Class 10 Mathematics: Coordinate Geometry important questions

Coordinate Geometry is a chapter of Class 10 Mathematics for the CBSE. These 16 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. The vertices of a parallelogram PQRS taken in order are P(3, 4), Q(−2, 3) and R(−3, −2). The fourth vertex S is: (a) (−2, −1) (b) (−2, −3) (c) (2, −1) (d) (1, 2)

   (c) (2, −1) The diagonals bisect each other, so S = P + R − Q = (3 − 3 + 2, 4 − 2 − 3) = (2, −1).

2. Practice · 1 marks

   2. Assertion (A): If the midpoints of the sides AB and AC of ΔABC are D(3, 5) and E(−3, −3), then BC = 20 units. Reason (R): The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.

   A. Both are true, and R explains A: DE = √(36 + 64) = 10, so BC = 2 × 10 = 20.

3. Practice · 1 marks

   3. In what ratio is the line segment joining (2, −3) and (5, 6) divided by the x-axis? (a) 1 : 2 (b) 2 : 1 (c) 2 : 5 (d) 5 : 2

   (a) 1 : 2 For ratio k : 1, the y-coordinate (6k − 3)/(k + 1) = 0, so k = 1/2.

4. Practice · 1 marks

   4. A point (x, y) is at a distance of 5 units from the origin. How many such points lie in the third quadrant? (a) 0 (b) 1 (c) 2 (d) infinitely many

   (d) Infinitely many: every point of the circle x² + y² = 25 that lies in the third quadrant.

5. Very Important · 4 marks

   5. Football players' positions are shown on a coordinate plane. (i) A(1, 2), B(4, 3), C(6, 6) and the central midfielder D form a parallelogram ABCD. Find D. (ii) Check whether G(−3, 5), H(3, 1) and K(0, 3) lie on a straight line. OR Check whether J(5, −3) and I(−4, 6) are equidistant from C(0, 1), and whether C is the midpoint of IJ. (iii) A(1, 4), B(2, −3) and E(a, b) lie on a straight line, and B is equidistant from A and E. Find E.

   (i) The diagonals bisect each other: midpoint of AC = (7/2, 4) = midpoint of BD, so D = (3, 5). (ii) GH = √52 = 2√13, GK = √13, HK = √13. GK + HK = GH, so G, H, K are collinear. OR CJ = √(25 + 16) = √41 and CI = √(16 + 25) = √41, so they are equidistant. The midpoint of IJ is (1/2, 3/2) ≠ C, so C is not the midpoint. (iii) B is the midpoint of AE: (1 + a)/2 = 2 and (4 + b)/2 = −3, so E = (3, −10).

6. Practice · 1 marks

   6. The point on the x-axis nearest to the point (−4, −5) is: (A) (0, 0) (B) (−4, 0) (C) (−5, 0) (D) (√41, 0)

   (B) (−4, 0), the foot of the perpendicular from the point to the x-axis.

7. Practice · 1 marks

   7. A point on the x-axis divides the line segment joining A(2, −3) and B(5, 6) in the ratio 1 : 2. The point is: (A) (4, 0) (B) (7/2, 3/2) (C) (3, 0) (D) (0, 3)

   (C) (3, 0) x = (1×5 + 2×2)/3 = 3, y = (1×6 + 2×(−3))/3 = 0.

8. Important · 2 marks

   8. Find the point(s) on the x-axis at a distance of √41 units from the point (8, −5).

   Let the point be (x, 0). (8 − x)² + 25 = 41, so (8 − x)² = 16 and 8 − x = ±4. x = 4 or 12: the points are (4, 0) and (12, 0).

9. Important · 2 marks

   9. Show that the points A(−5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles triangle.

   AB = √(8² + 6²) = 10, BC = √(6² + 8²) = 10, AC = √(14² + 2²) = 10√2. AB = BC, so ΔABC is isosceles.

10. Practice · 1 marks

    10. The shortest distance (in units) of the point (2, 3) from the y-axis is: (A) 2 (B) 3 (C) 5 (D) 1

    (A) 2 The distance from the y-axis is the x-coordinate, 2.

11. Practice · 1 marks

    11. The points A(9, 0), B(9, −6), C(−9, 0) and D(−9, 6) are the vertices of a: (A) square (B) rectangle (C) parallelogram (D) trapezium

    (C) Parallelogram AB and CD are both vertical with length 6, and the diagonals AC and BD bisect each other at (0, 0), but the angles are not right angles.

12. Very Important · 4 marks

    12. On a picnic, three friends stand at P(2, 5), Q(4, 4) and R(8, 3). (i) Find the distance PR. (ii) Is Q the midpoint of PR? Justify. (iii) (A) Find the point on the x-axis equidistant from P and Q. OR (B) S divides PQ in the ratio 2 : 3. Find S.

    (i) PR = √[(8 − 2)² + (3 − 5)²] = √40 = 2√10 units. (ii) The midpoint of PR is (5, 4), not (4, 4), so Q is not the midpoint. (iii) (A) Let the point be (x, 0): (2 − x)² + 25 = (4 − x)² + 16, so 4x = 3 and the point is (3/4, 0). OR (B) S = ((2×4 + 3×2)/5, (2×4 + 3×5)/5) = (14/5, 23/5).

13. Practice · 1 marks

    13. Which point lies on the perpendicular bisector of the segment joining A(−3, −4) and B(3, 4)? (A) (0, 0) (B) (0, 3) (C) (3, 0) (D) (0, 4)

    (A) (0, 0) The midpoint of AB lies on its perpendicular bisector: ((−3 + 3)/2, (−4 + 4)/2) = (0, 0).

14. Practice · 1 marks

    14. PM is a median of ΔPQR with P(5, −6), Q(6, 4) and R(0, 0). The length of PM is: (A) 2√17 units (B) 2√15 units (C) √101 units (D) 10 units

    (A) 2√17 units M, the midpoint of QR, is (3, 2). PM = √[(5 − 3)² + (−6 − 2)²] = √68 = 2√17.

15. Practice · 1 marks

    15. Assertion (A): If three vertices of a parallelogram, in order, are (−1, −6), (2, −5) and (7, 2), then the fourth vertex is (4, 1). Reason (R): The diagonals of a parallelogram bisect each other. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.

    A. Both are true, and R explains A: the midpoint of (−1, −6) and (7, 2) is (3, −2), which is also the midpoint of (2, −5) and (4, 1).

16. Important · 3 marks

    16. In what ratio does the x-axis divide the segment joining (−4, −6) and (−1, 7)? Find the point of division.

    Let the ratio be k : 1. The point is ((−k − 4)/(k + 1), (7k − 6)/(k + 1)). It lies on the x-axis, so 7k − 6 = 0 and k = 6/7. The ratio is 6 : 7. x = (−6/7 − 4)/(6/7 + 1) = (−34/7)/(13/7) = −34/13. The point is (−34/13, 0).