Question

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

• A. $(15,11,4)$
• B. $(\frac{9}{5},-\frac{13}{5},4)$
• C. $(8,4,4)$
• D. None of these

Answer 1

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

We know that the image $(x,y,z)$ of point$(x_1,y_1{z}_1)$

In a plane $ax+by+cz+d=0$ is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2(a{x}_1+b{y}_1+c{z}_1+d)}{a^2+b^2+c^2}$

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

$\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2\left[1\right(-1)+(-2)3+0(4\left)\right]}{1+0+4}$

$\frac{x+1}{1}=\frac{y-3}{-2}=\frac{z-4}{0}=\frac{-2(-7)}{5}$

$x=\frac{14}{5}-1=\frac{9}{5};y=\frac{-28}{5}+3=\frac{-13}{5}$ & $z=4$

$\therefore$ Image of $(-1,3,4)is(\frac{9}{5},\frac{13}{5},4)$