Prove that the lines x=py+q,z=ry+s and x=p'y+q',z=r'y+s' are perpendicular, if pp'+n'+1=0.

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Solution

We have, $x = p y + q \Rightarrow y = \frac{x - q}{P}$ ...(i)
and $z = r y + s \Rightarrow y = \frac{z - s}{r}$ ...(ii)
$\Rightarrow \frac{x - q}{p} = \frac{y}{1} = \frac{z - s}{r}$ [using Eqs.(i) and (ii)] ...(iii)
Similarly, $\frac{x - q^{'}}{p^{'}} = \frac{y}{1} = \frac{z - s^{'}}{r^{'}}$ ...(iv)
From Eqs. (iii) and (iv), $a_{1} = p, b - 1 = 1, c_{1} = r$ and $a_{2} = p^{'}, b - 2 = 1, c_{2} = r^{'}$
If these given lines are perpendicular to each other, then
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \Rightarrow p p^{'} + 1 + n^{'} = 0$
which is the required condition.

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