Question 670452: what equasion is perpendicular to 3x+y=1 and passes through the point (-2,1)? Found 2 solutions by mananth, VirtualMathTutor:Answer by mananth(16946) (Show Source):
1 y = -3 x + 1
Divide by 1
y = -3 x + 1
Compare this equation with y=mx+b, m= slope & b= y intercept
slope m = -3
The slope of a line perpendicular to the above line will be the negative reciprocal 1/3
Because m1*m2 =-1
The slope of the required line will be 1/3
m= 1/3 ,point ( -2 , 1 )
Find b by plugging the values of m & the point in
y=mx+b
1 = - 2/ 3 + b
b= 5/3
m= 1/3
The required equation is y = 1/ 3 x + 5/ 3
m.ananth@hotmail.ca
You can put this solution on YOUR website! Two lines are perpendicular if the slope of one line is the negative reciprocal of the other slope of the other line.
First let's find the slope of the given line
3x + y = 1
y = 1 - 3x
slope = -3
the line perpendicular to the given line will then have a slope of
Use the formula y - y1 = m(x - x1)
x1 = -2
y1 = 1
y - 1 = (x - (-2))
y - 1 = (x + 2)
y - 1 = x +
Then add 1 to both sides
y = x +
