(i) The line 4x - 3y + 12 = 0 meets the x-axis at A. Write down the coordinates of A.

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

(i) When the line meets x-axis, its y-coordinate = 0.

So, putting y = 0 in 4x - 3y + 12 = 0, we get

⇒ 4x - 3(0) + 12 = 0
⇒ 4x = -12
⇒ x = -3.

Hence, the line meets the x-axis at A(-3, 0).

(ii) Converting 4x - 3y + 12 = 0, in the form y = mx + c.

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x + 4$

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

Slope (m₁) = $\dfrac{4}{3}$

Let the slope of the line perpendicular to the given line be m₂.

∴ m₁ × m₂ = -1

$\Rightarrow \dfrac{4}{3} \times m$2 = -1 \\[1em] \Rightarrow m2 = -\dfrac{3}{4}.

Equation of the line having slope = $-\dfrac{3}{4}$ and passing through (-3, 0) can be given by,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 0 = -\dfrac{3}{4}(x - (-3)) \\[1em] \Rightarrow 4y = -3(x + 3) \\[1em] \Rightarrow 4y = -3x - 9 \\[1em] \Rightarrow 4y + 3x + 9 = 0.

Hence, the equation of the line is 3x + 4y + 9 = 0.