Exercise 7.2 — Coordinate Geometry

1. 12 marksNCERT Cl-10 Maths, Ex 7.2, Q1

   Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

   Hint. Substitute into ((m₁x₂ + m₂x₁)/(m₁+m₂), (m₁y₂ + m₂y₁)/(m₁+m₂)). Watch which coordinate m₁ multiplies.

   Step 1 — Name the parts. A(x₁, y₁) = (−1, 7), B(x₂, y₂) = (4, −3), and m₁ : m₂ = 2 : 3.

   The ratio is AP : PB, so m₁ = 2 goes with the far point B — that is, m₁ multiplies x₂ and y₂.

   Step 2 — The x-coordinate. x = (2 × 4 + 3 × (−1)) / (2 + 3) = (8 − 3)/5 = 5/5 = 1

   Step 3 — The y-coordinate. y = (2 × (−3) + 3 × 7) / 5 = (−6 + 21)/5 = 15/5 = 3

   Sanity check. The ratio 2 : 3 puts P nearer A than B, and (1, 3) is indeed less than halfway from (−1, 7) to (4, −3) ✓

   ✦ Answer: the point is (1, 3).

   Where students slip. Pairing m₁ with x₁, giving (2×(−1) + 3×4)/5 = 2. That produces the point dividing in the ratio 3 : 2 instead — a different point on the same segment.

   Another way. Think of it as walking: P is 2/5 of the way from A to B. The x-step is 4 − (−1) = 5, so x = −1 + (2/5)(5) = 1; the y-step is −3 − 7 = −10, so y = 7 + (2/5)(−10) = 3.

2. 23 marksNCERT Cl-10 Maths, Ex 7.2, Q2

   Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).

   Hint. Two points are needed. The first divides in the ratio 1 : 2, the second in the ratio 2 : 1.

   Let A(4, −1) and B(−2, −3), and let P and Q be the points of trisection with AP = PQ = QB.

   Step 1 — P divides AB in the ratio 1 : 2, because it cuts off one of the three equal parts, leaving two. x = (1 × (−2) + 2 × 4)/3 = (−2 + 8)/3 = 6/3 = 2 y = (1 × (−3) + 2 × (−1))/3 = (−3 − 2)/3 = −5/3

   P = (2, −5/3)

   Step 2 — Q divides AB in the ratio 2 : 1, since it sits two parts from A and one from B. x = (2 × (−2) + 1 × 4)/3 = (−4 + 4)/3 = 0 y = (2 × (−3) + 1 × (−1))/3 = (−6 − 1)/3 = −7/3

   Q = (0, −7/3)

   Check. The x-values run 4 → 2 → 0 → −2, stepping down by 2 each time, and the y-values run −1 → −5/3 → −7/3 → −3, stepping down by 2/3 each time. Equal steps confirm equal parts ✓

   ✦ Answer: the points of trisection are (2, −5/3) and (0, −7/3).

   Where students slip. Using the ratios 1 : 3 and 2 : 3. Trisection makes three equal parts, so the ratios are part-to-remaining-parts — 1 : 2 and 2 : 1 — not part-to-whole.

   Another way. Find P first, then note that Q is simply the midpoint of P and B: ((2 + (−2))/2, (−5/3 + (−3))/2) = (0, −7/3). The textbook points this shortcut out in its own Example 8.

3. 34 marksNCERT Cl-10 Maths, Ex 7.2, Q3 (Fig. 7.12)

   On a rectangular school ground ABCD, chalk lines are drawn 1 m apart and 100 flower pots are placed 1 m apart along AD. Niharika runs ¼ of the distance AD along the 2nd line and posts a green flag; Preet runs ⅕ of AD along the 8th line and posts a red flag. What is the distance between the two flags? If Rashmi posts a blue flag exactly halfway between them, where should it go?

   Hint. Turn the description into coordinates: the line number is the x-coordinate and the distance run along AD is the y-coordinate, with AD = 100 m.

   Step 1 — Set up coordinates. There are 100 pots at 1 m spacing along AD, so AD = 100 m. Take the line number as x and the distance run along AD as y.

   Green flag: 2nd line, ¼ × 100 = 25 m along, so (2, 25) Red flag: 8th line, ⅕ × 100 = 20 m along, so (8, 20)

   Step 2 — Distance between the flags. d = √((8 − 2)² + (20 − 25)²) = √(36 + 25) = √61

   √61 ≈ 7.81, so the flags are about 7.81 m apart.

   Step 3 — Where the blue flag goes. "Exactly halfway between" means the midpoint: ((2 + 8)/2, (25 + 20)/2) = (5, 22.5)

   ✦ Answer: the flags are √61 m ≈ 7.81 m apart, and Rashmi should post the blue flag on the 5th line, 22.5 m along AD.

   Where students slip. Reading "¼ the distance AD" as ¼ of a pot count or as 25 pots rather than 25 metres. The pots are 1 m apart precisely so that the two readings agree — but say the unit, because the answer is a length.

   Another way. You can avoid the midpoint formula for the last part: the blue flag must be as far from each flag as from the other, so it sits halfway along in both directions — 3 lines across from the 2nd line, and 2.5 m back from 25 m.

4. 43 marksNCERT Cl-10 Maths, Ex 7.2, Q4

   Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).

   Hint. The ratio is the unknown, so use the k : 1 form — one variable instead of two.

   Step 1 — Choose the k : 1 form. With A(−3, 10) and B(6, −8), a point dividing AB in the ratio k : 1 has coordinates ((6k − 3)/(k + 1), (−8k + 10)/(k + 1))

   Using k : 1 rather than m₁ : m₂ leaves one unknown, which is why this form exists.

   Step 2 — Match the x-coordinate to −1. (6k − 3)/(k + 1) = −1 6k − 3 = −(k + 1) = −k − 1 7k = 2, so k = 2/7

   Step 3 — Convert to a ratio. k : 1 = 2/7 : 1 = 2 : 7

   Step 4 — Verify with the y-coordinate, which has not been used yet. (−8 × 2/7 + 10)/(2/7 + 1) = (−16/7 + 70/7)/(9/7) = (54/7)/(9/7) = 6 ✓

   That the unused coordinate agrees is the real confirmation the point lies on the segment at all.

   ✦ Answer: the point (−1, 6) divides the segment in the ratio 2 : 7.

   Where students slip. Solving with the x-coordinate and stopping. The y-check is not optional decoration — if it failed, the given point would not lie on AB and the question would have no answer.

   Another way. Ratios of differences work too: along x the point has moved from −3 to −1, a step of 2, out of a total step of 9 from −3 to 6. So it is 2/9 of the way along, giving 2 : 7.

5. 53 marksNCERT Cl-10 Maths, Ex 7.2, Q5

   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.

   Hint. On the x-axis the y-coordinate is 0. Write the y-coordinate in k : 1 form and set it to zero.

   Step 1 — Write the dividing point in k : 1 form. x = (−4k + 1)/(k + 1), y = (5k − 5)/(k + 1)

   Step 2 — Use the defining property of the x-axis. Every point on the x-axis has y = 0, so (5k − 5)/(k + 1) = 0

   Step 3 — Solve. A fraction is zero exactly when its numerator is zero. 5k − 5 = 0, so k = 1

   The ratio is 1 : 1 — the x-axis happens to cut AB at its midpoint.

   Step 4 — Find the point by substituting k = 1. x = (−4 + 1)/2 = −3/2, y = 0

   Check. The midpoint of A(1, −5) and B(−4, 5) is ((1 − 4)/2, (−5 + 5)/2) = (−3/2, 0) ✓ — and it lands on the x-axis, as it must, because A and B are 5 below and 5 above it.

   ✦ Answer: the ratio is 1 : 1, and the point of division is (−3/2, 0).

   Where students slip. Setting the x-coordinate to zero instead of the y-coordinate. On the x-axis it is y that vanishes; x = 0 describes the y-axis.

   Another way. Spot it without algebra: A is 5 units below the axis and B is 5 units above it, so by symmetry the crossing point is exactly halfway — ratio 1 : 1, and the x-value is the average of 1 and −4.

6. 63 marksNCERT Cl-10 Maths, Ex 7.2, Q6

   If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

   Hint. "Taken in order" means the diagonals join the 1st to the 3rd vertex and the 2nd to the 4th. Diagonals of a parallelogram bisect each other.

   Step 1 — Identify the diagonals. With A(1, 2), B(4, y), C(x, 6), D(3, 5) taken in order, the diagonals are AC and BD.

   Step 2 — Use the defining property. The diagonals of a parallelogram bisect each other, which means they share a midpoint:

   midpoint of AC = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4) midpoint of BD = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2)

   Step 3 — Equate the x-coordinates. (1 + x)/2 = 7/2, so 1 + x = 7 and x = 6

   Step 4 — Equate the y-coordinates. 4 = (y + 5)/2, so y + 5 = 8 and y = 3

   Check. The vertices are (1, 2), (4, 3), (6, 6), (3, 5). Both diagonals now have midpoint (3.5, 4) ✓, and AB = √(9 + 1) = √10 matches DC = √(9 + 1) = √10 ✓

   ✦ Answer: x = 6 and y = 3.

   Where students slip. Pairing the wrong vertices as diagonals — for instance joining the 1st to the 2nd. "In order" fixes the labelling A, B, C, D around the shape, so the diagonals must be AC and BD.

   Another way. Equal opposite sides as vectors gives the same result: for a parallelogram, the step from A to B equals the step from D to C, so (4 − 1, y − 2) = (x − 3, 6 − 5), giving 3 = x − 3 and y − 2 = 1.

7. 73 marksNCERT Cl-10 Maths, Ex 7.2, Q7

   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).

   Hint. The centre of a circle is the midpoint of every diameter. Work the midpoint formula backwards.

   Step 1 — State the key fact. A diameter passes through the centre, and the centre is equidistant from both ends. So the centre is the midpoint of AB.

   Step 2 — Let A = (x, y) and write the midpoint condition. ((x + 1)/2, (y + 4)/2) = (2, −3)

   Step 3 — Solve the x-equation. (x + 1)/2 = 2, so x + 1 = 4 and x = 3

   Step 4 — Solve the y-equation. (y + 4)/2 = −3, so y + 4 = −6 and y = −10

   Check. The midpoint of (3, −10) and (1, 4) is ((3+1)/2, (−10+4)/2) = (2, −3) ✓ — and both radii come to √(1 + 49) = √50, as they must.

   ✦ Answer: A = (3, −10).

   Where students slip. Averaging the centre with B, giving (1.5, 0.5). That finds the midpoint of the *radius*, not the far end of the diameter — the unknown point is an endpoint, so it belongs inside the formula, not outside it.

   Another way. Step-counting is quicker: going from B(1, 4) to the centre (2, −3) is +1 across and −7 down, so continuing the same step from the centre reaches (2 + 1, −3 − 7) = (3, −10).

8. 83 marksNCERT Cl-10 Maths, Ex 7.2, Q8

   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.

   Hint. Convert the fraction of the whole into a ratio of the two parts before using the section formula.

   Step 1 — Turn AP = (3/7)AB into a ratio. If AP is 3 parts out of 7, then PB is the remaining 4 parts. So P divides AB internally in the ratio

   AP : PB = 3 : 4

   This conversion is the whole difficulty of the question — the section formula needs part-to-part, not part-to-whole.

   Step 2 — Apply the section formula with m₁ = 3, m₂ = 4. x = (3 × 2 + 4 × (−2))/7 = (6 − 8)/7 = −2/7 y = (3 × (−4) + 4 × (−2))/7 = (−12 − 8)/7 = −20/7

   Check. AB = √(16 + 4) = √20 ≈ 4.472, and (3/7) of that is ≈ 1.917. Meanwhile AP = √((−2/7 + 2)² + (−20/7 + 2)²) = √((12/7)² + (−6/7)²) = √(180/49) ≈ 1.917 ✓

   ✦ Answer: P = (−2/7, −20/7).

   Where students slip. Using the ratio 3 : 7 straight from the fraction. That would put P three parts from A and seven from B — ten parts in total, which is not the segment described.

   Another way. Work in steps: from A, move 3/7 of the way to B. The x-step is 2 − (−2) = 4, so x = −2 + (3/7)(4) = −2/7; the y-step is −4 − (−2) = −2, so y = −2 + (3/7)(−2) = −20/7.

9. 93 marksNCERT Cl-10 Maths, Ex 7.2, Q9

   Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.

   Hint. Three points are needed, dividing in the ratios 1 : 3, 2 : 2 and 3 : 1. The middle one is just the midpoint.

   Four equal parts need three dividing points; call them P, Q and R in order from A.

   Step 1 — P divides AB in the ratio 1 : 3. x = (1 × 2 + 3 × (−2))/4 = (2 − 6)/4 = −1 y = (1 × 8 + 3 × 2)/4 = (8 + 6)/4 = 14/4 = 7/2 P = (−1, 7/2)

   Step 2 — Q is the midpoint, since 2 : 2 reduces to 1 : 1. x = (−2 + 2)/2 = 0 y = (2 + 8)/2 = 5 Q = (0, 5)

   Step 3 — R divides AB in the ratio 3 : 1. x = (3 × 2 + 1 × (−2))/4 = (6 − 2)/4 = 1 y = (3 × 8 + 1 × 2)/4 = 26/4 = 13/2 R = (1, 13/2)

   Check. The x-values go −2, −1, 0, 1, 2 — steps of 1 — and the y-values go 2, 7/2, 5, 13/2, 8 — steps of 3/2. Uniform steps confirm four equal parts ✓

   ✦ Answer: the three points are (−1, 7/2), (0, 5) and (1, 13/2).

   Where students slip. Finding only the midpoint and calling the job done. Four parts require three cuts, so the answer is three points, not one.

   Another way. Halve twice. The midpoint of A and B is (0, 5); the midpoint of A and that is (−1, 7/2); the midpoint of that and B is (1, 13/2). No section formula needed at all.

10. 103 marksNCERT Cl-10 Maths, Ex 7.2, Q10

    Find the area of a rhombus whose vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1), taken in order. [Hint: area of a rhombus = ½ × product of its diagonals.]

    Hint. Taken in order means the diagonals join the 1st vertex to the 3rd and the 2nd to the 4th. You need only those two lengths.

    Step 1 — Work out which segments the diagonals are. With A(3, 0), B(4, 5), C(−1, 4), D(−2, −1) taken in order, the diagonals join opposite vertices, so they are AC and BD. The sides are not needed at all, since the hint's formula uses only the diagonals.

    Step 2 — Length of AC. AC = √((−1 − 3)² + (4 − 0)²) = √(16 + 16) = √32 = 4√2

    Step 3 — Length of BD. BD = √((−2 − 4)² + (−1 − 5)²) = √(36 + 36) = √72 = 6√2

    Step 4 — Apply the formula. Area = ½ × AC × BD = ½ × 4√2 × 6√2

    The two surds multiply to give √2 × √2 = 2, so Area = ½ × 24 × 2 = 24

    Worth confirming it really is a rhombus. Each side comes to √26: AB = √(1 + 25), BC = √(25 + 1), CD = √(1 + 25), DA = √(25 + 1) ✓ — four equal sides, and the diagonals differ, so rhombus rather than square.

    ✦ Answer: the area is 24 square units.

    Where students slip. Multiplying the diagonals as decimals (5.66 × 8.49) and rounding to something like 24.02. Keep the surds: 4√2 × 6√2 = 48 exactly, and half of that is 24.

    Another way. Because the diagonals of a rhombus meet at right angles, the shape is four congruent right triangles with legs 2√2 and 3√2. Each has area ½ × 2√2 × 3√2 = 6, and 4 × 6 = 24.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 7 now runs to two exercises — 7.1 and 7.2; the area of a triangle from coordinates, with old exercises 7.3 and 7.4, is no longer part of this chapter, and the summary lists only four results). Questions are referenced from the NCERT textbook for identification.