Question

The point on the line $\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+5}{2}$ at a distance of $6$ from the point $(2,-3,-5)$ is

• A. $(3,-5,-3)$
• B. $(4,-7,-9)$
• C. $(0,2,-1)$
• D. $(-3,5,3)$

Answer 1

The correct option is B $(4,-7,-9)$

Given, equation of line is
$\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+5}{-2}$
DR's of given line is $<1,-2,-2>$
Then, DC's of given line is $<\frac{1}{3},\frac{-2}{3},\frac{-2}{3}>$
Thus, the given equation can be rewritten as
$\frac{x-2}{\frac{1}{3}}=\frac{y+3}{\frac{-2}{3}}=\frac{z+5}{-\frac{2}{3}}=r$ [say]
$\therefore$ Any point on the line is $(2+\frac{r}{3},-3-\frac{2r}{3},-5-\frac{2r}{3})$
where, $r=\pm 6$
Hence, points are $(4,-7,-9)$ and $(0,1,-1)$.