89. The triangle is defined by points $A(5 ; 2), B(-1 ;-4), C(-5$; -3). Form the equation of the line passing through point $B$ and parallel to $A C$.
Solution. $1^{0}$. Choose a point $M(x ; y)$.

$2^{0}$. Find the vector $\overrightarrow{B M}=(x+1 ; y+4)$.
$3^{0}$. Find the vector defined by points $A(5 ; 2)$ and $C(-5 ;-3)$; we have $\overrightarrow{A C}=(-10 ;-5)$.

$4^{0}$. Since the desired line and line $A C$ are parallel, their direction vectors are collinear. Write the desired equation of the line:

$$
\frac{x+1}{-10}=\frac{y+4}{-5}
$$

from which $x+1=2 y+8 ; x-2 y-7=0$ - the desired equation in general form.