Question:
Let $a, b \in R$. If the mirror image of the point $P(a, 6,9)$ with respect to the line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to:
Correct Option: , 2
Solution:
$P(a, 6,9), Q(20, b,-a-9)$
mid point of $\mathrm{PQ}=\left(\frac{\mathrm{a}+20}{2}, \frac{\mathrm{b}+6}{2},-\frac{\mathrm{a}}{2}\right)$
lie on line
$\frac{\frac{a+20}{2}-3}{7}=\frac{\frac{b+6}{2}-2}{5}=\frac{-\frac{a}{2}-1}{-9}$
$\frac{a+20-6}{14}=\frac{b+6-4}{10}=\frac{-a-2}{-18}$
$\frac{\mathrm{a}+14}{14}=\frac{\mathrm{a}+2}{18}$
$18 a+252=14 a+28$
$a=-56$
$\frac{b+2}{10}=\frac{a+2}{18}$
$\frac{b+2}{10}=\frac{-54}{18}$
$\frac{b+2}{10}=-3 \Rightarrow b=-32$
$|a+b|=|-56-32|=88$