Question

Sol. [Key Concept: Image of a line in the mirror $y = x$: reflect two of its points (swap coordinates) and join them.]

Image of $(0,0)$ w.r.t. $L$ lies on $L_{1}^{'}$ Image of (0,0) w.r.t. $L$ is

$$\begin{matrix} \frac{x - 0}{2} & \ = \frac{y - 0}{- 1} = - 2\left( \frac{- 1}{5} \right) \\ P^{'} & \ \equiv (x,y) = \left( \frac{4}{5},\frac{- 2}{5} \right) \end{matrix}$$

Equation of $L_{1}^{'}$ which passes through $(1,1)$ and $\left( \frac{4}{5},\frac{- 2}{5} \right)$ is

$$\begin{matrix} y - 1 & \ = \frac{\frac{- 2}{5} - 1}{\frac{4}{5} - 1}(x - 1) = \frac{- 7}{- 1}(x - 1) \\ y & \ = 7x - 6 \end{matrix}$$