Let $a, b \in R$. If the mirror image of the point $\mathrm{p}(\mathrm{a}, 6,9)$ w.r.t. line…

Let $a, b \in R$. If the mirror image of the point $\mathrm{p}(\mathrm{a}, 6,9)$ w.r.t. line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to
  1. 88
  2. 86
  3. 90
  4. 84

Solution

Note that mid-point of the line joining points $(a, 6,9)$ and $(20, b,-a-9)$ lies on the given line. $\therefore \quad$ The midpoint is $\left(\frac{\mathrm{a}+20}{2}, \frac{6+\mathrm{b}}{2}, \frac{9-\mathrm{a}-9}{2}\right)$ Substituting this point in the equation of the given line, we get $\begin{array}{ll} & \frac{\frac{a+20}{2}-3}{7}=\frac{\frac{6+b}{2}-2}{5}=\frac{\frac{-a}{2}-1}{-9} \\ \therefore \quad & \frac{a+14}{14}=\frac{a+2}{18} \\ \therefore \quad & a=-56 \\ \therefore \quad & \frac{-56+4}{14}=\frac{2+b}{10} \\ \therefore \quad & b=-32 \\ \therefore \quad & |a+b|=88 \end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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