The line x - 4y = 6 is the perpendicular bisector of the line segment AB. If B = (1, 3); find the coordinates of point A.

Given, equation of line,

⇒ x - 4y = 6

⇒ x - 4y - 6 = 0

⇒ 4y = x - 6

⇒ y = $\dfrac{1}{4}x - \dfrac{6}{4}$.

Comparing with y = mx + c, we get :

Slope = $\dfrac{1}{4}$.

Since, given line and AB are perpendicular to each other, so their products = -1. Let slope of line AB be m₁.

$\therefore \dfrac{1}{4} \times m_1 = -1$

⇒ m₁ = -4.

Now, equation of AB can be found by point slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -4(x - 1)

⇒ y - 3 = -4x + 4

⇒ 4x + y - 3 - 4 = 0

⇒ 4x + y - 7 = 0.

Since, line x - 4y - 6 = 0 is perpendicular bisector of 4x + y - 7 = 0 hence solving them simultaneously to find point of intersection,

⇒ x - 4y = 6 ………(1)

⇒ 4x + y = 7 ……..(2)

Multiplying (2) with 4 and adding with (1), we get :

⇒ 4(4x + y) + x - 4y = 4 × 7 + 6

⇒ 16x + 4y + x - 4y = 28 + 6

⇒ 17x = 34

⇒ x = $\dfrac{34}{17}$

⇒ x = 2.

Putting value of x in (1),

⇒ 2 - 4y = 6

⇒ -4y = 4

⇒ y = -1.

Mid-point of AB = (2, -1). Let A be (a, b).

By mid-point formula, coordinates of mid-point of AB are $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

$\therefore (2, -1) = \Big(\dfrac{a + 1}{2}, \dfrac{b + 3}{2}\Big) \\[1em] \Rightarrow \dfrac{a + 1}{2} = 2 \text{ and } \dfrac{b + 3}{2} = -1 \\[1em] \Rightarrow a + 1 = 4 \text{ and } b + 3 = -2 \\[1em] \Rightarrow a = 3 \text{ and } b = -5.$

Hence, coordinates of A = (3, -5).