Question

Let the image of the point $P(1,0,0)$ in the line
$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ be $R$ $(x,y,-z)$. Find $x+y+z$ ?

Answer 1

Any point $Q$ on the given line is of the form $(2r+1,-3r-1,8r-10).$
Direction ratios of $PQ$ are
$2r+1-1,-3r-1-0,8r-10-0$
i.e., $2r,-3r-1,8r-10$
If $Q$ is the foot of perpendicular, then $Q$ is perpendicular to the line whose D.R.'s are $2,-3,8$
$\therefore 2\left(2r\right)-3(-3r-1)+8(8r-10)=0$
$\therefore 77r-77=0$

$\therefore r=1$
Putting $r=1$, the point $Q$ is $(3,-4,-2)$ and $P$ is $(1,0,0).$
If $R(x,y,-z)$ be the reflection(image) of $P$ in the line, then $Q$ is mid - point of $PR$.
$\therefore (3,-4,-2)=(\frac{x+1}{2},\frac{y}{2},\frac{-z}{2})$
Above gives $x=5,y=-8,z=4$
$\therefore$ $R$ is $(5,-8,4).$

Hence, $x+y+z=1$.