Question

The reflection of the point $P(1,0,0)$ in the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is

• A. $(3,-4,-2)$
• B. $(5,-8,-4)$
• C. $(1,-1,-10)$
• D. $(2,-3,8)$

Answer 1

Answer: (B)

$(5,-8,-4)$

Coordinates of any point $Q$ on the given line are $(2r+1,-3r-1,8r-10)$.
So the direction ratios of $PQ$ are $2r,-3r-1-,8r-10$.
Now $PQ$ is perpendicular to the given line.
If $2\left(2r\right)-3(-3r-1)+8(8r-10)=0$
$\Rightarrow$ $77r-77=0\Rightarrow r=1$
Thus the coordinates of $Q$, the foot of the perpendicular from P on the line are $(3,-4,-2)$.
Let $R(a,b,c)$ be the reflection of $P$ in the given line
then $Q$ is the mid-point of $PR$
$\Rightarrow$ $\frac{a+1}{2}=3,\frac{b}{2}=-4,\frac{c}{2}=-2$
$\Rightarrow$ $a=5,b=-8,c=-4$
Hence the coordinate of the required point are $(5,-8,-4)$.