Coordinate Geometry: Class 10 Important Questions with Answers

HOTS (higher-order) practice

H1. Find the ratio in which the point (x, 2) divides the join of (12, 5) and (4, –3). Also find x.

H2. Show that (1, 7), (4, 2), (–1, –1) and (–4, 4) are the vertices of a square.

H3. If A(5, 2), B(2, –2) and C(–2, t) are the vertices of a triangle right-angled at B, find t.

Case-based practice

20 MCQs (1 mark each)

1. The distance of (3, –4) from the origin is
   (a) 7 (b) 5 (c) 1 (d) 25
2. The distance of (–5, 2) from the y-axis is
   (a) 2 (b) –5 (c) 5 (d) $\sqrt{29}$
3. The mid-point of (2, –3) and (6, 7) is
   (a) (4, 2) (b) (8, 4) (c) (2, 5) (d) (4, 5)
4. The distance between (2, 3) and (5, 7) is
   (a) 5 (b) 25 (c) 7 (d) $\sqrt{7}$
5. The point (0, –4) lies on
   (a) x-axis (b) y-axis (c) quadrant III (d) quadrant IV
6. The point dividing (1, 2) and (4, 8) in 1 : 2 is
   (a) (2, 4) (b) (3, 6) (c) (2.5, 5) (d) (1, 4)
7. If (a, 2) is the mid-point of (4, 1) and (2, 3), then a is
   (a) 2 (b) 3 (c) 6 (d) 1
8. The x-axis divides the join of (2, –3) and (5, 6) in the ratio
   (a) 1 : 2 (b) 2 : 1 (c) 3 : 5 (d) 1 : 1
9. If the distance between (x, 0) and (0, 3) is 5, then x is
   (a) 4 only (b) ±4 (c) 2 (d) ±2
10. A(–1, 0), B(5, 0) are ends of a diameter. The centre is
    (a) (3, 0) (b) (2, 0) (c) (0, 2) (d) (4, 0)
11. The perimeter of the triangle with vertices (0, 0), (3, 0), (0, 4) is
    (a) 7 (b) 12 (c) 10 (d) 5
12. The points (0, 0), (2, 2), (4, 4) are
    (a) collinear (b) vertices of a right triangle (c) vertices of an isosceles triangle (d) none
13. The 4th vertex D of parallelogram ABCD with A(1, 2), B(3, 5), C(6, 7) is
    (a) (4, 4) (b) (8, 10) (c) (2, 0) (d) (4, 5)
14. If P(2, p) is the mid-point of (6, –5) and (–2, 11), p is
    (a) 6 (b) 3 (c) 2 (d) 4
15. The points equidistant from (2, 0) and (–2, 0) lie on
    (a) the y-axis (b) the x-axis (c) the line x = 2 (d) the line y = 2
16. The coordinates of a point on the x-axis 5 units from (3, 4) are
    (a) (3, 0) only (b) (0, 0) or (6, 0) (c) (8, 0) (d) (–2, 0)
17. The ratio in which (2, y) divides (–2, 3) and (6, 7) is
    (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1
18. The triangle with vertices (0, 0), (5, 0), (0, 5) is
    (a) equilateral (b) right isosceles (c) scalene (d) obtuse
19. The distance between (a cos θ, 0) and (0, a sin θ) is
    (a) a (b) $a^2$ (c) 2a (d) 0
20. The mid-point $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ of AB divides AB in the ratio
    (a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) 1 : 3

Answer key

1 (b) · 2 (c) · 3 (a) · 4 (a) · 5 (b) · 6 (a) · 7 (b) · 8 (a) · 9 (b) · 10 (b) · 11 (b) · 12 (a) · 13 (a) · 14 (b) · 15 (a) · 16 (b) · 17 (a) · 18 (b) · 19 (a) · 20 (c)

Hints: Q6 $\left(\frac{4+2}{3},\frac{8+4}{3}\right)$. Q8 $\frac{6k-3}{k+1}=0\Rightarrow k=\frac{1}{2}$. Q13 D = A + C − B. Q15 the y-axis is the perpendicular bisector of the two points. Q16 $(x-3{)}^2+16=25$. Q17 $x=2$ is the mid-point of −2 and 6.

Assertion–Reason (10)

(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.

1. A: The distance of (–3, 4) from the origin is 5. R: The distance of (x, y) from the origin is $\sqrt{x^2+y^2}$.
2. A: The point (0, 5) lies on the x-axis. R: Every point on the y-axis has abscissa 0.
3. A: (1, 1), (2, 2), (3, 3) are collinear. R: Three points are collinear if the sum of two distances equals the third.
4. A: The mid-point of (–4, 6) and (4, –6) is the origin. R: The mid-point of two points is the average of their coordinates.
5. A: The distance of (–7, 2) from the x-axis is 7. R: The distance of (x, y) from the x-axis is |y|.
6. A: (2, 3), (4, 1), (–2, –3) are vertices of a right triangle. R: A triangle is right-angled if the square of one side equals the sum of squares of the other two.
7. A: The x-axis divides the join of (1, –3) and (4, 6) in the ratio 1 : 2. R: A point on the x-axis has ordinate 0.
8. A: A(1, 3), B(–1, 2), C(2, 5) and D(4, 6) form a parallelogram. R: Diagonals of a parallelogram bisect each other.
9. A: The point equidistant from (0, 0) and (4, 0) lies on the line x = 2. R: The perpendicular bisector of a segment is the set of points equidistant from its ends.
10. A: A rhombus always has equal diagonals. R: All sides of a rhombus are equal.

Key: 1 (a) · 2 (d) · 3 (a) · 4 (a) · 5 (d) · 6 (d) · 7 (a) · 8 (a) · 9 (a) · 10 (d)

Q6: squares of the sides are 8, 52, 52; no side² equals the sum of the other two ⇒ A false (it is isosceles). Q7: $\frac{6k-3}{k+1}=0\Rightarrow k=\frac{1}{2}$, ratio 1 : 2 ✓. Q8: mid-point of AC $=\left(\frac{3}{2},4\right)$ = mid-point of BD ✓.

Very short answer (1 mark each)

1. Find the distance between (–2, 5) and (4, –3). Ans. $\sqrt{36+64}=10$
2. Write the mid-point of (a, b) and (–a, –b). Ans. (0, 0)
3. In which ratio does the y-axis divide the join of (–4, 1) and (2, 5)? Ans. 2 : 1
4. Find k if (k, 3) is equidistant from (2, 0) and (6, 0). Ans. k = 4
5. The centre of a circle is (1, 2) and one end of a diameter is (3, 5). Find the other end. Ans. (–1, –1)
6. Write the coordinates of the point where $2x+3y=12$ meets the y-axis. Ans. (0, 4)
7. Find the perimeter of the triangle with vertices (0, 0), (6, 0), (0, 8). Ans. 6 + 8 + 10 = 24
8. If the distance between (4, p) and (1, 0) is 5, find p. Ans. p = ±4

Short answer practice (2 marks each)

1. Find the point on the y-axis which is equidistant from A(6, 5) and B(–4, 3).

2. Find the ratio in which P(4, m) divides the join of A(2, 3) and B(6, –3). Hence find m.

3. If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find a and b.

Short answer practice (3 marks each)

4. Find the points of trisection of the line segment joining (–3, 6) and (6, –3).

5. Show that A(–3, 2), B(–5, –5), C(2, –3) and D(4, 4) are the vertices of a rhombus.

6. If the point P(x, y) is equidistant from A(a + b, b – a) and B(a – b, a + b), prove that bx = ay.

Long answer (5 marks)

7. The vertices of △ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to meet AB and AC at D and E such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}$. (i) Find D and E. (ii) Show that $DE=\frac{1}{4}BC$. (iii) Is DE parallel to BC? Why?