Question

If the foot of the perpendicular of a point $P(2,-3,1)$ with respect to line L is $(\frac{-22}{7},\frac{-3}{14},\frac{-13}{14})$, then the coordinates of it's image $I$ is .

• A. $\frac{-58}{7},\frac{36}{14},\frac{-40}{14}$
• B. $\frac{-44}{7},\frac{-6}{14},\frac{-26}{14}$
• C. $\frac{44}{7},\frac{6}{14},\frac{-30}{14}$
• D. $\frac{-44}{7},\frac{6}{14},\frac{-30}{14}$

Answer 1

The correct option is A $\frac{-58}{7},\frac{36}{14},\frac{-40}{14}$

The midpoint of $P$ and its $I$ (image) w.r.t $L$ is foot of the perpendicular $Q$ on $L$. Hence, if $(x,y,z)$ are the coordinates of $I$, then,

$\Rightarrow (\frac{x+2}{2},\frac{y-3}{2},\frac{z+1}{2})=(\frac{-22}{7},\frac{-3}{14},\frac{-13}{14})$

Lets compare the corresponding coordinates,

$\frac{x+2}{2}=\frac{-22}{7}$

$\Rightarrow x+2=\frac{-44}{7}$

$\Rightarrow x=\frac{-44}{7}-2$

$\Rightarrow x=\frac{-58}{7}$------(i)

$\frac{y-3}{2}=\frac{-3}{14}$

$\Rightarrow y-3=\frac{-3}{14}$

$\Rightarrow y=\frac{-6}{14}+3$

$\Rightarrow y=\frac{36}{14}$-----(ii)

$\frac{z+1}{2}=\frac{-13}{14}$

$\Rightarrow z+1=\frac{-26}{14}$

$\Rightarrow z=\frac{-26}{14}-1$

$\Rightarrow z=\frac{-40}{14}$----(iii)

$\Rightarrow I=(\frac{-58}{7},\frac{36}{14},\frac{-40}{14})$