Question

The point on the line 3x+2y+9=0 which is closest to the point (2, −1) is a) (1, −6) b)-2,3/2 c) (−1, −3) d) (1, −2). please give explanation of your answer​

Answer 1

Let the required point be (p, q).

The closest point (2, - 1) must lie on the perpendicular straight line joining the points (p, q) and (2, - 1).

Since (p, q) lies on 3x + 2y + 9 = 0,

3p + 2q + 9 = 0 ..... (i)

Gradient of the straight line joining the points (p, q) and (2, - 1) is (q + 1)/(p - 2), and that of the straight line 3x + 2y + 9 = 0 is (- 3/2).

Then,

(q + 1)/(p - 2) * (- 3/2) = - 1

or, 3 (q + 1) = 2 (p - 2)

or, 3q + 3 = 2p - 4

or, 2p - 3q - 7 = 0 ..... (ii)

We have two equations:

3p + 2q + 9 = 0 ..... (i)

2p - 3q - 7 = 0 ..... (ii)

Multiplying (i) by 3 and (ii) by 2, we get:

9p + 6q + 27 = 0

4p - 6q - 14 = 0

On adding, we get:

13p + 13 = 0

i.e., p = - 1

Then q = - 3

Therefore the closest point on the straight line 3x + 2y + 9 = 0 is (- 1, - 3).

Option (C) is correct.