## Task Condition

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

$M(2 ; 1 ; 0)$

$\frac{x-2}{0}=\frac{y+1.5}{-1}=\frac{z+0.5}{1}$
## Solution

We 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 its normal vector:

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

Then the equation of the desired plane is:

$0 \cdot(x-2)-1 \cdot(y-1)+1 \cdot(z-0)=0$

$-y+1+z=0$

$-y+z+1=0$

We find the point $M_{0}$ of intersection of the line and the plane. We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-2}{0}=\frac{y+1.5}{-1}=\frac{z+0.5}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2 \\
y=-1.5-t \\
z=-0.5+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$-(-1.5-t)+(-0.5+t)+1=0$

$1.5+t-0.5+t+1=0$

$2 t+2=0$

$t=-1$

We find the coordinates of the point of intersection of the line and the plane:

$\left\{\begin{array}{l}x=2 \\ y=-1.5-(-1)=-0.5 \\ z=-0.5+(-1)=-1.5\end{array}\right.$

We get:

$M_{0}(2 ;-0.5 ;-1.5)$

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

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

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

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

We get:

$M^{\prime}(2 ;-2 ;-3)$