Prove that the lines $x=p y+q, z=r y+s$ and $x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime}$ are perpendicular if $p p^{\prime}+r r^{\prime}+1=0$.

(A) The given equations of the first line are $x=p y+q$ and $z=r y+s$.
This can be rewritten as $\frac{x-q}{p} = y = \frac{z-s}{r}$.
Thus,the direction ratios of the first line are $(p, 1, r)$.
Similarly,the equations of the second line are $x=p^{\prime} y+q^{\prime}$ and $z=r^{\prime} y+s^{\prime}$.
This can be rewritten as $\frac{x-q^{\prime}}{p^{\prime}} = y = \frac{z-s^{\prime}}{r^{\prime}}$.
Thus,the direction ratios of the second line are $(p^{\prime}, 1, r^{\prime})$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the direction ratios,we get $(p)(p^{\prime}) + (1)(1) + (r)(r^{\prime}) = 0$.
This simplifies to $p p^{\prime} + r r^{\prime} + 1 = 0$.
Hence,the lines are perpendicular if $p p^{\prime} + r r^{\prime} + 1 = 0$.