## Task Condition

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

$M(2, -1, 1)$

$$
\frac{x-4.5}{1}=\frac{y+3}{-0.5}=\frac{z-2}{1}
$$
## 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:

$\vec{n}=\vec{s}=\{1 ;-0.5 ; 1\}$

Then the equation of the desired plane is:

$1 \cdot(x-2)-0.5 \cdot(y-(-1))+1 \cdot(z-1)=0$

$x-2-0.5 \cdot y-0.5+z-1=0$

$x-0.5 \cdot y+z-3.5=0$

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

We write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-4.5}{1}=\frac{y+3}{-0.5}=\frac{z-2}{1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=4.5+t \\
y=-3-0.5 t \\
z=2+t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(4.5+t)-0.5 \cdot(-3-0.5 t)+(2+t)-3.5=0$

$4.5+t+1.5+0.25 t+2+t-3.5=0$

$2.25 t+4.5=0$

$t=-2$

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

$$
\left\{\begin{array}{l}
x=4.5+(-2)=2.5 \\
y=-3-0.5 \cdot(-2)=-2 \\
z=2+(-2)=0
\end{array}\right.
$$

We get:

$$
M_{0}(2.5 ;-2 ; 0)
$$

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

$$
\begin{aligned}
& x_{M_{0}}=\frac{x_{M}+x_{M^{\prime}}}{2} \Rightarrow x_{M^{\prime}}=2 x_{M_{0}}-x_{M}=2 \cdot 2.5-2=3 \\
& y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-2)-(-1)=-3 \\
& z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 0-1=-1
\end{aligned}
$$

We get:

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