Question

what is an equation of the line that passes through the point (5, -1) and is perpendicular to the line 5x + 4y = 28?

Explanation:

Step1: Find the slope of the given line

Rewrite $5x + 4y=28$ in slope - intercept form $y = mx + b$ (where $m$ is the slope).
$4y=-5x + 28$, so $y=-\frac{5}{4}x+7$. The slope of this line $m_1 =-\frac{5}{4}$.

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the required line be $m_2$. Then $m_1\times m_2=-1$.
$-\frac{5}{4}\times m_2=-1$, so $m_2=\frac{4}{5}$.

Step3: Use the point - slope form to find the equation of the line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(5,-1)$ and $m = \frac{4}{5}$.
$y-(-1)=\frac{4}{5}(x - 5)$
$y + 1=\frac{4}{5}x-4$
$y=\frac{4}{5}x-5$

Answer:

$y=\frac{4}{5}x - 5$