|
Question 455145: find the slope of a line perpendicular to the line that passes through 3 9 and 7 15
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! x1 y1 x2 y2
3 9 7 15
slope m =(y2-y1)/(x2-x1)
(15-9)/(7-3)
(6/4)
m=1.5
Plug value of the slope and point (3, 9) in
Y =mx+b
9.00=4.5+b
b=9-4.5
b=4.5
So the equation will be
Y =1.5x +4.5
PERPENDICULAR TO A LINE
Compare this equation with y=mx+b
slope m = 1.50
The slope of a line perpendicular to the above line will be the negative reciprocal
m1*m2=-1
The slope of the required line will be =-0.67
m= -0.67 ,point (3,9)
Find b by plugging the values of m & the point in
y=mx+b
9=-2.00+b
b=11
m=-0.67
The required equation is y=- 2/3x+11
m.ananth@hotmail.ca
|
|
|
| |
