$\quad$ Given $x=p y+q \Rightarrow y=\frac{x-q}{p}$
and $z=r y+s \Rightarrow y=\frac{z-s}{r}$
From (i) \& (ii)

$\Rightarrow \quad \frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r}$
Similarly,
$$
\frac{\mathrm{x}-\mathrm{q}^{\prime}}{\mathrm{p}^{\prime}}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}-\mathrm{s}^{\prime}}{\mathrm{r}^{\prime}}
$$
From Eqs. (ii) and (iv),
$$
\mathrm{a}_1=\mathrm{p}, \mathrm{b}_1=1, \mathrm{c}_1=\mathrm{r}
$$
and $a_2=p^{\prime}, b_2=1, c_2=r^{\prime}$
the lines are perpendicular to each other if:
$$
\Rightarrow \begin{aligned}
& \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0 \\
\Rightarrow & \mathrm{pp}^{\prime}+1+\mathrm{rr}^{\prime}=0
\end{aligned}
$$
Hence proved