Given points $A(-2 ; 2), B(-2 ;-2)$ and $C(6 ; 6)$. Form the equations of the lines on which the sides of triangle $A B C$ lie.

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Since the abscissas of points $A(-2 ; 2)$ and $B(-2 ;-2)$ are equal, the equation of the line $A B$ is $x=-2$, or $x+2=0$.

If $x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$, then the equation of the line passing through points $M_{1}\left(x_{1} ; y_{1}\right)$ and $M_{2}\left(x_{2} ; y_{2}\right)$ can be written as

$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$

Therefore, the equation of the line $A C$ is

$$
\frac{y-2}{6-2}=\frac{x-(-2)}{6-(-2)} . \text { or } x-2 y+6=0
$$

Similarly, we find the equation of the line $B C$.
$A B: x+2=0, A C: x-2 y+6=0, B C: x-y=0$.