## Problem Statement

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

$M(1 ; 0 ;-1)$

$\frac{x-3.5}{2}=\frac{y-1.5}{2}=\frac{z}{0}$
## 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:

$\vec{n}=\vec{s}=\{2 ; 2 ; 0\}$

Then the equation of the desired plane is:

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

$2 x-2+2 y=0$

$x+y-1=0$

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

We write the parametric equations of the line.

$\frac{x-3.5}{2}=\frac{y-1.5}{2}=\frac{z}{0}=t \Rightarrow$

$\left\{\begin{array}{l}x=3.5+2 t \\ y=1.5+2 t \\ z=0\end{array}\right.$

Substitute into the equation of the plane:

$(3.5+2 t)+(1.5+2 t)-1=0$

$3.5+2 t+1.5+2 t-1=0$

$4+4 t=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=3.5+2 \cdot(-1)=1.5 \\
y=1.5+2 \cdot(-1)=-0.5 \\
z=0
\end{array}\right.
$$

We get:

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

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

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

We get:

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