find the slope of a line perpendicular to the line that passes through (3,9) and (7,15)

Question

slope of (3,9) and (7,15)

Answer 1

Say the points (x1,y1) = (3,9)

(x2,y2) = (7,15)

Slope = y2-y1/x2-x1

=(15-9)/(7-3)

= 6/4

= 3/2

Given line slope say m1 = 3/2.

Perpendicular line slope say m2.

We know that m1m2 = -1

3/2*m2 = -1

Multiple to each side by 2/3.

3/2*2/3*m2 = -1*2/3

m2 = -2/3

Perpendicular line equation is y-9 = -2/3(x-3)

Cross multiplication.

3(y-9) = -2(x-3)

3y-27 = -2x+6

Add 27 to each side.

3y-27+27 = -2x+6+27

3y = -2x+33

Divide to each side by 3.

3y/3 = (-2x+33)/3

y = (-2/3)x+33/3

y = (-2/3)x+11

Required line equation is y  = (-2/3)x+11