A light ray emerging from the point source placed at $\mathrm{P}(1,3)$ is reflected at a point $\mathrm{Q}$ in the axis of $x$. If the reflected ray passes through the point $R$ $(6,7)$, then the abscissa of $Q$ is:

Question

Aicharya · Accepted Answer

Let abcissa of $\mathrm{Q}=x$

$\begin{aligned} \therefore \quad & Q=(x, 0) \\ & \tan \theta=\frac{0-7}{x-6}, \tan \left(180^{\circ}-\theta\right)=\frac{0-3}{x-1}\end{aligned}$ Now, $\tan \left(180^{\circ}-\theta\right)=-\tan \theta$ $ \therefore \quad \frac{-3}{x-1}=\frac{-7}{x-6} \Rightarrow x=\frac{5}{2} $

A light ray emerging from the point source placed at $\mathrm{P}(1,3)$ is reflected at a point $\mathrm{Q}$…

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