Question 500172: Find an equation of the line that satisfies the given conditions.
Through
(−8, −8);
perpendicular to the line passing through
(−5, 4) and (−1, 2)
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! x1 y1 x2 y2
-5 4 -1 2
We find the equation of this line
slope m = (y2-y1)/(x2-x1)
( 2 - 4 )/( -1 - -5 )
( -2 / 4 )
m= - 1/ 2
Plug value of the slope and point ( -5 , 4 ) in
Y = m x + b
4.00 = 5/2 + b
b= 4 - 5/2
b= 3/ 2
So the equation will be
Y = - 1/2 x + 3/2
EQUATION OF LINE PERPENDICULAR TO THIS LINE
Find the slope of this line
1 y = -1/2 x + 3/2
Divide by 1
y = -1/2 x + 3/2
Compare this equation with y=mx+b
slope m = -1/2
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 2
m= 2 ,point ( -8 , -8 )
Find b by plugging the values of m & the point in
y=mx+b
-8 = -16.00 + b
b= 8
m= 2
The required equation is y = 2 x + 8
m.ananth@hotmail.ca
|
|
|
