Equation of a line AB is x + 2y + 6 = 0. A perpendicular PQ is dropped on AB from the point P(3, –2) meeting AB at Q. Find the:

(a) equation of PQ.

(b) coordinates of the point Q.

(a) Given,

⇒ x + 2y + 6 = 0

⇒ 2y = -x - 6

⇒ y = $\dfrac{-x - 6}{2}$

⇒ y = -$\dfrac{1}{2}x$ - 3

Comparing y = $-\dfrac{1}{2}x$ - 3 with y = mx + c, we get :

Slope (m_AB) = $-\dfrac{1}{2}$

Given,

PQ is perpendicular to AB.

∴ Product of their slopes = -1

⇒ m_PQ × m_AB = -1

⇒ m_PQ × $\Big(-\dfrac{1}{2}\Big)$ = -1

⇒ m_PQ = -1 × -2

⇒ m_PQ = 2.

By point-slope formula,

Equation of PQ : y - y₁ = m(x − x₁)

⇒ y - (-2) = 2(x - 3)

⇒ y + 2 = 2x - 6

⇒ y = 2x - 8.

Hence, the equation of PQ is y = 2x - 8.

(b) The point Q is the intersection of line AB and line PQ.

Equation of AB

⇒ x + 2y + 6 = 0 …..(1)

Equation of PQ

⇒ y = 2x − 8 ….(2)

Substituting the value of y from (2) in (1), we get :

⇒ x + 2(2x - 8) + 6 = 0

⇒ x + 4x - 16 + 6 = 0

⇒ 5x - 10 = 0

⇒ 5x = 10

⇒ x = $\dfrac{10}{5}$

⇒ x = 2.

Substituting the value of x in equation (2), we get :

⇒ y = 2x − 8

⇒ y = 2(2) − 8

⇒ y = 4 − 8

⇒ y = -4.

Q = (x, y) = (2, -4).

Hence, the coordinates of the point Q are (2, -4).