Write an equation for a line that passes through (3,-2) and is perpendicular to y=-5x + 1?

y - (-2) = (1/5)(x - 3)

y + 2 = (1/5)(x - 3) [point slope form]

y + 2 = (1/5)x - 3/5

y = (1/5)x - 13/5 [slope intercept]

5y = x - 13

-x + 5y = -13

x - 5y = 13 [standard]

x - 5y - 13 = 0 [general]

Perpendicular must have a slope that is the negative reciprocal of the one in the given line.

Here, 1/5

So you now have a point and a slope.

Point slope form

y-b = m(x-a) slope m and through point (a,b)

y+2 = 1/5 (x-3)