## Problem Statement

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

$M(-1 ; 0 ; 1)$

$\frac{x+0.5}{0}=\frac{y-1}{0}=\frac{z-4}{2}$
## 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:
$\vec{n}=\vec{s}=\{0 ; 0 ; 2\}$

Then the equation of the desired plane is:

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

$2(z-1)=0$

$z-1=0$

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

Write the parametric equations of the line.

$\frac{x+0.5}{0}=\frac{y-1}{0}=\frac{z-4}{2}=t \Rightarrow$

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

Substitute into the equation of the plane:

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

$3+2 t=0$

$t=-1.5$

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

$$
\left\{\begin{array}{l}
x=-0.5 \\
y=1 \\
z=4+2 \cdot(-1.5)=1
\end{array}\right.
$$

We get:

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

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

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

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

$z_{M_{0}}=\frac{z_{M}+z_{M^{\prime}}}{2} \Rightarrow z_{M^{\prime}}=2 z_{M_{0}}-z_{M}=2 \cdot 1-1=1$

We get:

$M^{\prime}(0 ; 2 ; 1)$