SOLUTION: We have just leaned about parallel and Perpendiculars mline and I am Strugling with them we are using a workbook for this lesson. At the top of the section it says Write an

Algebra ->  Graphs -> SOLUTION: We have just leaned about parallel and Perpendiculars mline and I am Strugling with them we are using a workbook for this lesson. At the top of the section it says Write an       Log On


   

Question 172498This question is from textbook Albegra Concepts and Applacations
: We have just leaned about parallel and Perpendiculars mline and I am Strugling with them we are using a workbook for this lesson.
At the top of the section it says
Write an equation in slope-intercept form of the line thast is perpendicular to the graph of each equation and passes through the given point.
The problem that I am stuck on is
y=-5x+1; (2,-1)
For my work I got
y-1=[-5x(-2)Then...
y=-5x-10 for my anwer but I d not think that it is right.

This question is from textbook Albegra Concepts and Applacations

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
y=-5x%2B1--------> This is in Slope-Intercept Form, y=mx%2Bb
where --->system%28Slope=m%5B1%5D=-5%2CYIntercept=b=1%29 --> Line 1
As wee see this line in the graph:

Now, I "think" you're looking the eqn of the line that is perpendicular to this and passing thru (2,-1)? Which is Line 2?
Remember for perpendicular lines, m%5B2%5D=-1%2Fm%5B1%5D
So, m%5B2%5D=-1%2F%28-5%29=1%2F5
Then thru points (2,-1):
Via Slope-Intercept Form, -1=%281%2F5%29%282%29%2Bb
b=-1-%282%2F5%29=%28-5-2%29%2F5=-7%2F5
Therefore Line 2 -----> highlight%28y=%281%2F5%29x-%287%2F5%29%29, Answer
We see the graph again,
----> See GREEN Line 2, perpendicular to Line 1 (RED)
Thank you,
Jojo
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