## Problem Statement

Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.

$M(1 ; 0 ;-1)$
$2 y+4 z-1=0$
## Solution

Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$.

Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:

$\vec{s}=\vec{n}=\{0 ; 2 ; 4\}$

Then the equation of the desired line is:

$\frac{x-1}{0}=\frac{y}{2}=\frac{z+1}{4}$

Let's find the point $M_{0}$ of intersection of the line and the plane.

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{0}=\frac{y}{2}=\frac{z+1}{4}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1 \\
y=2 t \\
z=-1+4 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$2(2 t)+4(-1+4 t)-1=0$

$4 t-4+16 t-1=0$

$20 t-5=0$

$t=\frac{1}{4}$

Find the coordinates of the point of intersection of the line and the plane:

$$
\left\{\begin{array}{l}
x=1 \\
y=2 \cdot \frac{1}{4}=\frac{1}{2}=0.5 \\
z=-1+4 \cdot \frac{1}{4}=0
\end{array}\right.
$$

We get:

$M_{0}(1 ; 0.5 ; 0)$

Since $M_{0}$ is the midpoint of the segment $M M^{\prime}$, then

$x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 1-1=1$

$$
\begin{aligned}
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot 0.5-0=1 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0-(-1)=1
\end{aligned}
$$

We get:

$M^{\prime}(1 ; 1 ; 1)$