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

Question

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

Accepted Answer

$$E{q^n}\,\,\,\,$$ of $$\,\,\,\,PN:  $$-

$${{x + 1} \over 1} = {{y - 3} \over { - 2}} = {{z - 4} \over 0} = \lambda $$

$$N\left( {\lambda  - 1, - 2\lambda  + 3 - 4} \right)$$

It lies on $x-2y=0$

$$ \Rightarrow \lambda  - 1 + 4\lambda  - 6 = 0$$

$$ \Rightarrow \lambda  = 7/5$$

$$N\left( {{2 \over 5},{1 \over 5},4} \right)$$

$N$ is mid point of $PP'$

$$\therefore$$ $$\alpha  - 1 = {4 \over 5},\beta  + 3 = {2 \over 5},r + 4 = 8$$

$$ \Rightarrow \alpha  = {9 \over 5},\beta  = {{ - 13} \over 5},r = 4$$

$$\therefore$$ Image is $$\left( {{9 \over 5},{{ - 13} \over 5},4} \right)$$

Solutions · 1