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

5x+y-8=0.

Step-by-step explanation:

Given the equation of the line:

`x-5y-8=0`

Expressing the line in slope-intercept form:

`\begin{gathered} 5y=x-8 \\ y=(1)/(5)x-(8)/(5) \\ \implies\text{Slope of the line }=(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.