Question 1003202: Hello, I am working on a series of questions with the equation y-3=3(x+1). I have the slope-intercept form y=3x+6, and the standard form 3x-y=-6? Now, I am trying to find an equation in standard form of a perpendicular line that passes through (5,-1). I took y=mx+b and solved for b=2/3, then y=-1/3x+2/3 and switched it around to y+x/3=2/3. Is this correct? I am confused because standard form is Ax+By=c, if my answer should be x/3+y=2/3?? Seems like that would be in the correct form. Thank you for your time!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I have the slope-intercept form y=3x+6, and the standard form 3x-y=-6?
------
Now, I am trying to find an equation in standard form of a perpendicular line that passes through (5,-1).
I took y=mx+b and solved for b=2/3, then y=-1/3x+2/3 and switched it around to y+x/3=2/3. Is this correct?
It's correct, tho you didn't show how you got it.
----
The slope is m = -1/3
Use y-y1 = m*(x-x1) where (x1,y1) is the point (5,-1)
y+1 = (-1/3)*(x-5)
3y+3 = -x+5
x + 3y = 2
==============================
I am confused because standard form is Ax+By=c, if my answer should be x/3+y=2/3?? Seems like that would be in the correct form.
------------
x/3+y=2/3
Standard form has integers for A, B & C
Multiply by 3
x + 3y = 2
|
|
|
