## Problem Statement

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

$M(-1 ; 2 ; 0)$

$4 x-5 y-z-7=0$
## 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:

$\vec{s}=\vec{n}=\{4 ;-5 ;-1\}$

Then the equation of the desired line is:

$\frac{x+1}{4}=\frac{y-2}{-5}=\frac{z}{-1}$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+1}{4}=\frac{y-2}{-5}=\frac{z}{-1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1+4 t \\
y=2-5 t \\
z=-t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$4(-1+4 t)-5(2-5 t)-(-t)-7=0$

$-4+16 t-10+25 t+t-7=0$

$42 t-21=0$

$t=0.5$

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

$$
\left\{\begin{array}{l}
x=-1+4 \cdot 0.5=1 \\
y=2-5 \cdot 0.5=-0.5 \\
z=-0.5
\end{array}\right.
$$

We get:

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

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

$$
\begin{aligned}
& x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 1-(-1)=3 \\
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-0.5)-2=-3 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot(-0.5)-0=-1
\end{aligned}
$$

We get:

$M^{\prime}(3 ;-3 ;-1)$