## Task Condition

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

$$
\begin{aligned}
& M(-1 ; 2 ; 0) \\
& \frac{x+0.5}{1}=\frac{y+0.7}{-0.2}=\frac{z-2}{2}
\end{aligned}
$$
## 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:

$\vec{n}=\vec{s}=\{1 ;-0.2 ; 2\}$

Then the equation of the desired plane is:

$1 \cdot(x-(-1))-0.2 \cdot(y-2)+2 \cdot(z-0)=0$

$x+1-0.2 \cdot y+0.4+2 z=0$

$x-0.2 \cdot y+2 z+1.4=0$

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

We write the parametric equations of the line.

$\frac{x+0.5}{1}=\frac{y+0.7}{-0.2}=\frac{z-2}{2}=t \Rightarrow$

$\left\{\begin{array}{l}x=-0.5+t \\ y=-0.7-0.2 t \\ z=2+2 t\end{array}\right.$

Substitute into the equation of the plane:

$(-0.5+t)-0.2 \cdot(-0.7-0.2 t)+2(2+2 t)+1.4=0$

$-0.5+t+0.14+0.04 t+4+4 t+1.4=0$

$5.04+5.04 t=0$

$t=-1$

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

$$
\left\{\begin{array}{l}
x=-0.5+(-1)=-1.5 \\
y=-0.7-0.2 \cdot(-1)=-0.5 \\
z=2+2 \cdot(-1)=0
\end{array}\right.
$$

We get:

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

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

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

$y_{M_{0}}=\frac{y_{M}+y_{M^{\prime}}}{2} \Rightarrow y_{M^{\prime}}=2 y_{M_{0}}-y_{M}=2 \cdot(-0.5)-2=-3$

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

We get:

$M^{\prime}(-2 ;-3 ; 0)$