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Question 800861: Helo I was wondering if you could help me with this question.
Find the equation of a line parallel and perpendicular to y=1/2x+3 through point (1,-6)
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! PERPENDICULAR TO A LINE
-1 x + 2 y = 6
Find the slope of this line
2 y = 1 x + 6
Divide by 2
y = 1/ 2 x + 3
Compare this equation with y=mx+b, m= slope & b= y intercept
slope m = 1/2
The slope of a line perpendicular to the above line will be the negative reciprocal -2
Because m1*m2 =-1
The slope of the required line will be -2
m= -2 ,point ( 1 , -6 )
Find b by plugging the values of m & the point in
y=mx+b
-6 = -2 + b
b= -4
m= -2
The required equation is y = -2 x + -4
m.ananth@hotmail.ca
PARALLEL
-1 x 2 y = 6
Find the slope of this line
make y the subject
2 y = 1 x + 6
Divide by 2
y = 1/2 x + 3
Compare this equation with y=mx+b
slope m = 1/2
The slope of a line parallel to the above line will be the same
The slope of the required line will be 1/2
m= 1/2 ,point ( 1 , -6 )
Find b by plugging the values of m & the point in
y=mx+b
-6 = 1/ 2 + b
b= -6.5
m= 1/2
Plug value of the slope and b in y = mx +b
The required equation is y = 1/2 x -6.5
m.ananth@hotmail.ca
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