SOLUTION: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x-5y=4 and passing through (-3, 5). I have done this every way that I can think

Algebra ->  Graphs -> SOLUTION: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x-5y=4 and passing through (-3, 5). I have done this every way that I can think       Log On


   

Question 68602: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x-5y=4 and passing through (-3, 5).
I have done this every way that I can think of and can not come up with it being perpendicular. I can only come up with it as a parallel.
I have:
x-5y=4
x-5y-x=4-x
-5y=-x+4
-5y/(-5)=-x/(-5)+4/(-5)
y=1/5x-4/5
y-y[sub 1]=m(x-x[sub 1])
y-5=1/5(x-(-3))
y-5=1/5x+3/5
y-5+5 = 1/5x + 3/5 + 5
y = 1/5x + 28/5
Both slopes are 1/5 which makes it parallel. Am I doing something wrong here?
Thank you, Rich
Answer by rkennedy(3) About Me  (Show Source):
You can put this solution on YOUR website!
Okay I forgot one thing. I figured it out. To find the perpendicular you have to flip the slope and change the sign. It would be a slope of -5/1 or -5
Then it would be y - 5 = -5 (x - (-3))
y - 5 = -5 (x + 3)
y - 5 = -5x -15
y - 5 + 5 = -5x - 15 + 5
y = -5x - 10