of a, if the circle passes through the point (11, -9) and has diameter \( 10\sqrt{2} \) units. 
Answer: Diameter of a circle \( = 10\sqrt{2} \text{ units} \)

\( \implies \text{Radius of a circle } = 5\sqrt{2} \text{ units} \)
Let the centre of a circle be O(2a, a - 7) which passes through the point P(11, -9).
OP is the radius of the circle.

\( \implies OP = 5\sqrt{2} \text{ units} \)

\( \implies OP^2 = (5\sqrt{2})^2 \)

\( \implies (11 - 2a)^2 + (-9 - a + 7)^2 = 50 \)

\( \implies 121 + 4a^2 - 44a + (-2 - a)^2 = 50 \)

\( \implies 121 + 4a^2 - 44a + 4 + a^2 + 4a = 50 \)

\( \implies 5a^2 - 40a + 125 = 50 \)

\( \implies 5a^2 - 40a + 75 = 0 \)

\( \implies a^2 - 8a + 15 = 0 \)

\( \implies a^2 - 5a - 3a + 15 = 0 \)

\( \implies a(a - 5) - 3(a - 5) = 0 \)

\( \implies (a - 5)(a - 3) = 0 \)

\( \implies a - 5 = 0 \text{ or } a - 3 = 0 \)

\( \implies a = 5 \text{ or } a = 3 \)

 

Question. Find the area of a triangle ABC with A(1, - 4) and mid-points of sides through A being (2, -1) and (0, -1). 
Answer: Let E be the midpoint of AB.
\( \therefore \frac{x+1}{2} = 2 \text{ or } x = 3 \)
and \( \frac{y+(-4)}{2} = -1 \text{ or, } y = 2 \)
or, B(3, 2)
Let F be the mid-point of AC. Then,
\( 0 = \frac{x_1+1}{2} \text{ or } x_1 = -1 \)
and \( \frac{y_1+(-4)}{2} = -1 \text{ or, } y_1 = 2 \)
or, C= (-1, 2)
Now the co-ordinates are A(1, -4), B(3, 2), C (-1, 2)
Area of triangle \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
\( = \frac{1}{2} [1(2 - 2) + 3(2 + 4) - 1(-4 - 2)] \)
\( = \frac{1}{2} [0 + 18 + 6] = 12 \text{ sq units.} \)

 

Question. Find the lengths of the medians of a \( \triangle ABC \) having vertices at A (0, -1), B (2, 1) and C (0, 3). 
Answer: Let A(0, -1), B(2, 1) and C(0, 3) be the given points. Let AD, BE and CF be the medians.
Coordinates of D are \( (\frac{2+0}{2}, \frac{1+3}{2}) = (1, 2) \)
Coordinates of E are \( (\frac{0}{2}, \frac{3-1}{2}) = (0, 1) \)
Coordinates of F are \( (\frac{2+0}{2}, \frac{1-1}{2}) = (1, 0) \)
Length of median \( AD = \sqrt{(1 - 0)^2 + (2 + 1)^2} = \sqrt{10} \text{ units} \)
Length of median \( BE = \sqrt{(2 - 0)^2 + (1 - 1)^2} = 2 \text{ units} \)
Length of median \( CF = \sqrt{(1 - 0)^2 + (0 - 3)^2} = \sqrt{10} \text{ units} \)

 

Question. If the centre of circle is (2a, a – 7) then find the values of a if the circle passes through the point (11, –9) and has diameter \( 10\sqrt{2} \) units. 
Answer: Let C(2a, a – 7) be the centre of the circle and it passes through the point P(11, –9).
\( \therefore PQ = 10\sqrt{2} \)

\( \implies CP = 5\sqrt{2} \)

\( \implies CP^2 = (5\sqrt{2})^2 = 50 \)

\( \implies (2a – 11)^2 + (a – 7 + 9)^2 = 50 \)

\( \implies (2a)^2 + (11)^2 – 2(2a) (11) + (a + 2)^2 = 50 \)

\( \implies 4a^2 + 121 – 44a + (a)^2 + (2)^2 + 2(a)(2) = 50 \)

\( \implies 5a^2 – 40a + 125 = 50 \)

\( \implies a^2 – 8a + 25 = 10 \)

\( \implies a^2 – 8a + 25 – 10 = 0 \)

\( \implies a^2 – 8a + 15 = 0 \)

\( \implies a^2 – 5a – 3a + 15 = 0 \)

\( \implies a(a – 5) – 3(a – 5) = 0 \)

\( \implies (a – 5) (a – 3) = 0 \)

\( \implies a – 5 = 0 \text{ or } a – 3 = 0 \)

\( \implies a = 5 \text{ or } a = 3 \)
Hence, the required values of a are 5 and 3.

 

Question. If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3,4), find the vertices of the triangle. 
Answer: Let \( A(x_1, y_1), B(x_2, y_2) \) and \( C(x_3, y_3) \) be the vertices of \( \triangle ABC \).
Let D(1, 1), E(2, -3) and F(3, 4) be the mid-points of sides BC, CA and AB respectively.
Since, D is the mid-point of BC
\( \therefore \frac{x_2 + x_3}{2} = 1 \text{ and } \frac{y_2 + y_3}{2} = 1 \)

\( \implies x_2 + x_3 = 2 \text{ and } y_2 + y_3 = 2 \) ...(i)
Similarly E and F are the mid-points of CA and AB respectively.
\( \therefore \frac{x_1 + x_3}{2} = 2 \text{ and } \frac{y_1 + y_3}{2} = -3 \)

\( \implies x_1 + x_3 = 4 \text{ and } y_1 + y_3 = -6 \) ...(ii)
and, \( \frac{x_1 + x_2}{2} = 3 \text{ and } \frac{y_1 + y_2}{2} = 4 \)

\( \implies x_1 + x_2 = 6 \text{ and } y_1 + y_2 = 8 \) ...(iii)
From (i), (ii) and (iii) we get
\( x_2 + x_3 + x_1 + x_3 + x_1 + x_2 = 2 + 4 + 6 \)
and, \( y_2 + y_3 + y_1 + y_3 + y_1 + y_2 = 2 + (-6) + 8 \)

\( \implies 2(x_1 + x_2 + x_3) = 12 \text{ and } 2(y_1 + y_2 + y_3) = 4 \)

\( \implies x_1 + x_2 + x_3 = 6 \text{ and } y_1 + y_2 + y_3 = 2 \) ...(iv)
From (i) and (iv) we get
\( x_1 + 2 = 6 \text{ and } y_1 + 2 = 2 \)

\( \implies x_1 = 6 - 2 \)

\( \implies y_1 = 2 - 2 \)

\( \implies x_1 = 4 \)

\( \implies y_1 = 0 \)
So the coordinates of A are (4, 0)
From (ii) and (iv) we get
\( x_2 + 4 = 6 \text{ and } y_2 + (-6) = 2 \)

\( \implies x_2 = 2 \)

\( \implies y_2 - 6 = 2 \)

\( \implies y_2 = 8 \)
So the coordinates of B are (2, 8)
From (iii) and (iv) we get
\( 6 + x_3 = 6 \text{ and } 8 + y_3 = 2 \)

\( \implies x_3 = 6 - 6 \)

\( \implies y_3 = 2 - 8 \)

\( \implies x_3 = 0 \)

\( \implies y_3 = -6 \)
So the coordinates of C are (0, -6)
Hence, the vertices of triangle ABC are: A(4, 0), B(2, 8) and C(0, -6).

 

Question. The centroid of a triangle whose vertices are (3, -7), ( -8, 6) and (5, 10) is 
(a) (0, 3)
(b) (1, 3)
(c) (3, 3)
(d) (0, 9)
Answer: (a) (0, 3)
Explanation: Given: \( (x_1, y_1) = (3, -7), (x_2, y_2) = (-8, 6) \) and \( (x_3, y_3) = (5, 10) \)
Coordinates of Centroid of triangle = \( x = \frac{x_1+x_2+x_3}{3} \) and \( y = \frac{y_1+y_2+y_3}{3} \)
\( \therefore x = \frac{3-8+5}{3} = \frac{0}{3} = 0 \)
and \( y = \frac{-7+6+10}{3} = \frac{9}{3} = 3 \)
Therefore, the coordinates of centroid of triangle are (0, 3).

 

Question. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then AP is equal to 
(a) \( AP = \frac{1}{4} AB \)
(b) \( AP = \frac{1}{2} AB \)
(c) \( AP = \frac{1}{3} AB \)
(d) AP = PB
Answer: (b) \( AP = \frac{1}{2} AB \)
Explanation: \( AP = \sqrt{(2 - 4)^2 + (1 - 2)^2} \)
\( = \sqrt{4 + 1} = \sqrt{5} \text{ units} \)
\( AB = \sqrt{(8 - 4)^2 + (4 - 2)^2} \)
\( = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \text{ units} \)
Here \( AB = 2 \times AP \)
\( \therefore AP = \frac{1}{2} AB \)

 

Question. If the points (x, y), (1, 2) and (7, 0) are collinear, then the relation between 'x' and 'y' is given by 
(a) 3x – y – 7 = 0
(b) 3x + y + 7 = 0
(c) x + 3y – 7 = 0
(d) x – 3y + 7 = 0
Answer: (c) x + 3y – 7 = 0
Explanation: Given: \( (x_1, y_1) = (x, y), (x_2, y_2) = (1, 2) \) and \( (x_3, y_3) = (7, 0) \) and these are collinear.
\( \therefore \frac{1}{2} |x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)| = 0 \)

\( \implies |x(2 - 0) + 1(0 - y) + 7(y - 2)| = 0 \)

\( \implies |2x - y + 7y - 14| = 0 \)

\( \implies 2x + 6y - 14 = 0 \)

\( \implies x + 3y - 7 = 0 \)

 

Question. The points A(1, 2), B(5, 4), C(3, 8) and D( – 1, 6) are the vertices of a 
(a) Rectangle
(b) Rhombus
(c) Square
(d) Parallelogram
Answer: (c) Square
Explanation: Given: The points A(1, 2), B(5, 4), C(3, 8) and D( – 1, 6)
\( \therefore AB = \sqrt{(5 - 1)^2 + (4 - 2)^2} = \sqrt{16 + 4} = 2\sqrt{5} \text{ units} \)
\( BC = \sqrt{(3 - 5)^2 + (8 - 4)^2} = \sqrt{4 + 16} = 2\sqrt{5} \text{ units} \)
\( CD = \sqrt{(-1 - 3)^2 + (6 - 8)^2} = \sqrt{16 + 4} = 2\sqrt{5} \text{ units} \)
\( AD = \sqrt{(-1 - 1)^2 + (6 - 2)^2} = \sqrt{4 + 16} = 2\sqrt{5} \text{ units} \)
Therefore the 4 sides AB, BC, CD and DA are equal
and the diagonal \( AC = \sqrt{(3 - 1)^2 + (8 - 2)^2} = \sqrt{4 + 36} = 2\sqrt{10} \text{ units} \)
and \( BD = \sqrt{(-1 - 5)^2 + (6 - 4)^2} = \sqrt{36 + 4} = 2\sqrt{10} \text{ units} \)
Therefore diagonals AC and BD are equal
Since, all 4 sides are equal and both diagonals are also equal.
Therefore, the given quadrilateral is a square.

 

Question. The triangle whose vertices are ( – 3, 0), (1, – 3) and (4, 1) is ___________ triangle. 
(a) Obtuse triangle
(b) equilateral
(c) right angled isosceles
(d) scalene
Answer: (c) right angled isosceles
Explanation: Let A (-3, 0), B(1, -3) and C (4, 1) are the vertices of a triangle ABC.
\( \therefore AB = \sqrt{(1 + 3)^2 + (-3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units} \)
\( BC = \sqrt{(4 - 1)^2 + (1 + 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units} \)
\( CA = \sqrt{(-3 - 4)^2 + (0 - 1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \text{ units} \)
Now, check if \( AC^2 = AB^2 + BC^2 \)

\( \implies (5\sqrt{2})^2 = (5)^2 + (5)^2 \)

\( \implies 50 = 50 \)
Therefore, \( \triangle ABC \) is a right-angled triangle. and also AB = BC = 5 units
Therefore triangle ABC is a right-angled isosceles triangle.

 

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

 

Question. Find the radius of the circle whose end points of diameter are (24,1) and (2,23)] 
Answer: \( (x_1, y_1) = (24,1) \) and \( (x_2, y_2) = (2,23) \)
Diameter of Circle = \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 24)^2 + (23 - 1)^2} \)
\( = \sqrt{(-22)^2 + (22)^2} = \sqrt{(22)^2(1 + 1)} \)
\( = 22\sqrt{2} \text{ units} \)
Therefore, Radius of circle, \( r = \frac{d}{2} = \frac{22\sqrt{2}}{2} = 11\sqrt{2} \text{ units} \)

 

Question. Find the distance between the following pairs of points: (2, 3), (4,1) 
Answer: Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
\( d = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ units} \)

 

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

 

Question. What is the distance between the points A(c,0) and B(0, - c)? 
Answer: \( AB = \sqrt{(0 - c)^2 + (-c - 0)^2} \)
\( = \sqrt{c^2 + c^2} \)
\( = \sqrt{2c^2} \)
\( = \sqrt{2}c \)

 

Question. Use distance formula to show that the points A (- 2,3), B (1, 2) and C (7,0) are collinear. 
Answer: \( AB = \sqrt{(1 + 2)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \)
\( BC = \sqrt{(7 - 1)^2 + (0 - 2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \)
\( AC = \sqrt{(7 + 2)^2 + (0 - 3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \)
Since AB + BC = \( \sqrt{10} + 2\sqrt{10} = (1 + 2)\sqrt{10} = 3\sqrt{10} = AC \)
Hence, the points A, B and C are colinear.

 

Question. If the mid-point of the line joining (3,4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of k. 
Answer: Since, (x, y) is the mid-point
\( x = \frac{3+k}{2}, y = \frac{4+7}{2} = \frac{11}{2} \)
Again,
2x + 2y + 1 = 0

\( \implies 2 \times \frac{(3+k)}{2} + 2 \times \frac{11}{2} + 1 = 0 \)

\( \implies 3 + k + 11 + 1 = 0 \)

\( \implies 3 + k + 12 = 0 \)

\( \implies k + 15 = 0 \)

\( \implies k = -15 \)

 

Question. Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line segment joining the points (8, 6) and (0, 10). 
Answer: Let A(5, 7), B(3, 9), C(8, 6) and D(0, 10) be the given points. Therefore, by mid-point formula, we have,
Coordinates of the mid-point of AB are \( \left( \frac{5+3}{2}, \frac{7+9}{2} \right) = (4, 8) \)
Coordinates of the mid-point of CD are \( \left( \frac{8+0}{2}, \frac{6+10}{2} \right) = (4, 8) \)
Therefore, the mid-point of AB = mid point of CD.

 

Question. If (5,2), (- 3,4) and (x, y ) are collinear, show that x + 4y - 13 = 0. 
Answer: Since the points are collinear
The area of triangle = 0
\( \therefore \text{Area of triangle } = 0 \)
\( \frac{1}{2} [5(4 - y) + (-3)(y - 2) + x(2 - 4)] = 0 \)
\( = \frac{1}{2} [20 - 5y - 3y + 6 + (-2x)] = 0 \)
\( \frac{1}{2} [-2x - 8y + 26] = 0 \)
x + 4y - 13 = 0
Hence Proved.

 

Question. If the point C(-1, 2) divides the line segment AB in the ratio 3 : 4, where the coordinates of A are (2, 5), find the coordinates of B. 
Answer: Given: A (2,5) and C(-1,2)
Let the coordinate of the point B be (a,b).
it is given that AC : BC = 3:4
Then, by section formula, coordinates of C are given by
- 1 = \( \frac{3a + 4 \times 2}{3+4} \) and 2 = \( \frac{3b + 4 \times 5}{3+4} \)

\( \implies - 7 = 3a + 8 \) and 14 = 3b + 20

\( \implies 3a = -15 \) and 3b = -6

\( \implies a = - 5 \) and b = - 2
Hence, coordinates of B are (-5,-2).

 

Question. The three vertices of a parallelogram ABCD taken in order are A (-1, 0), B(3, 1) and C(2, 2). Find the height of a parallelogram with AD as its base. 
Answer: Area of \( \triangle ABC \)
\( = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac{1}{2} [-1(1 - 2) + 3(2 - 0) + 2(0 - 1)] \)
\( = \frac{1}{2} [1 + 6 - 2] = \frac{5}{2} \text{ sq. units} \)
Area of ||gm = \( 2 \times \text{area of } \triangle ABC \)

\( \implies \text{Area of ||gm} = 2 \times \frac{5}{2} = 5 \text{ sq. units} \)
Let coordinates of D are (x, y)
Mid point of AC = \( \left( \frac{-1+2}{2}, \frac{0+2}{2} \right) = (\frac{1}{2}, 1) \)
Mid-point of BD = \( \left( \frac{3+x}{2}, \frac{1+y}{2} \right) \)
\( \therefore \) Diagonals of a ||gm bisect each other
\( \therefore \) Mid-point of BD = Mid-point of AC

\( \implies \left( \frac{3+x}{2}, \frac{1+y}{2} \right) = (\frac{1}{2}, 1) \)

\( \implies \frac{3+x}{2} = \frac{1}{2} \text{ and } \frac{1+y}{2} = 1 \)

\( \implies x = - 2 \)

\( \implies y = 1 \)
Now \( AD = \sqrt{(-1 + 2)^2 + (0 + 1)^2} = \sqrt{2} \)
Also area of ||gm = base \( \times \) height

\( \implies AD \times \text{height} = 5 \)

\( \implies \sqrt{2} \times \text{height} = 5 \)

\( \implies \text{height} = \frac{5}{\sqrt{2}} = \frac{5}{2}\sqrt{2} \text{ units.} \)

 

Question. Find the ratio in which the line segment joining the points A(3, - 3) and B(- 2,7) is divided by the x-axis. Also, find the coordinates of the point of division. 
Answer: According to the question,
A (3,-3) and B (- 2, 7)
On the x-axis, the y-coordinate is zero
So, let the point be (x, 0)
Let the ratio be k : 1
\( (x, 0) = \left( \frac{-2k+3}{k+1}, \frac{7k-3}{k+1} \right) \)

\( \implies \frac{7k-3}{k+1} = 0 \)

\( \implies 7k - 3 = 0 \)

\( \implies k = \frac{3}{7} \)
\( \therefore \) The line is divided in the ratio of 3:7

\( \implies \frac{-2k+3}{k+1} = x \)

\( \implies \frac{-2 \times \frac{3}{7} + 3}{\frac{3}{7} + 1} = x \)

\( \implies \frac{-\frac{6}{7} + 3}{\frac{10}{7}} = x \)

\( \implies \frac{\frac{-6+21}{7}}{\frac{10}{7}} = x \)

\( \implies \frac{15/7}{10/7} = x \)

\( \implies x = \frac{3}{2} \)
Coordinate of y is 0 at x-axis,
\( \therefore \) The coordinates of the point at which x axis divides AB is \( (\frac{3}{2}, 0) \) in ratio of 3:7.

 

Question. Find the point on the x-axis which is equidistant from (2,-5) and (-2,9) 
Answer: Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
That is PA = PB
Hence \( PA^2 = PB^2 \rightarrow (1) \)
Distance between two points is \( \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2]} \)
\( PA = \sqrt{[(2 - x)^2 + (-5 - 0)^2]} \)
\( PA^2 = 4 - 4x + x^2 + 25 \)
\( = x^2 - 4x + 29 \)
Similarly, \( PB^2 = x^2 + 4x + 85 \)
Equation (1) becomes
\( x^2 - 4x + 29 = x^2 + 4x + 85 \)
- 8x = 56
x = -7
Hence the point on x-axis is (-7, 0)

 

Question. The points A \( (x_1, y_1) \), B \( (x_2, y_2) \) and C \( (x_3, y_3) \) are the vertices of \( \triangle ABC \).
i. The median from A meets BC at D. Find the coordinates of the point D.
ii. Find the coordinates of the point P on AD such that AP : PD = 2:1.
iii. Find the points of coordinates Q and R on medians BE and CP respectively such that BQ : QE = 2 :1 and CR : RP = 2 :1.
iv. What are the coordinates of the centroid of the triangle ABC? (4)
Answer: \( A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) \) are the three vertices of \( \triangle ABC \).
i. Median from A meets BC at D.
D is the mid-point of BC.
Coordinates of \( D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \)
ii. P divides AD in the ratio 2 : 1.
Coordinates of \( P = \left( \frac{2 \times \frac{x_2 + x_3}{2} + 1 \times x_1}{2+1}, \frac{2 \times \frac{y_2 + y_3}{2} + 1 \times y_1}{2+1} \right) \)
\( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
iii. Median from B meet AC at E and median from C meets AB at F.
E is the mid-point of AC and F is the mid-point of AB.
Coordinates of \( E = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) \) and Coordinates of \( F = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Q divides BE in the ratio 2 : 1.
Coordinates of \( Q = \left( \frac{2 \times \frac{x_1 + x_3}{2} + 1 \times x_2}{2+1}, \frac{2 \times \frac{y_1 + y_3}{2} + 1 \times y_2}{2+1} \right) \)
\( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
R divides CF in the ratio 2 : 1.
Coordinates of \( R = \left( \frac{2 \times \frac{x_1 + x_2}{2} + 1 \times x_3}{2+1}, \frac{2 \times \frac{y_1 + y_2}{2} + 1 \times y_3}{2+1} \right) \)
\( = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
iv. Coordinates of centroid of \( \triangle ABC = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

 

Question. Find the coordinates of the points Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B. 
Answer: Let Q(x, 0) be a point on x-axis which lies on the perpendicular bisector of AB.
Therefore, QA = QB

\( \implies QA^2 = QB^2 \)

\( \implies (–5 – x)^2 + (–2 – 0)^2 = (4 – x)^2 + (–2 – 0)^2 \)

\( \implies (x + 5)^2 + (–2)^2 = (4 – x)^2 + (–2)^2 \)

\( \implies x^2 + 25 + 10x + 4 = 16 + x^2 – 8x + 4 \)

\( \implies 10x + 8x = 16 – 25 \)

\( \implies 18x = –9 \)

\( \implies x = \frac{-9}{18} = \frac{-1}{2} \)
Hence, the point Q is \( \left( \frac{-1}{2}, 0 \right) \).
Now, \( QA^2 = [-5 + \frac{1}{2}]^2 + [-2 - 0]^2 \)

\( \implies QA^2 = (\frac{-9}{2})^2 + \frac{4}{1} = \frac{81}{4} + \frac{4}{1} = \frac{81+16}{4} = \frac{97}{4} \)

\( \implies QA = \sqrt{\frac{97}{4}} = \frac{\sqrt{97}}{2} \text{ units} \)
Now, \( QB^2 = (4 + \frac{1}{2})^2 + (-2 - 0)^2 = (\frac{9}{2})^2 + (-2)^2 \)

\( \implies QB^2 = \frac{81}{4} + \frac{4}{1} = \frac{81+16}{4} = \frac{97}{4} \)

\( \implies QB = \sqrt{\frac{97}{4}} = \frac{\sqrt{97}}{2} \text{ units} \)
and AB = \( \sqrt{(4 + 5)^2 + [-2 - (-2)]^2} = \sqrt{(9)^2} = 9 \text{ units} \)

\( \implies AB = 9 \text{ units} \)
As QA = QB
So, \( \triangle QAB \) is an isosceles \( \triangle \).

 

 

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