OP Malhotra Class 10 Maths Solutions Chapter 11 Coordinate Geometry Exercise 11 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 11 Coordinate Geometry Ex 11(A)

 

Question 1. Find the mid-points of lines joining
(a) (5, 8), (9, 11)
(b) (0, 0), (8, -5)
(c) (-7, 0), (0, 10)
(d) (-4, 3), (6, -7)
Answer: We use the mid-point formula: \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
(a) For (5, 8) and (9, 11), the mid-point is \( \left(\frac{5+9}{2}, \frac{8+11}{2}\right) = \left(\frac{14}{2}, \frac{19}{2}\right) = \left(7, \frac{19}{2}\right) \).
(b) For (0, 0) and (8, -5), the mid-point is \( \left(\frac{0+8}{2}, \frac{0-5}{2}\right) = \left(\frac{8}{2}, \frac{-5}{2}\right) = \left(4, \frac{-5}{2}\right) \).
(c) For (-7, 0) and (0, 10), the mid-point is \( \left(\frac{-7+0}{2}, \frac{0+10}{2}\right) = \left(\frac{-7}{2}, \frac{10}{2}\right) = \left(\frac{-7}{2}, 5\right) \).
(d) For (-4, 3) and (6, -7), the mid-point is \( \left(\frac{-4+6}{2}, \frac{3-7}{2}\right) = \left(\frac{2}{2}, \frac{-4}{2}\right) = (1, -2) \).
In simple words: To find the middle point between two points, you just average their x-coordinates and average their y-coordinates. This gives you the exact center of the line segment connecting them.

🎯 Exam Tip: Remember the mid-point formula is a simple average of coordinates. Make sure to keep the negative signs correct during addition or subtraction.

 

Question 2. Find the mid-points of the sides of a triangle whose vertices are A (1, -1), B (4, -1), C (4, 3).
Answer: Let D, E, and F be the mid-points of sides BC, CA, and AB respectively.
The vertices of the triangle are A(1, -1), B(4, -1), and C(4, 3).

Using the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):
Co-ordinates of D (mid-point of BC):
\( D = \left(\frac{4+4}{2}, \frac{-1+3}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1) \)
Co-ordinates of E (mid-point of CA):
\( E = \left(\frac{4+1}{2}, \frac{3-1}{2}\right) = \left(\frac{5}{2}, \frac{2}{2}\right) = \left(\frac{5}{2}, 1\right) \)
Co-ordinates of F (mid-point of AB):
\( F = \left(\frac{1+4}{2}, \frac{-1-1}{2}\right) = \left(\frac{5}{2}, \frac{-2}{2}\right) = \left(\frac{5}{2}, -1\right) \)
The mid-points of the sides AB, BC, and CA are \( \left(\frac{5}{2}, -1\right) \), \( (4, 1) \), and \( \left(\frac{5}{2}, 1\right) \) respectively.
In simple words: To find the midpoints of a triangle's sides, you take each side, find the average of the x-coordinates of its two end points, and the average of their y-coordinates. This gives you three new points, one on each side.

🎯 Exam Tip: Labeling the vertices A, B, C, and their mid-points D, E, F helps in systematically applying the mid-point formula for each side and avoids confusion. A simple diagram is always helpful.

 

Question 3. Find the centre of a circle if the end points of a diameter are A (-5, 7) and C (3, -11).
Answer: Let O be the centre of the circle. The centre of a circle is the mid-point of its diameter.
The end points of the diameter are A(-5, 7) and C(3, -11).
Let the coordinates of O be (x, y).

Using the mid-point formula: \( x = \frac{x_1+x_2}{2} \) and \( y = \frac{y_1+y_2}{2} \)
\( x = \frac{-5+3}{2} \)
\( \implies x = \frac{-2}{2} \)
\( \implies x = -1 \)

\( y = \frac{7+(-11)}{2} \)
\( \implies y = \frac{7-11}{2} \)
\( \implies y = \frac{-4}{2} \)
\( \implies y = -2 \)
Therefore, the co-ordinates of O (the center of the circle) are (-1, -2). A diameter always passes through the center, which is its midpoint.
In simple words: The center of a circle is simply the middle point of its diameter. You can find this by averaging the x-coordinates and y-coordinates of the two ends of the diameter.

🎯 Exam Tip: This question tests the understanding that the center of a circle is the midpoint of its diameter. Be careful with signs when adding negative numbers.

 

Question 4. If M is the mid-point of AB, find the co-ordinates of :
(a) A if the coordinates of M and B are M (2, 8) and B (-4, 19)
(b) B if the co-ordinates of A and M are A (-1, 2), M (-2, 4).
Answer: M is the mid-point of AB. We use the mid-point formula: \( M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
(a) Co-ordinates of M are (2, 8) and of B are (-4, 19). Let the co-ordinates of A be \( (x_1, y_1) \).
So, \( (2, 8) = \left(\frac{x_1+(-4)}{2}, \frac{y_1+19}{2}\right) \)
\( (2, 8) = \left(\frac{x_1-4}{2}, \frac{y_1+19}{2}\right) \)
Comparing the coordinates, we get:
\( \frac{x_1-4}{2} = 2 \)
\( \implies x_1-4 = 4 \)
\( \implies x_1 = 4+4 \)
\( \implies x_1 = 8 \)
And
\( \frac{y_1+19}{2} = 8 \)
\( \implies y_1+19 = 16 \)
\( \implies y_1 = 16-19 \)
\( \implies y_1 = -3 \)
Therefore, the co-ordinates of A are (8, -3).
(b) Let the co-ordinates of B be \( (x_2, y_2) \). Co-ordinates of A are (-1, 2) and M are (-2, 4).
So, \( (-2, 4) = \left(\frac{-1+x_2}{2}, \frac{2+y_2}{2}\right) \)
Comparing the coordinates, we get:
\( \frac{-1+x_2}{2} = -2 \)
\( \implies -1+x_2 = -4 \)
\( \implies x_2 = -4+1 \)
\( \implies x_2 = -3 \)
And
\( \frac{2+y_2}{2} = 4 \)
\( \implies 2+y_2 = 8 \)
\( \implies y_2 = 8-2 \)
\( \implies y_2 = 6 \)
Therefore, the co-ordinates of B are (-3, 6). Finding an endpoint when the midpoint is known requires reversing the midpoint formula.
In simple words: When you know the middle point and one end of a line, you can find the other end. You just double the middle point's coordinates and then subtract the known end point's coordinates to find the missing one.

🎯 Exam Tip: When given a midpoint and one endpoint, think of the midpoint formula as an equation to solve for the unknown endpoint's coordinates. Be careful with algebraic manipulation, especially with negative numbers.

 

Question 5. Find the coordinates of the point which divides internally the join of the points
(a) (8, 9) and (-7, 4) in the ratio 2 : 3
(b) (1, -2) and (4, 7) in the ratio 1 : 2.
Answer: If a point (x, y) divides a line segment with end points \( (x_1, y_1) \) and \( (x_2, y_2) \) internally in the ratio \( m_1 : m_2 \), then the coordinates are given by the section formula:
\( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \) and \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
(a) The points are (8, 9) and (-7, 4), and the ratio \( m_1 : m_2 \) is 2 : 3.
Here, \( x_1=8, y_1=9, x_2=-7, y_2=4, m_1=2, m_2=3 \).
\( x = \frac{2 \times(-7) + 3 \times 8}{2+3} = \frac{-14+24}{5} = \frac{10}{5} = 2 \)
\( y = \frac{2 \times 4 + 3 \times 9}{2+3} = \frac{8+27}{5} = \frac{35}{5} = 7 \)
The co-ordinates of the required point P are (2, 7).
(b) The points are (1, -2) and (4, 7), and the ratio \( m_1 : m_2 \) is 1 : 2.
Here, \( x_1=1, y_1=-2, x_2=4, y_2=7, m_1=1, m_2=2 \).
\( x = \frac{1 \times 4 + 2 \times 1}{1+2} = \frac{4+2}{3} = \frac{6}{3} = 2 \)
\( y = \frac{1 \times 7 + 2 \times(-2)}{1+2} = \frac{7-4}{3} = \frac{3}{3} = 1 \)
The co-ordinates of the required point are (2, 1). This formula helps find a point that divides a line segment into a specific proportion.
In simple words: To find a point that divides a line in a certain ratio, you use a special formula. You multiply the coordinates of each end point by the opposite part of the ratio, add them, and then divide by the total ratio.

🎯 Exam Tip: Clearly identify \( x_1, y_1, x_2, y_2, m_1, m_2 \) before applying the section formula to prevent calculation errors. Pay close attention to negative signs.

 

Question 6. Find the coordinates of the points of trisection of the line joining the points (2, 3) and (6, 5).
Answer: Let A(2, 3) and B(6, 5) be the given points. Trisection means dividing the line segment into three equal parts.
Let P and Q be the points that trisect the line segment AB.
So, AP = PQ = QB.
For point P, it divides AB in the ratio 1 : 2 (\( m_1=1, m_2=2 \)).
Using the section formula \( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \) and \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \):
\( x_P = \frac{1 \times 6 + 2 \times 2}{1+2} = \frac{6+4}{3} = \frac{10}{3} \)
\( y_P = \frac{1 \times 5 + 2 \times 3}{1+2} = \frac{5+6}{3} = \frac{11}{3} \)
The co-ordinates of P are \( \left(\frac{10}{3}, \frac{11}{3}\right) \).
For point Q, it divides AB in the ratio 2 : 1 (\( m_1=2, m_2=1 \)).
\( x_Q = \frac{2 \times 6 + 1 \times 2}{2+1} = \frac{12+2}{3} = \frac{14}{3} \)
\( y_Q = \frac{2 \times 5 + 1 \times 3}{2+1} = \frac{10+3}{3} = \frac{13}{3} \)
The co-ordinates of Q are \( \left(\frac{14}{3}, \frac{13}{3}\right) \). These two points break the line into three equal pieces.
In simple words: To split a line into three equal parts, you find two points. The first point is 1/3 of the way from the start, and the second point is 2/3 of the way from the start. You use the section formula twice, with ratios 1:2 and then 2:1.

🎯 Exam Tip: Remember that trisection involves two points. Point P divides the line in ratio 1:2, and point Q divides it in ratio 2:1. Or, Q can be found as the midpoint of P and B. Double-check your ratios and calculations.

 

Question 7. In what ratio is the line joining the points (2, -3) and (5, 6) divided by the x-axis, and (3, -6) and (-6, 8) divided by the y-axis?
Answer:
(a) For the line joining (2, -3) and (5, 6), let the x-axis divide it in the ratio \( m_1 : m_2 \).
A point on the x-axis has y-coordinate 0, so the coordinates of the division point are (x, 0).
Using the y-coordinate of the section formula: \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
\( 0 = \frac{m_1 \times 6 + m_2 \times (-3)}{m_1+m_2} \)
\( \implies 0 = 6m_1 - 3m_2 \)
\( \implies 6m_1 = 3m_2 \)
\( \implies \frac{m_1}{m_2} = \frac{3}{6} \)
\( \implies \frac{m_1}{m_2} = \frac{1}{2} \)
So the ratio is 1 : 2. This means the x-axis cuts the line segment closer to the point (2, -3).
(b) For the line joining (3, -6) and (-6, 8), let the y-axis divide it in the ratio \( m_1 : m_2 \).
A point on the y-axis has x-coordinate 0, so the coordinates of the division point are (0, y).
Using the x-coordinate of the section formula: \( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \)
\( 0 = \frac{m_1 \times (-6) + m_2 \times 3}{m_1+m_2} \)
\( \implies 0 = -6m_1 + 3m_2 \)
\( \implies 6m_1 = 3m_2 \)
\( \implies \frac{m_1}{m_2} = \frac{3}{6} \)
\( \implies \frac{m_1}{m_2} = \frac{1}{2} \)
So the ratio is 1 : 2. Similarly, the y-axis divides this line segment closer to (3, -6).
In simple words: If a line crosses the x-axis, its y-coordinate is zero. If it crosses the y-axis, its x-coordinate is zero. You use the section formula with this zero coordinate to find the ratio in which the axis cuts the line.

🎯 Exam Tip: Remember that any point on the x-axis has y-coordinate 0, and any point on the y-axis has x-coordinate 0. Use this fact to set up the section formula and solve for the ratio.

 

Question 8. Find the centroid of the triangle whose angular points are (-4, 6), (2, -2) and (2, 5) respectively.
Answer: Let the vertices of the triangle be A(-4, 6), B(2, -2), and C(2, 5).
The centroid (G) of a triangle is the point where the medians meet. Its coordinates are the average of the x-coordinates and the average of the y-coordinates of the vertices.
The formula for the centroid \( G(x, y) \) is \( \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \).
\( x = \frac{-4+2+2}{3} \)
\( \implies x = \frac{0}{3} \)
\( \implies x = 0 \)

\( y = \frac{6+(-2)+5}{3} \)
\( \implies y = \frac{6-2+5}{3} \)
\( \implies y = \frac{9}{3} \)
\( \implies y = 3 \)
Therefore, the co-ordinates of the centroid are (0, 3). This central point represents the triangle's center of mass.
In simple words: The centroid is like the balancing point of a triangle. You find it by adding all the x-coordinates together and dividing by three, and doing the same for all the y-coordinates.

🎯 Exam Tip: The centroid formula is straightforward. Ensure careful addition and division, especially with negative numbers. It's distinct from the circumcenter or incenter.

 

Question 9. If \( (x_1, y_1) = (2, 3) \); \( x_2 = 3 \) and \( y_3 = -2 \) and G is (0, 0), find \( y_2 \) and \( x_3 \).
Answer: The centroid G is (0, 0). The vertices are A(2, 3), B(\( x_2 \), \( y_2 \)), and C(\( x_3 \), \( y_3 \)).
We are given: \( x_1=2, y_1=3 \), \( x_2=3 \), and \( y_3=-2 \).
The centroid formula is \( G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \).
We know G = (0, 0). So we can set up two equations:
For the x-coordinate:
\( 0 = \frac{x_1+x_2+x_3}{3} \)
\( \implies 0 = \frac{2+3+x_3}{3} \)
\( \implies 0 = \frac{5+x_3}{3} \)
\( \implies 0 = 5+x_3 \)
\( \implies x_3 = -5 \)
For the y-coordinate:
\( 0 = \frac{y_1+y_2+y_3}{3} \)
\( \implies 0 = \frac{3+y_2+(-2)}{3} \)
\( \implies 0 = \frac{3+y_2-2}{3} \)
\( \implies 0 = \frac{y_2+1}{3} \)
\( \implies 0 = y_2+1 \)
\( \implies y_2 = -1 \)
Therefore, \( x_3 = -5 \) and \( y_2 = -1 \). This calculation demonstrates how to find missing vertex coordinates when the centroid is known.
In simple words: If you know the middle point (centroid) of a triangle and some parts of its corners, you can use the centroid formula backward to figure out the missing numbers for the corners.

🎯 Exam Tip: This question reverses the centroid problem. Remember to set each coordinate of the centroid equal to the average of the respective vertex coordinates and solve the resulting simple linear equations carefully.

 

Question 10. The mid-points of the sides of a triangle are at (1, 4), (4, 8) and (5, 6). Find the coordinates of the vertices of the triangle.
Answer: Let the vertices of the triangle be A(\( x_1, y_1 \)), B(\( x_2, y_2 \)), and C(\( x_3, y_3 \)).
Let the mid-points of the sides AB, BC, and CA be F(5, 6), D(1, 4), and E(4, 8) respectively.
Using the mid-point formula for each side:
For AB, with midpoint F(5, 6):
\( \frac{x_1+x_2}{2} = 5 \implies x_1+x_2 = 10 \) ... (i)
\( \frac{y_1+y_2}{2} = 6 \implies y_1+y_2 = 12 \) ... (v)
For BC, with midpoint D(1, 4):
\( \frac{x_2+x_3}{2} = 1 \implies x_2+x_3 = 2 \) ... (ii)
\( \frac{y_2+y_3}{2} = 4 \implies y_2+y_3 = 8 \) ... (vi)
For CA, with midpoint E(4, 8):
\( \frac{x_3+x_1}{2} = 4 \implies x_3+x_1 = 8 \) ... (iii)
\( \frac{y_3+y_1}{2} = 8 \implies y_3+y_1 = 16 \) ... (vii)
Now we solve these systems of linear equations.
Adding (i), (ii), and (iii) for x-coordinates:
\( (x_1+x_2) + (x_2+x_3) + (x_3+x_1) = 10+2+8 \)
\( 2(x_1+x_2+x_3) = 20 \)
\( x_1+x_2+x_3 = 10 \) ... (iv)
Subtracting (ii) from (iv): \( x_1 = 10-2 = 8 \)
Subtracting (iii) from (iv): \( x_2 = 10-8 = 2 \)
Subtracting (i) from (iv): \( x_3 = 10-10 = 0 \)
So, the x-coordinates of the vertices are \( x_1=8, x_2=2, x_3=0 \).
Adding (v), (vi), and (vii) for y-coordinates:
\( (y_1+y_2) + (y_2+y_3) + (y_3+y_1) = 12+8+16 \)
\( 2(y_1+y_2+y_3) = 36 \)
\( y_1+y_2+y_3 = 18 \) ... (viii)
Subtracting (vi) from (viii): \( y_1 = 18-8 = 10 \)
Subtracting (vii) from (viii): \( y_2 = 18-16 = 2 \)
Subtracting (v) from (viii): \( y_3 = 18-12 = 6 \)
So, the y-coordinates of the vertices are \( y_1=10, y_2=2, y_3=6 \).
Therefore, the co-ordinates of the vertices of the triangle are A(8, 10), B(2, 2), and C(0, 6). This process shows how knowing midpoints allows us to reconstruct the original shape.
In simple words: If you know the middle points of all sides of a triangle, you can find the actual corner points (vertices) by setting up a few simple equations. Add all the midpoint sums together to get a total, then subtract the individual sums to find each corner's coordinates.

🎯 Exam Tip: This problem requires solving a system of linear equations. Be systematic: first set up equations for all midpoints, then add them up, and finally subtract to find each vertex's coordinate. Keep your equations clearly labeled.

 

Question 11. The vertices of a triangle are A (-2, 2), B (4, 4) and C (8, 2). Find the length of the medians to the sides (i) BC, (ii) AC and (iii) AB.
Answer: The vertices of triangle ABC are A(-2, 2), B(4, 4), and C(8, 2).
Let D, E, and F be the mid-points of sides BC, CA, and AB respectively. The medians are AD, BE, and CF.

First, find the co-ordinates of the mid-points using the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \).
Mid-point D of BC (to find median AD):
\( D = \left(\frac{4+8}{2}, \frac{4+2}{2}\right) = \left(\frac{12}{2}, \frac{6}{2}\right) = (6, 3) \)
Mid-point E of AC (to find median BE):
\( E = \left(\frac{-2+8}{2}, \frac{2+2}{2}\right) = \left(\frac{6}{2}, \frac{4}{2}\right) = (3, 2) \)
Mid-point F of AB (to find median CF):
\( F = \left(\frac{-2+4}{2}, \frac{2+4}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3) \)
Now, calculate the length of each median using the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
(i) Length of median AD (from A(-2, 2) to D(6, 3)):
\( AD = \sqrt{(6-(-2))^2+(3-2)^2} = \sqrt{(6+2)^2+(1)^2} = \sqrt{(8)^2+1^2} = \sqrt{64+1} = \sqrt{65} \)
(ii) Length of median BE (from B(4, 4) to E(3, 2)):
\( BE = \sqrt{(3-4)^2+(2-4)^2} = \sqrt{(-1)^2+(-2)^2} = \sqrt{1+4} = \sqrt{5} \)
(iii) Length of median CF (from C(8, 2) to F(1, 3)):
\( CF = \sqrt{(1-8)^2+(3-2)^2} = \sqrt{(-7)^2+(1)^2} = \sqrt{49+1} = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \)
The lengths of the medians AD, BE, and CF are \( \sqrt{65} \), \( \sqrt{5} \), and \( 5\sqrt{2} \) respectively. Medians connect a vertex to the midpoint of the opposite side.
In simple words: To find the length of a median, first find the midpoint of the side opposite to the vertex. Then, calculate the distance between that vertex and the midpoint you just found using the distance formula. Do this for all three medians.

🎯 Exam Tip: This question combines midpoint and distance formulas. First, find all midpoints. Then, correctly pair each midpoint with its opposite vertex to calculate the median length. Pay attention to negative signs in the distance formula.

 

Question 12. Calculate the coordinates of the point P which divides the line joining A (- 1, 3) and B (5, 9) in the ratio 1:2.
Answer: Let the co-ordinates of the point P be (x, y). The line joins A(-1, 3) and B(5, 9), and P divides it in the ratio \( m_1 : m_2 = 1 : 2 \).
Using the section formula: \( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \) and \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
For x-coordinate:
\( x = \frac{1 \times 5 + 2 \times (-1)}{1+2} \)
\( \implies x = \frac{5-2}{3} \)
\( \implies x = \frac{3}{3} \)
\( \implies x = 1 \)
For y-coordinate:
\( y = \frac{1 \times 9 + 2 \times 3}{1+2} \)
\( \implies y = \frac{9+6}{3} \)
\( \implies y = \frac{15}{3} \)
\( \implies y = 5 \)
The co-ordinates of point P are (1, 5). This point is one-third of the way from A to B.
In simple words: To find a point that splits a line segment in a certain ratio, you use the section formula. Just plug in the coordinates of the two end points and the given ratio to get the coordinates of the dividing point.

🎯 Exam Tip: Clearly label \( x_1, y_1, x_2, y_2, m_1, m_2 \) before substituting into the section formula. Double-check your arithmetic, especially when dealing with negative numbers.

 

Question 13. The mid-point of the line joining A (2,p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Answer: Let P(3, 5) be the mid-point of the line joining A(2, p) and B(q, 4).
Using the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):
For the x-coordinate:
\( \frac{2+q}{2} = 3 \)
\( \implies 2+q = 3 \times 2 \)
\( \implies 2+q = 6 \)
\( \implies q = 6-2 \)
\( \implies q = 4 \)
For the y-coordinate:
\( \frac{p+4}{2} = 5 \)
\( \implies p+4 = 5 \times 2 \)
\( \implies p+4 = 10 \)
\( \implies p = 10-4 \)
\( \implies p = 6 \)
Therefore, the numerical values are \( p = 6 \) and \( q = 4 \). This shows how to find unknown coordinates of endpoints if the midpoint is given.
In simple words: If you know the middle point of a line and one or both of its end points have missing numbers, you can use the midpoint formula like a puzzle. Set up equations for the x and y values and solve for the missing letters.

🎯 Exam Tip: Treat the midpoint coordinates as given values in the midpoint formula. This allows you to set up two separate equations (one for x, one for y) and solve for the unknown variables.

 

Question 14.
(a) A is the point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Calculate the length of AB.
(b) The mid-point of the line joining (a, 2) and (3, 6) is (2, b). Find the numerical values of a and b.
Answer:
(a) Point A is on the y-axis, so its x-coordinate (abscissa) is 0. Its ordinate is 5, so A = (0, 5).
Point B is (-3, 1).
To find the length of AB, use the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
\( AB = \sqrt{(-3-0)^2+(1-5)^2} \)
\( \implies AB = \sqrt{(-3)^2+(-4)^2} \)
\( \implies AB = \sqrt{9+16} \)
\( \implies AB = \sqrt{25} \)
\( \implies AB = 5 \)
The length of AB is 5 units. A key step here is correctly identifying the coordinates of point A.
(b) The mid-point of the line joining (a, 2) and (3, 6) is (2, b).
Using the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):
For the x-coordinate:
\( \frac{a+3}{2} = 2 \)
\( \implies a+3 = 4 \)
\( \implies a = 4-3 \)
\( \implies a = 1 \)
For the y-coordinate:
\( \frac{2+6}{2} = b \)
\( \implies \frac{8}{2} = b \)
\( \implies b = 4 \)
Therefore, the numerical values are \( a = 1 \) and \( b = 4 \).
In simple words: (a) To find the distance between two points, first figure out their exact coordinates, especially if one is on an axis. Then use the distance formula. (b) If you know the middle point of a line and one or both ends have missing numbers, use the midpoint formula to find those missing values.

🎯 Exam Tip: For part (a), correctly determining the coordinates of a point on an axis (x=0 for y-axis, y=0 for x-axis) is crucial. For part (b), remember to equate the x-coordinates and y-coordinates separately using the midpoint formula.

 

Question 15. The line joining A (2, 3) and B (6, -5) meets the x-axis at P. Write down the y-coordinates of P. Hence find the ratio AP : PB.
Answer: The line segment joins A(2, 3) and B(6, -5).
Point P lies on the x-axis, which means its y-coordinate is 0. So the coordinates of P are (x, 0).
To find the ratio AP : PB, let P divide AB in the ratio \( m_1 : m_2 \).
Using the y-coordinate of the section formula: \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
\( 0 = \frac{m_1(-5) + m_2(3)}{m_1+m_2} \)
\( \implies 0 = -5m_1 + 3m_2 \)
\( \implies 5m_1 = 3m_2 \)
\( \implies \frac{m_1}{m_2} = \frac{3}{5} \)
The ratio AP : PB is 3 : 5. This means P is closer to point A than to B.
In simple words: When a line crosses the x-axis, its y-coordinate is always zero. You can use this fact with the section formula to find out how the x-axis splits the line into two parts.

🎯 Exam Tip: The key insight here is that any point on the x-axis has a y-coordinate of 0. Use this to set up the section formula equation and solve for the ratio \( m_1 : m_2 \). Be careful with the algebraic manipulation.

 

Question 16. In the given figure P (2, 3) is the mid-point of the line AB. Write down the coordinates of A and B.
Answer: From the figure, point A lies on the x-axis and point B lies on the y-axis. The mid-point of AB is P(2, 3).
Since A is on the x-axis, its y-coordinate is 0. Let A = (x, 0).
Since B is on the y-axis, its x-coordinate is 0. Let B = (0, y).

Using the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \):
For the x-coordinate of P:
\( \frac{x+0}{2} = 2 \)
\( \implies \frac{x}{2} = 2 \)
\( \implies x = 2 \times 2 \)
\( \implies x = 4 \)
For the y-coordinate of P:
\( \frac{0+y}{2} = 3 \)
\( \implies \frac{y}{2} = 3 \)
\( \implies y = 2 \times 3 \)
\( \implies y = 6 \)
Therefore, the co-ordinates of A are (4, 0) and B are (0, 6). The midpoint formula helps determine the endpoints when the midpoint and their axis locations are known.
In simple words: Since A is on the x-axis, its y-value is zero. Since B is on the y-axis, its x-value is zero. We use the midpoint formula with these facts to solve for the missing x and y coordinates of A and B.

🎯 Exam Tip: Visualizing the points on the coordinate plane can help. Remember the properties of points on the x-axis (y=0) and y-axis (x=0) when setting up the midpoint equations.

 

Question 17. The line segment joining A (2, 3) and B (6, -5) is intersected by the x-axis at a point K. Write down the ordinate of the point K. Hence find the ratio in which K divides AB.
Answer: The line segment joins A(2, 3) and B(6, -5).
Point K lies on the x-axis.
The ordinate (y-coordinate) of any point on the x-axis is 0.
So, the y-coordinate of point K is 0.
Let K divide AB in the ratio \( m_1 : m_2 \). The coordinates of K are (x, 0).
Using the y-coordinate of the section formula: \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
\( 0 = \frac{m_1(-5) + m_2(3)}{m_1+m_2} \)
\( \implies 0 = -5m_1 + 3m_2 \)
\( \implies 5m_1 = 3m_2 \)
\( \implies \frac{m_1}{m_2} = \frac{3}{5} \)
Therefore, the ratio in which K divides AB is 3 : 5. Point K cuts the line segment closer to A.
In simple words: When a point is on the x-axis, its y-coordinate is always zero. Using this, we can set up an equation with the section formula to find out how the x-axis splits the line into parts.

🎯 Exam Tip: A common mistake is to confuse ordinate with abscissa. Ordinate is the y-coordinate. Always clearly state this property of points on the x-axis or y-axis before applying the formula.

 

Question 18.
(a) Coordinates of A and B are (-3, a) and (1, a + 4). The mid-point of AB is (-1, 1). Find the value of a.
(b) Calculate the ratio in which the line segment joining (3, 4) and (-2, 1) is divided by the y-axis.
Answer:
(a) The coordinates of A are (-3, a) and B are (1, a+4). The mid-point P is (-1, 1).
Using the mid-point formula for the y-coordinate: \( \frac{y_1+y_2}{2} = y_P \)
\( \frac{a+(a+4)}{2} = 1 \)
\( \implies \frac{2a+4}{2} = 1 \)
\( \implies a+2 = 1 \)
\( \implies a = 1-2 \)
\( \implies a = -1 \)
The value of a is -1. Using the y-coordinate of the midpoint is sufficient to solve for 'a'.
(b) Let the line segment joining A(3, 4) and B(-2, 1) be divided by the y-axis in the ratio \( m_1 : m_2 \).
A point on the y-axis has an x-coordinate (abscissa) of 0. So the division point P has coordinates (0, y).
Using the x-coordinate of the section formula: \( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \)
\( 0 = \frac{m_1(-2) + m_2(3)}{m_1+m_2} \)
\( \implies 0 = -2m_1 + 3m_2 \)
\( \implies 2m_1 = 3m_2 \)
\( \implies \frac{m_1}{m_2} = \frac{3}{2} \)
The ratio is 3 : 2. This means the y-axis cuts the line segment closer to point B.
In simple words: (a) When you have a midpoint and endpoints with unknown parts, use the midpoint formula for the y-coordinates to find the missing number. (b) If a line crosses the y-axis, its x-coordinate is zero. Use this to find the ratio in which the y-axis divides the line using the section formula.

🎯 Exam Tip: For part (a), you only need to use one coordinate (x or y) to solve for 'a'. Choose the one where 'a' appears. For part (b), correctly identify that the x-coordinate is 0 for a point on the y-axis. Remember to correctly set up the ratio equation.

 

Question 19.
(a) P divides the distance between A (-2, 1) and B (1, 4) in the ratio 2 : 1. Calculate the coordinates of the point P.
(b) A (-5, 4), B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.
Answer:
(a) The line segment joins A(-2, 1) and B(1, 4). P divides AB in the ratio \( m_1 : m_2 = 2 : 1 \).
Using the section formula: \( x = \frac{m_1 x_2+m_2 x_1}{m_1+m_2} \) and \( y = \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \)
For x-coordinate of P:
\( x = \frac{2 \times 1 + 1 \times (-2)}{2+1} \)
\( \implies x = \frac{2-2}{3} \)
\( \implies x = \frac{0}{3} \)
\( \implies x = 0 \)
For y-coordinate of P:
\( y = \frac{2 \times 4 + 1 \times 1}{2+1} \)
\( \implies y = \frac{8+1}{3} \)
\( \implies y = \frac{9}{3} \)
\( \implies y = 3 \)
The co-ordinates of P are (0, 3). This point is two-thirds of the way from A to B.
(b) The vertices of the triangle are A(-5, 4), B(-1, -2), and C(5, 2). We need to find the coordinates of D(a, b) such that ABCD forms a square.

First, let's verify if ABC is an isosceles right-angled triangle by calculating the lengths of its sides using the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
\( AB^2 = (-1 - (-5))^2 + (-2 - 4)^2 = (4)^2 + (-6)^2 = 16 + 36 = 52 \)
\( BC^2 = (5 - (-1))^2 + (2 - (-2))^2 = (6)^2 + (4)^2 = 36 + 16 = 52 \)
\( AC^2 = (5 - (-5))^2 + (2 - 4)^2 = (10)^2 + (-2)^2 = 100 + 4 = 104 \)
Since \( AB^2 + BC^2 = 52 + 52 = 104 = AC^2 \), triangle ABC is a right-angled triangle with the right angle at B. Also, \( AB^2 = BC^2 \), so AB = BC, making it an isosceles triangle. This confirms the given information.
For ABCD to be a square, its diagonals must bisect each other. This means the midpoint of AC must be the same as the midpoint of BD.
Midpoint of AC, let's call it O:
\( O = \left(\frac{-5+5}{2}, \frac{4+2}{2}\right) = \left(\frac{0}{2}, \frac{6}{2}\right) = (0, 3) \)
Now, O(0, 3) is also the midpoint of BD. Let D = (a, b). B = (-1, -2).
Using the midpoint formula for BD:
\( \frac{-1+a}{2} = 0 \)
\( \implies -1+a = 0 \)
\( \implies a = 1 \)
\( \frac{-2+b}{2} = 3 \)
\( \implies -2+b = 6 \)
\( \implies b = 6+2 \)
\( \implies b = 8 \)
Therefore, the co-ordinates of D are (1, 8). This completes the square ABCD.
In simple words: (a) To find a point dividing a line in a ratio, use the section formula directly. (b) To complete a square when three corners are known, remember that diagonals of a square cut each other in half. Find the middle of the known diagonal, then use that middle point and the third known corner to find the missing fourth corner.

🎯 Exam Tip: For problems involving squares, remember key properties: diagonals bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other. Use this to set up equations for the unknown coordinates.

 

Question 20. A (2, 2), B (-2, 4), C (2, 6) are the vertices of a triangle ABC. Prove that ABC is an isosceles triangle.
Answer: The vertices of triangle ABC are given as A(2, 2), B(-2, 4), and C(2, 6). To prove that triangle ABC is an isosceles triangle, we need to show that at least two of its sides have equal lengths. We will use the distance formula, \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \), to calculate the square of the length of each side. Calculating the square of the lengths first simplifies the calculations by avoiding square roots until the final comparison. First, calculate the square of the length of side AB: \( AB^2 = (-2 - 2)^2 + (4 - 2)^2 \) \( AB^2 = (-4)^2 + (2)^2 \) \( AB^2 = 16 + 4 \) \( AB^2 = 20 \) Next, calculate the square of the length of side BC: \( BC^2 = [2 - (-2)]^2 + (6 - 4)^2 \) \( BC^2 = (2 + 2)^2 + (2)^2 \) \( BC^2 = (4)^2 + (2)^2 \) \( BC^2 = 16 + 4 \) \( BC^2 = 20 \) Finally, calculate the square of the length of side AC: \( AC^2 = (2 - 2)^2 + (2 - 6)^2 \) \( AC^2 = (0)^2 + (-4)^2 \) \( AC^2 = 0 + 16 \) \( AC^2 = 16 \) From these calculations, we see that \( AB^2 = 20 \) and \( BC^2 = 20 \).
\( \implies \) This means the lengths of sides AB and BC are equal (\( AB = BC \)). A triangle having two sides of equal length is defined as an isosceles triangle. Therefore, based on our findings, triangle ABC is an isosceles triangle.
In simple words: We measured how long each side of the triangle ABC is. We found that two sides, AB and BC, have the same length. Since an isosceles triangle has two equal sides, triangle ABC is an isosceles triangle.

🎯 Exam Tip: To prove a triangle is isosceles, always calculate the lengths of all three sides. Showing that any two sides are equal is sufficient. Remember to apply the distance formula correctly for each pair of vertices.