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Question 705313: I am trying to understand the operation of this equation.
The instructions say to:
Write the equation of a line perpendicular to the given line but passing through the given points.
y=-2/3x + 2 points (9,-3)
Now from what I have read, a perpendicular slop is the opposite reciprocal.
Am I right so far??
If this is true then the reciprocal would be this:
y=3/2x + 2
Correct or have I misunderstood??
Anyways, I'm curious as to how to work the rest of the equation.
I have the points to plug in but do I use those numbers or is there a different
execution of this equation??
I have tried to follow the examples that I was given but none have a fraction in it so I can understand it better.
Any type of guidance to let me know I am on the right track, would greatly help.
I know how to work and get the line parallel to a line that is passing through given points. If it is similar to that, which I am sure it's not, then let me know.
Thank you
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! You are correct that the line perpendicular to y = -2/3x + 2 has the slope m = 3/2.
So the equation for this line is y = 3/2 x + b.
Now use the point on the line (9,-3) to find the y-intercept, b.
-3 = (3/2)9 + b
b = -3 - 27/2 = -33/2
So the equation is y = (3/2)x - 33/2
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