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(A) Let the point be $P(a, 6, 9)$ and its mirror image be $P'(20, b, -a-9)$.The midpoint $M$ of $PP'$ is $\left(\frac{a+20}{2}, \frac{6+b}{2}, \frac{9

Vedclass · Accepted Answer

$88$

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(A) Let the point be $P(a, 6, 9)$ and its mirror image be $P'(20, b, -a-9)$.The midpoint $M$ of $PP'$ is $\left(\frac{a+20}{2}, \frac{6+b}{2}, \frac{9-a-9}{2}\right) = \left(\frac{a+20}{2}, \frac{6+b}{2}, -\frac{a}{2}\right)$.Since $M$ lies on the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$,we have:$\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{6+b}{2}-2}{5} = \frac{-\frac{a}{2}-1}{-9} = k$.From the first and third parts: $\frac{a+14}{14} = \frac{a+2}{18} \implies 18a + 252 = 14a + 28 \implies 4a = -224 \implies a = -56$.Substituting $a = -56$ into the line equation: $\frac{-56+14}{14} = \frac{6+b-4}{10} \implies -3 = \frac{b+2}{10} \implies b+2 = -30 \implies b = -32$.Thus,$|a+b| = |-56 - 32| = |-88| = 88$.

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

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