Question

Image point of $(1,3,4)$ in the plane $2x-y+z+3=0$ is

• A. $(-3,5,2)$
• B. $(3,5,-2)$
• C. $(3,-5,3)$
• D. None of these

Answer 1

The correct option is A $(-3,5,2)$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(1,3,4)$, so midpoint of $PQ$ lie on the given plane
$\Rightarrow 2\left(\frac{\alpha +1}{2}\right)-\left(\frac{\beta +3}{2}\right)+\left(\frac{\gamma +4}{2}\right)+3=0$
$\therefore 2\alpha -\beta +\gamma +9=0$ ...(i)
Also $PQ$ is perpendicular to the normal to the plane,
$\Rightarrow \frac{\alpha -1}{2}=\frac{\beta -3}{-1}=\frac{\gamma -4}{1}$ . . . (ii)
Solving (i) and (ii), we get
$\alpha =-3,\beta =5,\gamma =2.$
Therefore, image is $(-3,5,2)$