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Question 788773: what is the equations of the lines that pass through the point (3,3 ) and are parallel to and perpendicular to the line with equation y +5 x=8.
Parallel: y=
Perpendicular: y=
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! PERPENDICULAR TO A LINE
5 x + 1 y = 8
Find the slope of this line
1 y = -5 x + 8
Divide by 1
y = -5 x + 8
Compare this equation with y=mx+b, m= slope & b= y intercept
slope m = -5
The slope of a line perpendicular to the above line will be the negative reciprocal 1/5
Because m1*m2 =-1
The slope of the required line will be 1/5
m= 1/5 ,point ( 3 , 3 )
Find b by plugging the values of m & the point in
y=mx+b
3 = 3/ 5 + b
b= 12/5
m= 1/5
The required equation is y = 1/ 5 x + 12/ 5
PARALLEL
5 x 1 y = 8
Find the slope of this line
make y the subject
1 y = -5 x + 8
Divide by 1
y = -5 x + 8
Compare this equation with y=mx+b
slope m = -5
The slope of a line parallel to the above line will be the same
The slope of the required line will be -5
m= -5 ,point ( 3 , 3 )
Find b by plugging the values of m & the point in
y=mx+b
3 = -15 + b
b= 18 - 1/4
m= -5 1 3/4
Plug value of the slope and b in y = mx +b
The required equation is y = -5 x + 18
m.ananth@hotmail.ca
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