How do you find an equation of a line perpendicular to y=3x-5 that contains the points (-1,5)?

Answer 1

y=3x-5

The slope of the perpendicular linewill be the negativereciprocal of the initial line.

m=1 for the given line (y is multiplied by 1).

given pt= -1, 5

point slope=y-y1=m(x-x1)

(y-5)=1(x-(-1))

=(x+1)

Answer 2

y = mx + b type of problem.

You start with knowing what slopes of perpendicular lines represent. It is the negative reciprocal of each other.

y = 3x - 5, the negative reciprocal slope would be (-1/3) for "m"

pop that into y = m(x) + b and then plug in the x and y values from the coordinate (-1,5) and solve for b.

5 = (-1/3)(-1) + b

5 = 1/3 + b

5 - (1/3) = b

14/3 = b

plug that back into the equation and you get:

y = (-1/3)x + 14/3