## Task Condition

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

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

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

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

Then the equation of the desired plane is:

$-1 \cdot(x-(-1))+0 \cdot(y-0)+1 \cdot(z-(-1))=0$

$-x-1+z+1=0$

$-x+z=0$

Find the point $M_{0}$ of intersection of the line and the plane.

Write the parametric equations of the line.
$\frac{x}{-1}=\frac{y-1.5}{0}=\frac{z-2}{1}=t \Rightarrow$

$\left\{\begin{array}{l}x=-t \\ y=1.5 \\ z=2+t\end{array}\right.$

Substitute into the equation of the plane:

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

$t+2+t=0$

$2 t+2=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=-(-1)=1 \\
y=1.5 \\
z=2+(-1)=1
\end{array}\right.
$$

We get:

$M_{0}(1 ; 1.5 ; 1)$

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

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

We get:

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