Question

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

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

Answer 1

Answer: (B)

$(3,-7,-9)$

Let any point on the line be $(t+1,-2t-3,-2t-5)$

Its distance from $(1,-3,-5)$ is 6

Therefore, $\sqrt{t^2+4t^2+4t^2}=6$

$\Rightarrow t=\pm 2$

The point is $(3,-7,-9)$