The line 4x - 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.

Determine the equation of line through A and perpendicular to 4x - 3y + 12 = 0.

At A,

y co-ordinate = 0 as it lies on x-axis.

A = (x, 0)

Substituting y = 0 in equation 4x - 3y + 12 = 0 we get,

⇒ 4x - 3(0) + 12 = 0

⇒ 4x + 12 = 0

⇒ 4x = -12

⇒ x = -3.

A = (x, 0) = (-3, 0).

Given,

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x$ + 4

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

m = $\dfrac{4}{3}$.

Let slope of line perpendicular to 4x - 3y + 12 = 0 be m₁.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

⇒ $\dfrac{4}{3} \times m_1 = -1$

⇒ $m_1 = -\dfrac{3}{4}$.

By point slope form,

Equation of line through A and slope = $-\dfrac{3}{4}$,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\dfrac{3}{4}$[x - (-3)]

⇒ 4y = -3(x + 3)

⇒ 4y = -3x - 9

⇒ 3x + 4y + 9 = 0.

Hence, equation of line through A and perpendicular to 4x - 3y + 12 = 0 is 3x + 4y + 9 = 0.