Question

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. $(2,-9,-1)$

Answer 1

Answer: (D)

$(2,-9,-1)$

Answer 2

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.
$\therefore$ $\frac{{{x}_{1}}+6}{2}=4,\,\frac{{{y}_{1}}+3}{2}=-3,\,\frac{{{z}_{1}}+9}{2}=4$
$\Rightarrow$ ${{x}_{1}}=2,\,{{y}_{1}}=-9,\,{{z}_{1}}=-1$