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NCERT Class 10 Mathematics Exercise 7.2 Solutions: Coordinate Geometry

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 7.2 in NCERT's Class 10 Mathematics book, chapter “Coordinate Geometry”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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Ex 7.2 Q1
Find the point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
1. Identify the coordinates of the two points as A(−1, 7) and B(4, −3).
2. Note that the point P divides the segment AB in the ratio m : n = 2 : 3.
3. Apply the section formula for the x-coordinate: x = (m x₂ + n x₁) / (m + n).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
4. Substitute the values into the x-coordinate formula: x = (2 × 4 + 3 × (−1)) / (2 + 3).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Calculate the numerator and denominator for x: x = (8 − 3) / 5 = 5 / 5 = 1.
6. Apply the section formula for the y-coordinate: y = (m y₂ + n y₁) / (m + n).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
7. Substitute the values into the y-coordinate formula: y = (2 × (−3) + 3 × 7) / (2 + 3).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
8. Calculate the numerator and denominator for y: y = (−6 + 21) / 5 = 15 / 5 = 3.
9. Combine the calculated coordinates to find the point P(1, 3).
Answer: (1, 3)
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Ex 7.2 Q2
Find the points of trisection of the segment joining (4, −1) and (−2, −3).
1. Let A be (4, -1) and B be (-2, -3).
2. The points of trisection divide AB in the ratio 1:2 and 2:1.
3. Let P be the point closer to A, so AP:PB = 1:2.
4. Using the section formula, the x-coordinate of P is (1 × (-2) + 2 × 4) / (1 + 2) = 6/3 = 2.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
5. The y-coordinate of P is (1 × (-3) + 2 × (-1)) / (1 + 2) = -5/3.
6. So, P is (2, -5/3).
7. Let Q be the point closer to B, so AQ:QB = 2:1.
8. Using the section formula, the x-coordinate of Q is (2 × (-2) + 1 × 4) / (2 + 1) = 0/3 = 0.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
9. The y-coordinate of Q is (2 × (-3) + 1 × (-1)) / (2 + 1) = -7/3.
10. So, Q is (0, -7/3).
Answer: The points of trisection are (2, -5/3) and (0, -7/3).
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Ex 7.2 Q3
On a school ground with lines 1 m apart and 100 pots 1 m apart along AD (Fig. 7.12), Niharika posts a green flag 1/4 of AD along the 2nd line and Preet a red flag 1/5 of AD along the 8th line. How far apart are the flags, and where should Rashmi post a blue flag halfway between them?
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the length of AD be 100 m, so the 2nd line is at x = 2 and the 8th line is at x = 8.
2. Niharika's green flag is at 1/4 of AD along the 2nd line, so its coordinates are (2, 25).
3. Preet's red flag is at 1/5 of AD along the 8th line, so its coordinates are (8, 20).
4. The distance between the flags is √[(8 − 2)² + (20 − 25)²] = √[36 + 25] = √61 m.
5. The midpoint for Rashmi's blue flag is ((2 + 8)/2, (25 + 20)/2) = (5, 22.5).
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
Answer: The flags are √61 m apart, and Rashmi should post the blue flag at (5, 22.5).
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Ex 7.2 Q4
In what ratio does (−1, 6) divide the join of (−3, 10) and (6, −8)?
1. Let the point (−1, 6) divide the segment joining (−3, 10) and (6, −8) in the ratio k : 1.
2. Using the section formula for the x-coordinate: (k × 6 + 1 × (−3)) / (k + 1) = −1.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
3. Simplifying the numerator gives 6k − 3, so (6k − 3) / (k + 1) = −1.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. Multiplying both sides by (k + 1) yields 6k − 3 = −(k + 1).
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
5. Expanding the right side gives 6k − 3 = −k − 1.
  why?
  Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.
6. Adding k to both sides results in 7k − 3 = −1.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
7. Adding 3 to both sides gives 7k = 2.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
8. Dividing by 7 gives k = 2/7.
9. Thus, the ratio is 2/7 : 1, which simplifies to 2 : 7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: 2 : 7
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Ex 7.2 Q5
In what ratio does the x-axis divide the join of A(1, −5) and B(−4, 5)? Find the point of division.
1. Let the x-axis divide the line segment AB at point P in the ratio k : 1.
2. Since P lies on the x-axis, its y-coordinate is 0.
3. Using the section formula for the y-coordinate: 0 = (k(5) + 1(-5)) / (k + 1).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
4. Solving the equation: 5k - 5 = 0, which gives k = 1.
5. So, the ratio in which the x-axis divides AB is 1 : 1.
6. Now, find the x-coordinate of P using the section formula: x = (1(-4) + 1(1)) / (1 + 1).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
7. Calculating x: x = (-4 + 1) / 2 = -3/2.
8. Thus, the point of division is (-3/2, 0).
Answer: The x-axis divides AB in the ratio 1 : 1 and the point of division is (-3/2, 0).
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Ex 7.2 Q6
(1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram in order. Find x and y.
1. Let the vertices of the parallelogram be A(1, 2), B(4, y), C(x, 6), and D(3, 5).
2. In a parallelogram, the diagonals bisect each other, so the midpoint of diagonal AC is equal to the midpoint of diagonal BD.
3. The midpoint of AC is ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4).
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
4. The midpoint of BD is ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2).
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
5. Equating the x-coordinates of the midpoints: (1 + x)/2 = 7/2.
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
6. Multiplying both sides by 2 gives 1 + x = 7.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
7. Solving for x gives x = 6.
8. Equating the y-coordinates of the midpoints: 4 = (y + 5)/2.
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
9. Multiplying both sides by 2 gives 8 = y + 5.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
10. Solving for y gives y = 3.
Answer: x = 6, y = 3
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Ex 7.2 Q7
AB is a diameter of a circle with centre (2, −3) and B = (1, 4). Find A.
1. The centre of the circle is the midpoint of its diameter AB.
2. Let the coordinates of point A be (x, y).
3. Using the midpoint formula, the x-coordinate of the centre is (x + 1) ÷ 2.
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
4. Set (x + 1) ÷ 2 equal to the x-coordinate of the centre, which is 2.
5. Solving x + 1 = 4 gives x = 3.
6. Using the midpoint formula, the y-coordinate of the centre is (y + 4) ÷ 2.
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
7. Set (y + 4) ÷ 2 equal to the y-coordinate of the centre, which is −3.
8. Solving y + 4 = −6 gives y = −10.
Answer: A = (3, -10)
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Ex 7.2 Q8
A = (−2, −2), B = (2, −4). Find P on segment AB with AP = (3/7)AB.
1. The point P divides the segment AB in the ratio AP : PB = 3 : 4.
2. Using the section formula, the x-coordinate of P is (4 × (−2) + 3 × 2) / (4 + 3).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
3. Simplifying the x-coordinate gives (−8 + 6) / 7 = −2/7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. Using the section formula, the y-coordinate of P is (4 × (−2) + 3 × (−4)) / (4 + 3).
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
5. Simplifying the y-coordinate gives (−8 − 12) / 7 = −20/7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: (-2/7, -20/7)
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Ex 7.2 Q9
Find the points that divide the join of A(−2, 2) and B(2, 8) into four equal parts.
1. The points dividing the segment AB into four equal parts are the trisection points of the first half and the second half, or simply points at 1/4, 2/4, and 3/4 of the way from A to B.
2. Let P, Q, and R be the points dividing AB in the ratios 1:3, 1:1, and 3:1 respectively.
3. Using the section formula for P dividing AB in ratio 1:3, x-coordinate is (1×2 + 3×(−2)) / (1+3) = (2 − 6) / 4 = −1.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
4. The y-coordinate of P is (1×8 + 3×2) / (1+3) = (8 + 6) / 4 = 14 / 4 = 7/2.
5. So, the first point P is (−1, 7/2).
6. Using the section formula for Q dividing AB in ratio 1:1 (midpoint), x-coordinate is (2 + (−2)) / 2 = 0 / 2 = 0.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
7. The y-coordinate of Q is (8 + 2) / 2 = 10 / 2 = 5.
8. So, the second point Q is (0, 5).
9. Using the section formula for R dividing AB in ratio 3:1, x-coordinate is (3×2 + 1×(−2)) / (3+1) = (6 − 2) / 4 = 4 / 4 = 1.
  why?
  The point dividing the join of (x₁, y₁) and (x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — a weighted average of the two end points.
10. The y-coordinate of R is (3×8 + 1×2) / (3+1) = (24 + 2) / 4 = 26 / 4 = 13/2.
11. So, the third point R is (1, 13/2).
Answer: The points are (-1, 7/2), (0, 5), and (1, 13/2).
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Ex 7.2 Q10
Find the area of the rhombus with vertices (3, 0), (4, 5), (−1, 4) and (−2, −1) in order (area = half the product of the diagonals).
1. Let the vertices of the rhombus be A(3, 0), B(4, 5), C(−1, 4), and D(−2, −1).
2. The diagonals of the rhombus are AC and BD.
3. Calculate the length of diagonal AC using the distance formula: AC = √[(−1 − 3)² + (4 − 0)²].
  why?
  The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.
4. AC = √[(−4)² + (4)²] = √[16 + 16] = √32.
5. Calculate the length of diagonal BD using the distance formula: BD = √[(−2 − 4)² + (−1 − 5)²].
  why?
  The distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]: the two differences are the legs of a right triangle and the distance is its hypotenuse.
6. BD = √[(−6)² + (−6)²] = √[36 + 36] = √72.
7. The area of a rhombus is given by the formula: Area = ½ × (product of diagonals).
8. Area = ½ × AC × BD = ½ × √32 × √72.
9. Area = ½ × √(32 × 72) = ½ × √2304.
10. Area = ½ × 48 = 24.
Answer: 24 square units
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Exercise 7.1 — 10 questions

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NCERT Class 10 Mathematics Exercise 7.2 Solutions: Coordinate Geometry

Ex 7.2 Q1

Ex 7.2 Q2

Ex 7.2 Q3

Ex 7.2 Q4

Ex 7.2 Q5

Ex 7.2 Q6

Ex 7.2 Q7

Ex 7.2 Q8

Ex 7.2 Q9

Ex 7.2 Q10