Question

The coordinates of a point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4\sqrt{1}4$ from the point $(1,-1,0)$ nearer the origin are

• A. $(9,-13,4)$
• B. $(8\sqrt{1}4,12,-1)$
• C. $(-8\sqrt{1}4,12,1)$
• D. $(-7,11,-4)$

Answer 1

Answer: (A)

$(9,-13,4)$

$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-0}{1}=r$

$x=2r+1,y=-3r-1,z=r$

General point on the line is $P(2r+1,-3r-1,r)$

And given point, $A(1,-1,0)$

$PA=4\sqrt{14}$

By distance formula

$\sqrt{(2r+1-1{)}^2+(-3r-1+1{)}^2+(r-0{)}^2}=4\sqrt{14}$

$r=4$

So, the required point

$(9,-13,4)$