SOLUTION: A line perpendicular to x + 3y = 5 passes through (1, -1). What is the equation of the line in standard form? The way I tried is x + 3y = 5 x + 5 = 3y y = 1x + 5/3 Sl

Question 503027: A line perpendicular to x + 3y = 5 passes through (1, -1). What is the equation of the line in standard form?
The way I tried is
x + 3y = 5
x + 5 = 3y
y = 1x + 5/3
Slope is 1.
y - 1 = 1(x + 1)
y = 1
Am I doing this right?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
x+3y=5
1 x + 3 y = 5
Find the slope of this line

3 y = -1 x + 5
Divide by 3
y = -0.333 x + 1.67
Compare this equation with y=mx+b
slope m = -0.33

The slope of a line perpendicular to the above line will be the negative reciprocal 3
m1*m2=-1
The slope of the required line will be 3.00

m= 3 ,point ( 1 , -1 )
Find b by plugging the values of m & the point in
y=mx+b
-1 = 3.00 + b
b= -4
m= 3
The required equation isy = 3 x -4
m.ananth@hotmail.ca

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