Find the equation of the line perpendicular to x−5y=15 that passes through the point (0,1).

Answer 1

First write the equation x- 5y = 15 in slope intercept form y=mx+b

x- 5y = 15

-5y = -x+15

y= 1/5 x -3

m1= 1/5

for perpendicular lines, the slopes are negative reciprocal of each other,

so the slope of the second line is

m2= -5, and it passes through (0,1)

so its equation is

y=mx +b

y= -5x +b, substitute (0,1) to get b

1= -5(0) +b giving b= 1

so the equation becomes: y = -5x + 1

Answer 2

y = mx + b

The product of slopes of perpendicular lines is -1, so

m = -1 / [-1/(-5)] = -5

y = -5x + b

Use the given point to find b.

1 = -5(0) + b

b = 1

y = -5x + 1