Answer
$y=\dfrac{5}{2}x+\dfrac{1}{2}$, or, in general form, $5x-2y+1=0$
Work Step by Step
Through $(-1,-2);$ perpendicular to the line $2x+5y+8=0$
A point through which the line passes and the fact that the line is perpendicular to the line $2x+5y+8=0$ are known.
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.
Solve the equation of the given line for $y$. The resulting expression will be in the slope-intercept form and the slope can be identified as the number in front of $x$:
$2x+5y+8=0$
$5y=-2x-8$
$y=-\dfrac{2}{5}x-\dfrac{8}{5}$
The slope of the given line is $-\dfrac{2}{5}$
Since the slopes of perpendicular lines are negative reciprocals, the slope of line whose equation must be found is:
$m=-\dfrac{1}{\Big(-\dfrac{2}{5}\Big)}=\dfrac{5}{2}$
Substitute $(x_{1},y_{1})$ and $m$ into the point-slope form of the equation of a line formula and simplify to obtain the answer:
$y-y_{1}=m(x-x_{1})$
$y+2=\dfrac{5}{2}(x+1)$
$y+2=\dfrac{5}{2}x+\dfrac{5}{2}$
$y=\dfrac{5}{2}x+\dfrac{5}{2}-2$
$y=\dfrac{5}{2}x+\dfrac{1}{2}$, or, in general form, $5x-2y+1=0$
