Let a, b in R. If the mirror image of the poi | vedclass

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(A) Let the point be $P(a, 6, 9)$ and its image be $Q(20, b, -a-9)$.Let the line be $L: \frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$.The midpoi

Vedclass · Accepted Answer

$88$

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(A) Let the point be $P(a, 6, 9)$ and its image be $Q(20, b, -a-9)$.Let the line be $L: \frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$.The midpoint $M$ of $PQ$ 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 $L$,we have $\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{6+b}{2}-2}{5} = \frac{-\frac{a}{2}-1}{-9} = k$.Simplifying,we get $\frac{a+14}{14} = \frac{b+2}{10} = \frac{a+2}{18} = k$.From $\frac{a+14}{14} = \frac{a+2}{18}$,we get $18(a+14) = 14(a+2) \Rightarrow 18a + 252 = 14a + 28 \Rightarrow 4a = -224 \Rightarrow a = -56$.Substituting $a = -56$ into $\frac{a+14}{14} = k$,we get $k = \frac{-56+14}{14} = \frac{-42}{14} = -3$.Now,$\frac{b+2}{10} = -3 \Rightarrow b+2 = -30 \Rightarrow b = -32$.Thus,$|a+b| = |-56 - 32| = |-88| = 88$.

Column-$I$	Column-$II$
$(A)$ Linnaeus	$(i)$ First to discover viruses
$(B)$ Ivanowsky	$(ii)$ First to discover lichens
$(C)$ Iyengar	$(iii)$ Father of Indian Bryology
$(D)$ Tulasne	$(iv)$ Father of Taxonomy
$(E)$ Shiv Ram Kashyap	$(v)$ Father of modern Phycology in India

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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