Explanation

(a)
The given equation are
$3x + 4y - 5 = 0$ …(i)
and $4x - 3y - 15 = 0$ …(ii)
Since, these lines are perpendicular to each other so $\mathcal{\angle Q}PR$ is right angle and $P\mathcal{Q =}PR$. Hence, $\mathrm{\Delta}P\mathcal{Q}R$ is a right angle isosceles triangle.
$\angle\ P\mathcal{Q}R = \angle PR\mathcal{Q} = 45{^\circ}$
Slope of $P\mathcal{Q = -}\frac{3}{4}$ and slope of $PR = \frac{4}{3}$
Let slope of $\mathcal{Q}R = m$
$\therefore\tan{45{^\circ} = \pm}\left| \frac{\frac{4}{3} - m}{1 + \frac{4}{3}m} \right|$
$\Rightarrow m = \frac{1}{7},\ - 7$