Question Bank — Coordinate Geometry

NCERT Class 10 Mathematics · Chapter 7 · CBSE Board Pattern

Class 10 Mathematics Chapter 7 Question Bank CBSE Board Pattern

Section A — MCQs (10 questions, 1 mark each)

  1. The distance between the points (2, 3) and (4, 1) is
    (a) \(\sqrt{8}\) (b) \(\sqrt{4}\) (c) 4 (d) 2

  2. The coordinates of the midpoint of the line segment joining A(–1, 7) and B(4, –3) are
    (a) (1.5, 2) (b) (3, 2) (c) (1.5, 5) (d) (2, 5)

  3. If the point P(x, y) divides the line segment joining A(4, –3) and B(8, 5) in the ratio 3 : 1 internally, then the x-coordinate of P is
    (a) 7 (b) 6 (c) 5 (d) 4

  4. The distance of the point (–5, 12) from the origin is
    (a) 13 (b) 7 (c) 17 (d) \(\sqrt{119}\)

  5. Assertion (A): The points (1, 7), (4, 2) and (–1, –1) are collinear.
    Reason (R): Three points are collinear if the sum of the distances between any two pairs equals the third distance.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  6. Assertion (A): The point (0, 9) is equidistant from A(6, 5) and B(–4, 3).
    Reason (R): A point on the y-axis is of the form (0, y) and the distance formula is applied to verify equality.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  7. If P divides the join of (–3, 10) and (6, –8) in the ratio k : 1 and the x-coordinate of P is –1, then the value of k is
    (a) 2 (b) 3 (c) 4 (d) 5

  8. The distance between the points (a, b) and (–a, –b) is
    (a) \(\sqrt{a^2 + b^2}\) (b) 2\(\sqrt{a^2 + b^2}\) (c) \(\sqrt{2(a^2 + b^2)}\) (d) a + b

  9. The coordinates of the point which divides the line segment joining (–6, 10) and (3, –8) in the ratio 2 : 7 are
    (a) (–4, 6) (b) (–3, 4) (c) (–5, 8) (d) (–2, 2)

  10. If the points (3, 2), (–2, –3) and (2, 3) form a triangle, then it is
    (a) isosceles (b) right-angled (c) equilateral (d) scalene

Section B — Very Short Answer (6 questions, 2 marks each)

  1. Find the distance between the points (–5, 7) and (–1, 3).
  2. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
  3. Determine whether the points (1, 5), (2, 3) and (–2, –11) are collinear.
  4. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
  5. Find the values of y for which the distance between P(2, –3) and Q(10, y) is 10 units.
  6. Find the ratio in which the point (–4, 6) divides the line segment joining A(–6, 10) and B(3, –8).

Section C — Short Answer (5 questions, 3 marks each)

  1. Show that the points (1, 7), (4, 2), (–1, –1) and (–4, 4) are the vertices of a square.
  2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
  3. 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.
  4. Find the values of x and y if (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order.
  5. 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).

Section D — Long Answer (3 questions, 5 marks each)

  1. Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts. Justify each division using the section formula.
  2. Show that the points (–1, –2), (1, 0), (–1, 2) and (–3, 0) form a rhombus. Verify by calculating all side lengths and one diagonal using the distance formula, and confirm one angle is 90° using the converse of Pythagoras theorem.
  3. A point P divides the line segment joining A(6, 1) and B(8, 2) in the ratio k : 1. If the coordinates of P are used to form a parallelogram with points C(9, 4) and D(p, 3) taken in order, find k and p. Provide a multi-step justification using both section and midpoint formulas.

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1: In a school ground marked with chalk lines 1 m apart, 100 flower pots are placed along side AD at 1 m intervals. Niharika places a green flag after running one-fourth the distance AD on the second line. Preet places a red flag after running one-fifth the distance AD on the eighth line. Rashmi wants to place a blue flag exactly at the midpoint of the segment joining the two flags.

(i) Taking A as origin and 1 m as unit, assign coordinates to the green and red flags.
(ii) Calculate the distance between the green and red flags using the distance formula.
(iii) Find the coordinates where Rashmi should place the blue flag.
(iv) Verify that the blue flag point lies on the line joining the two flags using the section formula.

Case 2: Towns A and B are located such that B is 36 km east and 15 km north of A. A relay tower P is to be placed on AB so that the distance from B to P is twice the distance from A to P.

(i) Represent the positions of A and B on the coordinate plane with A at the origin.
(ii) Find the ratio in which P divides AB.
(iii) Using the section formula, determine the coordinates of P.
(iv) Verify that the coordinates satisfy the given distance condition AP : PB = 1 : 2.

▶ Answer Key Attempt all questions first,
then tap to reveal

Section A

  1. (a) \(\sqrt{8}\) — Apply distance formula: \(\sqrt{(4-2)^2 + (1-3)^2} = \sqrt{4+4} = \sqrt{8}\) (1 mark).
  2. (a) (1.5, 2) — Mid-point formula: \(\left(\frac{-1+4}{2}, \frac{7-3}{2}\right) = (1.5, 2)\) (1 mark).
  3. (a) 7 — Section formula: \(x = \frac{3 \cdot 8 + 1 \cdot 4}{3+1} = 7\) (1 mark).
  4. (a) 13 — Distance from origin: \(\sqrt{(-5)^2 + 12^2} = 13\) (1 mark).
  5. (d) — Points are not collinear; distances PQ ≈ 7.07, QR ≈ 7.21, PR ≈ 1.41 do not satisfy collinearity condition (1 mark).
  6. (a) — Both true; verification: AP = BP = 5 units (1 mark).
  7. (a) 2 — Solve \(\frac{6k-3}{k+1} = -1\) gives k = 2 (1 mark).
  8. (b) 2\(\sqrt{a^2 + b^2}\) — Distance = \(\sqrt{(a-(-a))^2 + (b-(-b))^2} = 2\sqrt{a^2+b^2}\) (1 mark).
  9. (a) (–4, 6) — Section formula with ratio 2 : 7 yields (–4, 6) (1 mark).
  10. (b) right-angled — PQ² + PR² = QR² (1 mark).

Section B

  1. Distance = \(\sqrt{(-1+5)^2 + (3-7)^2} = \sqrt{32} = 4\sqrt{2}\) units (correct formula 1 mark, substitution & answer 1 mark).
  2. Coordinates = \(\left(\frac{2\cdot4 + 3\cdot(-1)}{5}, \frac{2\cdot(-3) + 3\cdot7}{5}\right) = (1, 3)\) (section formula 1 mark, calculation 1 mark).
  3. AB + BC = 10, AC = \(\sqrt{(-2-1)^2 + (-11-5)^2} = 10\sqrt{2} \neq\) AB + BC; not collinear (distance calculations 1 mark, conclusion 1 mark).
  4. Let point be (x, 0). Solve \(\sqrt{(x-2)^2 + 25} = \sqrt{(x+2)^2 + 81}\) → x = –7 (equation setup 1 mark, solution 1 mark).
  5. \(\sqrt{(10-2)^2 + (y+3)^2} = 10\) → 64 + (y+3)² = 100 → y = 3 or –9 (correct equation 1 mark, solutions 1 mark).
  6. Ratio = 2 : 7 (section formula equation –4 = (3m₁ – 6m₂)/(m₁ + m₂) yields m₁ : m₂ = 2 : 7) (setup 1 mark, ratio 1 mark).

Section C

  1. All sides = \(\sqrt{34}\), diagonals = \(\sqrt{68}\); AD² + DC² = AC² confirms 90° at D; hence square (side equality 1 mark, diagonal equality 1 mark, Pythagoras verification 1 mark).
  2. Points of trisection: (2, –5/3) and (0, –7/3) (apply section formula twice with ratios 1 : 2 and 2 : 1; each ratio 1 mark, coordinates 1 mark).
  3. Ratio 1 : 1; point (–1.5, 0) (x-axis intersection sets y = 0; solve k = 1) (ratio 1 mark, coordinates 1 mark, verification 1 mark).
  4. x = 6, y = 3 (mid-point of AC = mid-point of BD) (mid-point equality 1 mark, equations 1 mark, values 1 mark).
  5. A = (3, –7) (mid-point of AB equals centre (2, –3)) (mid-point formula 1 mark, solving 2 marks).

Section D

  1. Division points: (–1, 3.5), (0, 5), (1, 6.5) (four equal parts → ratios 1 : 3, 1 : 1, 3 : 1; each application of section formula 1 mark, final coordinates 2 marks).
  2. All sides = \(\sqrt{8}\); diagonals equal; AC² = 8 + 8 confirms 90°; rhombus (side calculations 2 marks, diagonal & angle verification 2 marks, conclusion 1 mark).
  3. k = 2, p = 7 (parallelogram diagonals bisect; equate mid-points after section formula on AB; multi-step equations 3 marks, values 2 marks).

Section E

Case 1

(i) Green (2, 25), Red (8, 20) (1 mark).
(ii) Distance = \(\sqrt{(8-2)^2 + (20-25)^2} = \sqrt{61}\) m (formula 1 mark, answer 1 mark).
(iii) Blue (5, 22.5) (mid-point 1 mark).
(iv) Ratio 1 : 1 confirms midpoint lies on segment (1 mark).

Case 2

(i) A(0,0), B(36,15) (1 mark).
(ii) Ratio 1 : 2 (given condition) (1 mark).
(iii) P(12,5) (section formula) (1 mark).
(iv) AP = 13, PB = 26 satisfies 1 : 2 (1 mark).

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All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.