Coordinate Geometry — Practice

Question 1

The distance between the points (2, 3) and (5, 7) is:
(a) 4 (b) 5 (c) 6 (d) 7

Accepted Answer

(b) 5.

√((5−2)² + (7−3)²) = √(9 + 16) = √25 = 5 (a 3-4-5 triple).

Question 2

The midpoint of the segment joining (−2, 4) and (6, −2) is:
(a) (2, 1) (b) (4, 2) (c) (2, 2) (d) (1, 2)

Accepted Answer

(a) (2, 1).

((−2+6)/2, (4−2)/2) = (2, 1).

Question 3

The point (−3, 5) lies in which quadrant?
(a) I (b) II (c) III (d) IV

Accepted Answer

(b) II.

Negative x, positive y is the second quadrant.

Question 4

The point dividing the join of (1, 2) and (4, 5) internally in the ratio 2:1 is:
(a) (2, 3) (b) (3, 4) (c) (3, 3) (d) (2, 4)

Accepted Answer

(b) (3, 4).

((2·4 + 1·1)/3, (2·5 + 1·2)/3) = (9/3, 12/3) = (3, 4).

Question 5

The slope of the line passing through (1, 2) and (4, 8) is:
(a) 1 (b) 2 (c) 3 (d) 6

Accepted Answer

(b) 2.

(8 − 2)/(4 − 1) = 6/3 = 2.

Question 6

The centroid of the triangle with vertices (0, 0), (6, 0) and (3, 9) is:
(a) (3, 3) (b) (3, 9) (c) (2, 3) (d) (3, 6)

Accepted Answer

(a) (3, 3).

((0+6+3)/3, (0+0+9)/3) = (3, 3).

Question 7

The area of the triangle with vertices (0, 0), (4, 0) and (0, 3) is:
(a) 6 (b) 7 (c) 12 (d) 5

Accepted Answer

(a) 6.

Legs on the axes: ½ × base × height = ½ × 4 × 3 = 6.

Question 8

The value of k for which the points (2, 3), (4, k) and (6, −3) are collinear is:
(a) 0 (b) 1 (c) −1 (d) 3

Accepted Answer

(a) 0.

Slopes must match: (k − 3)/(4 − 2) = (−3 − 3)/(6 − 2) = −6/4 = −3/2. So (k − 3)/2 = −3/2 → k − 3 = −3 → k = 0.

Question 9

The distance of the point (3, 4) from the line 3x + 4y − 5 = 0 is:
(a) 2 (b) 3 (c) 4 (d) 5

Accepted Answer

(c) 4.

|3·3 + 4·4 − 5| / √(3² + 4²) = |20| / 5 = 4.

Practice — Coordinate Geometry