Question

The point P is the foot of perpendicular from A(−5, 7) to the line whose equation is 2x – 3y + 18 = 0. Determine :

1. the equation of the line AP.
2. the co-ordinates of P.

Answer 1

i. The given equation is

2x − 3y + 18 = 0

3y = 2x + 18

`y = 2/3 x + 6`

Slope of this line = `2/3`

Slope of a line perpendicular to this line

= `(-1)/(2/3)`

= `(-3)/2`

(x_1, y₁₎ = (−5, 7)

The required equation of the line AP is given by

y − y₁ = m(x − x₁)

`y - 7 = (-3)/2 (x + 5)`

2y − 14 = −3x − 15

3x + 2y + 1 = 0

ii. P is the foot of perpendicular from point A.

So P is the point of intersection of the lines 2x − 3y + 18 = 0 and 3x + 2y + 1 = 0.

2x − 3y + 18 = 0

`=>` 4x − 6y + 36 = 0

3x + 2y + 1 = 0

`=>` 9x + 6y + 3 = 0

Adding the two equations, we get,

13x + 39 = 0

x = −3

∴ 3y = 2x + 18

= −6 + 18

= 12

y = 4

Thus, the co-ordinates of the point P are (−3, 4).