Aline has a slope of –5 and a y-intercept of (0, 3). what is the equation of the line that is perpendicular to the first line and passes through the point (3, 2)?

Slope-intercept form:
y=mx+b
m=slope
b=y-intercept

Data of the first line:
m=-5
b=y-intercept=3  (y-intercept=it is the value of "y" when x=0)

y=-5x+3

A line perpendicular to the line y=mx+b will have the following slope:
m`=-1/m

Therefore: the line perpendicular to the line y=-5x+3 will have the following slope:
m´=-1/(-5)=1/5

Point-slope form of a line: we need a point (x₀,y₀) and the slope (m):
y-y₀=m(x-x₀)

We know, the slope (m=1/5) and we have a point (3,2) therefore:
y-y₀=m(x-x₀)
y-2=1/5(x-3)     (point-slope form)
y-2=(1/5)x-3/5
y=(1/5)x-3/5+2
y=(1/5)x+7/5    (slope-intercept form)

the line perpendicular to the first line will be: y=(1/5)x+7/5