## Problem Statement

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

$$
\begin{aligned}
& M(-2 ;-3 ; 0) \\
& \frac{x+0.5}{1}=\frac{y+1.5}{0}=\frac{z-0.5}{1}
\end{aligned}
$$
## Solution

We find the equation of the plane that is perpendicular to the given line and passes through point $M$. Since the plane is perpendicular to the given line, we can use the direction vector of the line as its normal vector:

$\vec{n}=\vec{s}=\{1 ; 0 ; 1\}$

Then the equation of the desired plane is:

$$
\begin{aligned}
& 1 \cdot(x-(-2))+0 \cdot(y-(-3))+1 \cdot(z-0)=0 \\
& x+2+z=0 \\
& x+z+2=0
\end{aligned}
$$

We find the point $M_{0}$ of intersection between the line and the plane.

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+0.5}{1}=\frac{y+1.5}{0}=\frac{z-0.5}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-0.5+t \\
y=-1.5 \\
z=0.5+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$$
\begin{aligned}
& (-0.5+t)+(0.5+t)+2=0 \\
& -0.5+t+0.5+t+2=0 \\
& 2 t+2=0 \\
& t=-1
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
x=-0.5+(-1)=-1.5 \\
y=-1.5 \\
z=0.5+(-1)=-0.5
\end{array}\right.
$$

We get:

$M_{0}(-1.5 ;-1.5 ;-0.5)$

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

$$
\begin{aligned}
& x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot(-1.5)-(-2)=-1 \\
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-1.5)-(-3)=0 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot(-0.5)-0=-1
\end{aligned}
$$

We get:

$M^{\prime}(-1 ; 0 ;-1)$