If the line x - 4y - 6 = 0 is the perpendicular bisector of the line segment PQ and the coordinates of P are (1, 3), find the coordinates of Q.

Given, equation of line,

⇒ 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 PQ are perpendicular so their products will be equal to -1. Let slope of PQ be m₁,

$\therefore \dfrac{1}{4} \times m$1 = -1 \\[1em] \Rightarrow m1 = -4.

Hence, slope of PQ = -4.

Now equation of PQ can be found by point slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 3 = -4(x - 1) \\[1em] \Rightarrow y - 3 = -4x + 4 \\[1em] \Rightarrow 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 ……(i)
⇒ 4x + y = 7 ……(ii)

Multiplying (ii) with 4 and adding with (i) we get,

⇒ 16x + 4y + x - 4y = 28 + 6
⇒ 17x = 34
⇒ x = 2.

Putting value of x = 2 in (i),

⇒ 2 - 4y = 6
⇒ -4y = 4
⇒ y = -1.

Hence, the point of intersection which is the mid-point of PQ is (2, -1).

Let coordinates of Q be (a, b).

By mid-point formula, coordinates of mid-point of PQ are

$\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{1 + a}{2}, \dfrac{3 + b}{2}\Big) \\[1em]

Equating with mid-point of PQ (2, -1) we get,

$2 = \dfrac{1 + a}{2} \text{ and } -1 = \dfrac{3 + b}{2} \\[1em] 4 = 1 + a \text{ and } -2 = 3 + b \\[1em] a = 4 - 1 \text{ and } b = -2 - 3 \\[1em] a = 3 \text{ and } b = -5$

Hence, the coordinates of Q are (3, -5).