Answer
$y=\dfrac{5}{2}x + \dfrac{1}{2}$
Work Step by Step
RECALL:
(1) The slope-intercept form of a line's equation is $y=mx+b$ where m = slope and b = y-intercept.
(2) Perpendicular lines have slopes that are negative reciprocals of each other (product is $-1$).
Transform the line $2x+5y+8-0$ to slope-intercept form to obtain:
$5y=-2x-8
\\\dfrac{5y}{5} = \dfrac{-2x-8}{5}
\\y=-\dfrac{2}{5}x - \dfrac{8}{5}$
The slope of this line is $-\dfrac{2}{5}$.
The line we are looking for the equation of is perpendicular to the line above.
This means that its slope is the negative reciprocal of $-\dfrac{2}{5}$, which is $\dfrac{5}{2}$.
Thus, the tentative equation of the line we are looking for is:
$y=\dfrac{5}{2}x+b$
The line passes through the point $(-1, -2)$.
To find the value of $b$, substitute the x and y coordinates of this point into the tentative equation above to obtain:
$y=\dfrac{5}{2}x + b
\\-2 = \dfrac{5}{2}(-1) + b
\\-2 = -\dfrac{5}{2} + b
\\-2+\dfrac{5}{2}=b
\\-\dfrac{4}{2} + \dfrac{5}{2} = b
\\\dfrac{1}{2} = b$
Therefore, the equation of the line we are looking for is:
$y=\dfrac{5}{2}x + \dfrac{1}{2}$
