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Question 833193: Given y=-2/3x+1, write the equation of the line that is perpendicular and passes through the point (6,2) in slope intercept form.
Answer by LinnW(1048)   (Show Source):
You can put this solution on YOUR website! y=-2/3x+1
let m = slope
To be perpendicular we need slope to be negative inverse
so (-2/3)*m = -1
To solve for m, multiply each side by -3/2
(-3/2)(-2/3)*m=(-3/2)(-1)
m = 3/2
Another way to think about this is to invert and change the sign.
Using y = mx + b as a template, we want to solve for b.
Since we want to go through (6,2), x=6 and y=2.
y = mx + b becomes
2 = (3/2)(6) + b
2 = 9 + b
subtract 9 from each side
-7 = b
Our equation is y = (3/2)x - 7
To verify, substitute 6 for x
y = (3/2)(6) - 7
y = 9 - 7
y = 2
So for the point y = (3/2)x -7 passes through (6,2)
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