The line 3x - 4y + 12 = 0 meets x-axis at point A and y-axis at point B. Find :

(i) the coordinates of A and B.

(ii) equation of perpendicular bisector of line segment AB.

(i) We know that,

y-coordinate of any point on x-axis = 0.

So, let point A be (a, 0)

⇒ 3a - 4(0) + 12 = 0

⇒ 3a + 12 = 0

⇒ 3a = -12

⇒ a = $-\dfrac{12}{3}$ = -4.

A = (-4, 0).

x-coordinate of any point on y-axis = 0.

So, let point B be (0, b).

⇒ 3(0) - 4b + 12 = 0

⇒ 4b = 12

⇒ b = $\dfrac{12}{4}$ = 3.

B = (0, 3).

Hence, point A = (-4, 0) and B = (0, 3).

(ii) By formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Mid-point of AB = $\Big(\dfrac{-4 + 0}{2}, \dfrac{0 + 3}{2}\Big) = \Big(\dfrac{-4}{2}, \dfrac{3}{2}\Big)$ = (-2, 1.5).

By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Slope of AB = $\dfrac{3 - 0}{0 - (-4)} = \dfrac{3}{4}$.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of AB × Slope of perpendicular line = -1

⇒ $\dfrac{3}{4} \times$ Slope of perpendicular line = -1

⇒ Slope of perpendicular line (m) = $-\dfrac{4}{3}$.

By point-slope form, equation of line :

⇒ y - y₁ = m(x - x₁)

⇒ y - 1.5 = $-\dfrac{4}{3}$[x - (-2)]

⇒ 3(y - 1.5) = -4[x + 2]

⇒ 3y - 4.5 = -4x - 8

⇒ 4x + 3y - 4.5 + 8 = 0

⇒ 4x + 3y + 3.5 = 0

⇒ 4x + 3y + $\dfrac{35}{10}$ = 0

⇒ 4x + 3y + $\dfrac{7}{2}$ = 0

⇒ $\dfrac{8x + 6y + 7}{2}$ = 0

⇒ 8x + 6y + 7 = 0.

Hence, equation of perpendicular bisector of AB is 8x + 6y + 7 = 0.