Answer
$y=-8x-11$ or, in general form, $8x+y+11=0$
Work Step by Step
Through $(-2,5)$ and $(-1,-3)$
Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, where $(x_{1},y_{1})$ is a point through which the line passes and $m$ is the slope.
Two point through which the lines passes are given. Use them to fin de the slope of the line:
$m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\dfrac{-3-5}{-1-(-2)}=\dfrac{-8}{-1+2}=\dfrac{-8}{1}=-8$
Substitute $m$ and whichever of the points given into the formula and simplify to obtain the equation of this line:
$y-y_{1}=m(x-x_{1})$
$y-5=-8(x+2)$
$y-5=-8x-16$
$y=-8x-16+5$
$y=-8x-11$ or, in general form, $8x+y+11=0$
