Question

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

• A. $(1,1,-6)$
• B. $(-1,-1,6)$
• C. $(-1,1,-6)$
• D. None of these

Answer 1

The correct option is B $(-1,-1,6)$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(1,3,4)$, so midpoint of $PQ$ lie on the given plane
$\Rightarrow \left(\frac{\alpha +1}{2}\right)+2\left(\frac{\beta +3}{2}\right)-\left(\frac{\gamma +4}{2}\right)+3=0$
$\therefore \alpha +2\beta -\gamma +9=0:...\left(i\right)$
Also $PQ$ is perpendicular to the normal to the plane,
$\Rightarrow \frac{\alpha -1}{1}=\frac{\beta -3}{2}=\frac{\gamma -4}{-1}...\left(ii\right)$
Solving (i) and (ii), we get
$\alpha =-1,\beta =-1,\gamma =6.$
Therefore, image is $(-1,-1,6)$