Question

The two lines $x=my+n,z=py+q$ and $x=m^{\prime }y+n^{\prime },z=p^{\prime }y+q^{\prime }$ are perpendicular to each other, if

• A. $m{m}^{\prime }+p{p}^{\prime }=1$
• B. $\frac{m}{m^{\prime }}+\frac{p}{p^{\prime }}=-1$
• C. $\frac{m}{m^{\prime }}+\frac{p}{p^{\prime }}=1$
• D. $m{m}^{\prime }+p{p}^{\prime }=-1$

Answer 1

Answer: (D)

$m{m}^{\prime }+p{p}^{\prime }=-1$

Given lines are
$x=my+n,z=py+q$
and $x=m^{\prime }y+n^{\prime },z=p^{\prime }y+q^{\prime }$

$⟹\frac{x-n}{m}=y,\frac{z-q}{p}=y$

and $\frac{x-n^{\prime }}{m^{\prime }}=y,\frac{z-q^{\prime }}{p^{\prime }}=y$

Above equations can be rewritten as

$\frac{x-n}{m}=\frac{y-0}{1}=\frac{z-q}{p}$

and $\frac{x-n^{\prime }}{m^{\prime }}=\frac{y-0}{1}=\frac{z-q^{\prime }}{p^{\prime }}$

Lines will be perpendicular, if

$m{m}^{\prime }+1+p{p}^{\prime }=0$

$⟹m{m}^{\prime }+p{p}^{\prime }=-1$

Hence, option D is correct.