Question

how do I solve this equation passing through (2,-2) and perpendicular to the line whose equation is x-5y-8=0 in general form ?

Answer 1

Answer:

5x+y-8=0.

Explanation:

Given the equation of the line:

$x-5y-8=0$

Expressing the line in slope-intercept form:

$\begin{gathered} 5y=x-8 \\ y=\frac{1}{5}x-\frac{8}{5} \\ \implies\text{Slope of the line }=\frac{1}{5} \\ \implies\text{Slope of the perpendicular line }=-5 \end{gathered}$

If the perpendicular line passes through (2,-2), using the slope-point form:

$y-y_1=m(x-x_1)$

This gives:

$\begin{gathered} y-(-2)=-5(x-2) \\ y+2=-5x+10 \\ 5x+y+2-10=0 \\ 5x+y-8=0 \end{gathered}$

The equation of the perpendicular line is 5x+y-8=0.