Question

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

• A. $(-\frac{17}{3},-\frac{19}{3},4)$
• B. $(15,11,4)$
• C. $(-\frac{17}{3},-\frac{19}{3},1)$
• D. $(\frac{9}{5},-\frac{13}{5},4)$

Answer 1

The correct option is D $(\frac{9}{5},-\frac{13}{5},4)$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(-1,3,4)$, so midpoint of $PQ$ lie on the given plane

$\Rightarrow \frac{\alpha -1}{2}-2\left(\frac{\beta +3}{2}\right)=0$

$\therefore \alpha -1-2\beta -6=0\Rightarrow \alpha -2\beta =7...\left(i\right)$

Also $PQ$ is perpendicular to the normal to the plane,

$\Rightarrow \frac{\alpha +1}{1}=\frac{\beta -3}{-2}=\frac{\gamma -4}{0}...\left(ii\right)$

Solving (i) and (ii), we get

$\alpha =\frac{9}{5},\beta =-\frac{13}{5},\gamma =4.$

Therefore, image is $(\frac{9}{5},-\frac{13}{5},4)$