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

(i) the equation of the line AP

(ii) the co-ordinates of P

(i) Given,

⇒ 2x - 3y + 18 = 0

⇒ 3y = 2x + 18

⇒ y = $\dfrac{2}{3}x + 6$

Comparing above equation with y = mx + c we get,

m = $\dfrac{2}{3}$.

Since, AP is perpendicular to 2x - 3y + 18 = 0.

∴ Product of slope of AP and 2x - 3y + 18 = 0 will be -1.

Let slope of AP = m₁.

∴ m × m₁ = -1

⇒ $\dfrac{2}{3} \times m_1 = -1$

⇒ $m_1 = -\dfrac{3}{2}$.

By point-slope form, equation of AP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = $-\dfrac{3}{2}$[x - (-5)]

⇒ 2(y - 7) = -3(x + 5)

⇒ 2y - 14 = -3x - 15

⇒ 3x + 2y - 14 + 15 = 0

⇒ 3x + 2y + 1 = 0.

Hence, equation of AP is 3x + 2y + 1 = 0.

(ii) P is the point where AP and 2x - 3y + 18 = 0 meets,

Solving 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0 simultaneously,

⇒ 3x + 2y + 1 = 0

⇒ 2y = -3x - 1

⇒ y = $\dfrac{-3x - 1}{2}$ ………(1)

Substituting value of y in 2x - 3y + 18 = 0 we get,

$\Rightarrow 2x - 3\Big(\dfrac{-3x - 1}{2}\Big) + 18 = 0 \\[1em] \Rightarrow 2x - \dfrac{-9x - 3}{2} + 18 = 0 \\[1em] \Rightarrow 2x + \dfrac{9x + 3}{2} + 18 = 0 \\[1em] \Rightarrow \dfrac{4x + 9x + 3 + 36}{2} = 0 \\[1em] \Rightarrow 13x + 39 = 0 \\[1em] \Rightarrow 13x = -39 \\[1em] \Rightarrow x = -3.$

Substituting x = -3 in (1) we get,

$y = \dfrac{-3 \times -3 - 1}{2} \\[1em] = \dfrac{9 - 1}{2} \\[1em] = \dfrac{8}{2} \\[1em] = 4.$

P = (-3, 4).

Hence, co-ordinates of P = (-3, 4).