Question

what is an equation of the line that s perpendicullar to y-3=-4(x+2) and passes through the point (-5,7)

Answer 1

Answer:

4y=x+33

y=x/4+33/4 (slope-intercept form)

Step-by-step explanation:

y-3 = -4(x+2)

y-3 -4x-8

y= -4x-8+3

y= -4x-5

m1 = -4

For perpendicularity

m2= -1/-4 = 1/4

The equation is

y-y1 = m2(x-x1)

y-7 = 1/4(x-(-5))

y-7 = x/4+5/4

Multiply through by 4

4y-28=x+5

4y=x+5+28

4y=x+33

Divide through by 4

y=x/4+33/4 (slope-intercept form)