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

The correct option is B $(5,-8,-4)$

Coordinates of any point $Q$ on the given line are $(2r+1,-3r-1,8r-10)$

So, the direction cosines of $PQ$ are $2r,-3r-1,8r-10$

Now, $PQ$ is perpendicular to the given line, if

$2\left(2r\right)-3(-3r+1)+8(8r-10)=0\Rightarrow 77r-77=0\Rightarrow r=1$

So, 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 $PR$

$\Rightarrow \frac{a+1}{2}=3,\frac{b}{2}=-4,\frac{c}{2}=-2\Rightarrow a=5,b=-8,c=-4$

So, the coordinates of the required point are $(5,-8,-4)$