question_answer

The image of the point \[(6,3,9)\]in the straight line \[x-2=\frac{1-y}{2}=\frac{z}{2}\]is

A)  \[\left( \frac{28}{9},\frac{83}{9},\frac{11}{9} \right)\]

B)  \[(28,83,11)\]

C)  \[(4,-3,4)\]

D)  \[(2,-9,-1)\]

Correct Answer: D

Solution :

Given equation of line is \[\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z-0}{2}=k\] [say] Any point on the line is \[Q(k+2,\,\,-2k+1,2k)\] Direction ratios of PQ are \[(k+2-6,-2k+1-3,2k-9)\] ie,  \[(k-4,\,-2k-2,\,2k-9),\] Since,  the line \[PQ\] is perpendicular to \[AB\]. \[\therefore \]   \[1(k-4)-2(-2k-2)+2(2k-9)=0\] \[\Rightarrow \] \[k-4+4k+4+4k-18=0\] \[\Rightarrow \] \[9\,k=18\] \[\Rightarrow \] \[k=2\] \[\therefore \] Point Q is \[(4,-3,4)\]. Let the image of P about line AB is \[R({{x}_{1}},{{y}_{1}},{{z}_{1}}),\] Where Q is the mid point of PR. \[\therefore \]   \[\frac{{{x}_{1}}+6}{2}=4,\,\frac{{{y}_{1}}+3}{2}=-3,\,\frac{{{z}_{1}}+9}{2}=4\] \[\Rightarrow \]  \[{{x}_{1}}=2,\,{{y}_{1}}=-9,\,{{z}_{1}}=-1\]