SOLUTION: Find the point slope form of the equation of the line that is perpendicular to 2x-3y=4 and also contains the point (-2,1).

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Question 1004367: Find the point slope form of the equation of the line that is perpendicular to
2x-3y=4 and also contains the point (-2,1).



Found 2 solutions by Boreal, Cromlix:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
-3y=-2x+4 rewriting the formula
y=(2x/3)-(4/3)
slope is 2/3
perpendicular is negative reciprocal or -3/2 slope
point slope formula is
y-y1=m(x-x1)
y-1=(-3/2)(x+2)= (-3x/2)-3
y=(-3x/2)-2
graph%28300%2C200%2C-10%2C10%2C-8%2C8%2C%282x%2F3%29-%284%2F3%29%2C%28-3x%2F2%29-2%29

Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
2x-3y=4
Put in y = mx + c form
-3y = -2x + 4
3y = 2x - 4
y = 2/3x - 4/3
Slope = 2/3
Lines that are perpendicular
to one another have slopes that
multiply together to give -1
m1 x m2 = -1
2/3 x m2 = -1
m2 = -3/2
Using line equation
y - b = m(x - a)
m = -3/2 and (a,b) = (-2,1)
y - 1 = -3/2(x -(-2))
y - 1 = -3/2x + 4
y - 1 = -3/2x + 4
y = -3/2x + 4 + 1
y = -3/2x + 5
Hope this helps :-)