This page solves exercise 7.2 class 10 maths ncert solutions for Coordinate Geometry — all ten questions from Exercise 7.2 of the NCERT textbook (NCERT, p. 112), worked one step at a time. Every answer opens with the principle behind the section formula, shows the full substitution, and ends with the common error students make on that exact question.
Exercise 7.2 is the most exam-weighted part of Chapter 7 because it applies the section formula to real situations: dividing a segment in a given ratio, finding points of trisection, discovering an unknown ratio, and using the mid-point as a special case.
Work through the ten questions in order, then use the five-minute method recap at the bottom to lock in the workflow before a test.
Exercise 7.2 Class 10 Maths NCERT Solutions
This exercise trains you to apply the section formula — the tool that finds the coordinates of a point dividing a line segment internally in a given ratio.
The ten questions move from a direct substitution (Q1), through points of trisection (Q2, Q9), a figure-based word problem (Q3), unknown ratios (Q4, Q5), the mid-point as a 1 : 1 case (Q6, Q7), a fractional-distance trap (Q8), and finally the diagonal property of a rhombus (Q10).
Each question below is reproduced exactly as printed in the textbook, then solved with the reasoning shown. A quick map of which tool each question needs:
| Question | What it tests | Tool |
|---|---|---|
| Q1 | Direct division in ratio 2 : 3 | Section formula |
| Q2 | Points of trisection | Ratios 1 : 2 then 2 : 1 |
| Q3 | Flags in a physical ground | Figure reading + distance formula |
| Q4 | Find an unknown ratio | k : 1 form |
| Q5 | Division by the x-axis | x-axis means y = 0 |
| Q6 | Missing parallelogram vertices | Diagonals share a mid-point |
| Q7 | Diameter of a circle | Centre = mid-point of AB |
| Q8 | Fractional distance AP = 3/7 AB | Ratio 3 : 4, not 3 : 7 |
| Q9 | Four equal parts | Ratios 1 : 3, 1 : 1, 3 : 1 |
| Q10 | Area of a rhombus | Half the product of diagonals |
Section formula. The point P(x, y) dividing the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ is \[ \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \]
The formula comes from the AA similarity criterion of Chapter 6 — the perpendicular distances scale in the same ratio as the segment parts (NCERT, p. 107).
- k : 1 form: when the ratio is written k : 1, the coordinates become \( \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right) \).
- Mid-point formula: the 1 : 1 case, giving \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
- Axis rule: a point on the x-axis has y-coordinate 0; a point on the y-axis has x-coordinate 0.
- Trisection: cutting into three equal parts, so the two points use ratios 1 : 2 and 2 : 1.
- Parallelogram property: the diagonals bisect each other, so the two mid-points are equal.
- Rhombus area: \( \frac{1}{2} \) (product of the diagonals), with diagonals found by the distance formula.
Edge case: the formula in this chapter covers internal division only — P lying between A and B. External division, where P lies outside the segment, is studied in higher classes (NCERT, p. 113).
Question 1: Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
The section formula weighs each endpoint by the opposite part of the ratio. Because m₁ = 2 is attached to B and m₂ = 3 to A, the point is pulled more strongly toward A — exactly what a 2 : 3 split means (two parts from A, three parts from B).
- Step 1: Label A(−1, 7) = (x₁, y₁) and B(4, −3) = (x₂, y₂), with m₁ = 2 and m₂ = 3.
- Step 2: Apply the x-coordinate formula.
\[ x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 – 3}{5} = 1 \]
Step 3: Apply the y-coordinate formula.
\[ y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = 3 \]
Final answer: The dividing point is (1, 3).
Common error: swapping m₁ and m₂, which lies the point toward the wrong end. Check: the point must sit between (−1, 7) and (4, −3), and it must be closer to A because the ratio 2 : 3 gives the shorter part to A. (1, 3) is only 2 units along from (−1, 7) — correct.
Question 2: Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Trisection means cutting the segment into three equal parts, so there are two dividing points, not one. The first point P cuts off one part of three, so it divides AB in the ratio 1 : 2; the second point Q leaves one part remaining, so it divides AB in the ratio 2 : 1. The ratios are not equal.
Step 1 (Point P, ratio 1 : 2): A(4, −1), B(−2, −3), m₁ = 1, m₂ = 2.
\[ x = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = 2, \quad y = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-5}{3} \]
Step 2 (Point Q, ratio 2 : 1): m₁ = 2, m₂ = 1.
\[ x = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = 0, \quad y = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-7}{3} \]
Final answer: The points of trisection are \( \left( 2, -\frac{5}{3} \right) \) and \( \left( 0, -\frac{7}{3} \right) \).
Common error: finding only one point, or using 1 : 2 for both. Check: having found P, Q is the mid-point of PB (NCERT, p. 110, Example 8 note) — averaging P(2, −5/3) and B(−2, −3) gives (0, −7/3), matching.
Question 3: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Read the figure as a coordinate plane with A as the origin. The numbered chalk lines run vertically: the 2nd line has x-coordinate 2 and the 8th line has x-coordinate 8. The 100 pots spaced 1 m apart along AD make AD = 100 m, so the fraction of AD a student runs becomes the y-coordinate.
Step 1: Green flag = 2nd line, 1/4 of 100 = 25 m along AD, so G(2, 25).
Red flag = 8th line, 1/5 of 100 = 20 m along AD, so R(8, 20).
Step 2: Distance between the flags using the distance formula.
\[ GR = \sqrt{(8 – 2)^2 + (20 – 25)^2} = \sqrt{36 + 25} = \sqrt{61} \]
Step 3: The blue flag is the mid-point of G and R.
\[ \left( \frac{2 + 8}{2}, \frac{25 + 20}{2} \right) = (5, 22.5) \]
Final answer: The flags are \( \sqrt{61} \) m apart. The blue flag goes on the 5th line, 22.5 m from A along AD.
Common error: swapping which value is the line number (x) and which is the fraction of AD (y), or treating 1/4 of AD as 1/4 of one line. Check: the x-coordinates are the small line numbers, while the y-coordinates are distances up to 100, so (2, 25) and (8, 20) are the only sensible reading. Keep units in metres.
Question 4: Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
When the ratio is unknown, write it as k : 1 and set the section formula equal to the given point. One coordinate gives the equation that fixes k; the other coordinate confirms it — you must verify both before trusting the answer.
- Step 1: A(−3, 10), B(6, −8), ratio k : 1, point (−1, 6).
- Step 2: Set the x-coordinate equal to −1.
\[ -1 = \frac{k(6) + 1(-3)}{k + 1} \;\Rightarrow\; -k – 1 = 6k – 3 \;\Rightarrow\; 7k = 2 \;\Rightarrow\; k = \frac{2}{7} \]
Step 3: Verify in the y-coordinate: \( \frac{k(-8) + 10}{k+1} = \frac{-\frac{16}{7} + 10}{\frac{9}{7}} = 6 \), which matches.
Final answer: The point (−1, 6) divides AB in the ratio 2 : 7.
Common error: solving only the x-coordinate and never checking y, or a sign slip when cross-multiplying — the −1 on the left carries its negative sign through \(-1(k+1)\). This mirrors Example 7 of the textbook (NCERT, p. 109). Check: substitute k = 2/7 back into both coordinates; if only one works, the point does not lie on the segment.
Question 5: Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Any point on the x-axis has y-coordinate 0. So set the section-formula y-coordinate equal to zero; the equation then reveals the ratio. Do not set x = 0 — that is the condition for the y-axis, not the x-axis.
- Step 1: A(1, −5), B(−4, 5), ratio k : 1.
- Step 2: Set the y-coordinate to 0.
\[ \frac{k(5) + 1(-5)}{k + 1} = \frac{5k – 5}{k + 1} = 0 \;\Rightarrow\; k = 1 \]
Step 3: With k = 1 the ratio is 1 : 1, so the point is the mid-point.
\[ x = \frac{-4 + 1}{2} = -\frac{3}{2} \]
Final answer: The x-axis divides AB in the ratio 1 : 1 at the point \( \left( -\frac{3}{2}, 0 \right) \).
Common error: writing “on the x-axis” and setting x = 0. Remember the axis rule: x-axis → y = 0, y-axis → x = 0. Check: the answer’s y-coordinate is 0 and it lies between y = −5 and y = 5, so it is genuinely on the segment. This is the same logic the textbook uses in Example 9 on the y-axis (NCERT, p. 110).
Question 6: If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
The diagonals of a parallelogram bisect each other, so the two diagonals share the same mid-point. “Taken in order” tells you the vertices are A(1, 2), B(4, y), C(x, 6), D(3, 5), so the diagonals are AC and BD. Equating their mid-points gives one equation for x and one for y.
Step 1: Mid-point of AC = mid-point of BD.
\[ \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) \]
Step 2: Equate x-coordinates.
\[ \frac{1 + x}{2} = \frac{7}{2} \;\Rightarrow\; x = 6 \]
Step 3: Equate y-coordinates.
\[ \frac{8}{2} = \frac{y + 5}{2} \;\Rightarrow\; y = 3 \]
Final answer: x = 6 and y = 3.
Common error: pairing the wrong opposite vertices (e.g. AB and CD) as diagonals. Check: with x = 6 and y = 3, the mid-point of AC is (7/2, 4) and the mid-point of BD is (7/2, 4) — identical. This is the pattern of Example 10 (NCERT, p. 111).
Question 7: Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
The centre of a circle is the mid-point of every diameter. So the point (2, −3) is the mid-point of A and B(1, 4). Apply the mid-point formula in reverse: each coordinate of the mid-point is the average of the two endpoint coordinates, so set up and solve two equations.
Step 1: Let A = (x₁, y₁).
Centre = mid-point of A and B.
- Step 1: x-coordinate: \( \frac{x_1 + 1}{2} = 2 \;\Rightarrow\; x_1 = 3 \).
- Step 2: y-coordinate: \( \frac{y_1 + 4}{2} = -3 \;\Rightarrow\; y_1 = -10 \).
Final answer: A = (3, −10).
Common error: solving only one coordinate, or forgetting that the centre is the mid-point at all. Check: the mid-point of (3, −10) and (1, 4) is ((3+1)/2, (−10+4)/2) = (2, −3), the given centre — so (3, −10) is correct.
Question 8: If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
This is the classic trap of the exercise. AP = 3/7 AB means AP is 3 parts out of the whole AB of 7 parts, so the remaining part PB is 7 − 3 = 4 parts. The division ratio is therefore AP : PB = 3 : 4, never 3 : 7. The two ratio parts must always sum to the whole.
- Step 1: A(−2, −2), B(2, −4), ratio AP : PB = 3 : 4, so m₁ = 3, m₂ = 4.
- Step 2: Apply the section formula.
\[ x = \frac{3(2) + 4(-2)}{3 + 4} = \frac{6 – 8}{7} = -\frac{2}{7} \]
\[ y = \frac{3(-4) + 4(-2)}{3 + 4} = \frac{-12 – 8}{7} = -\frac{20}{7} \]
Final answer: \( P = \left( -\frac{2}{7}, -\frac{20}{7} \right) \).
Common error: writing the ratio as 3 : 7 because 7 appears in the denominator. Remember: the parts must sum to the whole, so 3 : 4 with 3 + 4 = 7. Check: P should lie between −2 and 2 in x, and between −4 and −2 in y — both hold for (−2/7, −20/7).
Question 9: Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Four equal parts need three dividing points, and the ratio changes at each one. The first point P₁ cuts one part of four, so it divides AB in the ratio 1 : 3. The second point P₂ is the exact middle, ratio 1 : 1 (the mid-point). The third point P₃ leaves one part remaining, ratio 3 : 1.
Step 1 (P₁, ratio 1 : 3): m₁ = 1, m₂ = 3.
\[ x = \frac{1(2) + 3(-2)}{4} = -1, \quad y = \frac{1(8) + 3(2)}{4} = \frac{7}{2} \]
Step 2 (P₂, ratio 1 : 1): the mid-point of A and B.
\[ \left( \frac{-2 + 2}{2}, \frac{2 + 8}{2} \right) = (0, 5) \]
Step 3 (P₃, ratio 3 : 1): m₁ = 3, m₂ = 1.
\[ x = \frac{3(2) + 1(-2)}{4} = 1, \quad y = \frac{3(8) + 1(2)}{4} = \frac{13}{2} \]
Final answer: The three points are \( \left( -1, \frac{7}{2} \right) \), (0, 5) and \( \left( 1, \frac{13}{2} \right) \).
Common error: finding only the middle point and stopping, or using 1 : 4 for the first point. Check: P₂ is also the mid-point of P₁ and P₃ — averaging (−1, 7/2) and (1, 13/2) gives (0, 5), confirming all three, as noted in Example 8 (NCERT, p. 110).
Question 10: Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint : Area of a rhombus = 1/2 (product of its diagonals)]
A rhombus’s area is half the product of its diagonals — not side length squared, which only works for a square. So find the two diagonals with the distance formula, then halve their product. The vertices taken in order give the diagonals as the joins of opposite vertices.
Step 1: Vertices in order are A(3, 0), B(4, 5), C(−1, 4), D(−2, −1).
Diagonals are AC and BD.
Step 2: Length of AC.
\[ d_1 = \sqrt{(3 – (-1))^2 + (0 – 4)^2} = \sqrt{16 + 16} = 4\sqrt{2} \]
Step 3: Length of BD.
\[ d_2 = \sqrt{(4 – (-2))^2 + (5 – (-1))^2} = \sqrt{36 + 36} = 6\sqrt{2} \]
Step 4: Area = half the product of the diagonals.
\[ \text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 48 = 24 \]
Final answer: The area of the rhombus is 24 square units.
Common error: pairing adjacent vertices as the diagonals, or computing side lengths and treating the rhombus as a square. Check: (3, 0) and (−1, 4) are opposite corners, as are (4, 5) and (−2, −1) — those pairs must be the diagonals. The hint in the question tells you the formula to use.
Section Formula Method Recap: A Five-Minute Workflow
Before a test, run this sequence on any division problem:
- Identify the two endpoints and decide which ratio part belongs to which endpoint (smaller part goes to the nearer endpoint).
- Apply the section formula \( \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \).
- If the ratio is unknown, write k : 1, set the formula equal to the given point, and solve — then verify the second coordinate.
- If the point lies on an axis, set the right coordinate to 0 (x-axis → y = 0, y-axis → x = 0).
- Verify the answer lies between the two endpoints and that the ratio parts sum to the whole.
Here is a fresh example with new numbers, not from the book: divide the segment joining (0, 0) and (6, 9) in the ratio 2 : 1.
\[ x = \frac{2(6) + 1(0)}{3} = 4, \quad y = \frac{2(9) + 1(0)}{3} = 6 \]
So the point is (4, 6), closer to (6, 9) because the larger part (2) belongs to that end.
| Tool | Formula | When to use |
|---|---|---|
| Section formula | \( \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \) | Dividing a segment internally in ratio m₁ : m₂ (Q1, Q2, Q8, Q9) |
| k : 1 form | \( \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right) \) | Finding an unknown ratio (Q4, Q5) |
| Mid-point formula | \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \) | Ratio 1 : 1 — centres, diameters, parallelogram diagonals (Q6, Q7) |
| Distance formula | \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \) | Lengths — flags apart, rhombus diagonals (Q3, Q10) |
| Rhombus area | \( \frac{1}{2} \) (product of diagonals) | Area of a rhombus from its vertices (Q10) |
The section formula’s derivation rests on the AA similarity criterion of Triangles, so a quick review of Triangles class 10 notes helps you see why the ratio of distances equals the ratio of the parts.
For the board, most questions here are 2–3 mark problems: a full-marks answer must state the formula, substitute with the ratio labels clear, and give the final coordinates or ratio with units where they apply.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing ratio 3 : 7 when AP = 3/7 AB | Ratio parts must sum to the whole: AP : PB = 3 : 4 | 3 + 4 = 7, matching the denominator 7 |
| Swapping m₁ and m₂ | The larger part belongs to the endpoint the point sits closer to | Your point must be nearer the endpoint whose ratio part is bigger |
| Setting x = 0 for “on the x-axis” | On the x-axis, y = 0; on the y-axis, x = 0 | Substitute back — the coordinate you set to 0 must actually be 0 |
| Finding only the mid-point in trisection | Two points: ratios 1 : 2 and 2 : 1 (or 1 : 3, 1 : 1, 3 : 1 for four parts) | The middle dividing point is the mid-point of the two outer ones |
For more chapter-level practice, browse the full Class 10 Maths notes collection, and if you need the distance-formula side of coordinate geometry, the related work in Introduction to Trigonometry notes builds on the same right-triangle ideas.
All solutions on this page are grounded in the official NCERT Class 10 Mathematics textbook chapter 7 PDF — open it to check any formula or figure against the source pages, no sign-up needed.
Frequently Asked Questions on Exercise 7.2
How do you find the ratio in which a point divides a line segment?
Take the point as P and the segment as AB. Write the ratio as k : 1 and set the section formula equal to P’s coordinates. Using the x-coordinate usually gives \( k = \frac{x_P – x_1}{x_2 – x_P} \) type equation; solve it, then verify the y-coordinate gives the same k.
For example, dividing (−3, 10) to (6, −8) at (−1, 6), the x-coordinate gives k = 2/7, and y confirms it, so the ratio is 2 : 7.
What are the points of trisection of a line segment and how do you find them?
Trisection splits a segment into three equal parts, so there are two points. The first point divides the segment in the ratio 1 : 2 and the second in the ratio 2 : 1 — never equal ratios. Apply the section formula twice, or find the first point and take the mid-point of it and B for the second (NCERT, p. 110).
Why is the ratio 3 : 4 in question 8 when it says AP = 3/7 AB?
Because 7 is the whole, not one part. AP is 3 parts of the whole AB; the rest, PB, is 7 − 3 = 4 parts. A division ratio is a comparison of the two pieces, so it is 3 : 4. Writing 3 : 7 would compare one piece to the whole, which is a length statement, not a division ratio.
How do I know which endpoint is (x1, y1) and which is (x2, y2) in the section formula?
By the order you choose, but you must stay consistent. If A is (x₁, y₁) and B is (x₂, y₂), then m₁ is the part touching A and m₂ the part touching B. The formula \( \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \) pairs m₁ with B’s coordinate and m₂ with A’s.
Swapping either the labels or the ratio parts lies the answer to the wrong end.
How does the midpoint formula help find a missing vertex of a parallelogram?
The diagonals of a parallelogram bisect each other, so they share the same mid-point. Write the mid-points of the two diagonals using the mid-point formula and equate them coordinate by coordinate. In Exercise 7.2 Q6 that gives x = 6 and y = 3 directly — one equation for each unknown.
For more support across chapters, the Class 10 NCERT notes hub and the wider CBSE Class notes pages organise every chapter’s essentials in one place.
Reference: NCERT Class 10 Mathematics textbook, chapter 7 (Coordinate Geometry).
Explore Class 10 Maths NCERT Solutions
More for this chapter:
Class 10 Maths on LearnCBSE:
Related chapters: