SOLUTION: find the equation , in standard form, of the line perpendicular to 2x-3y=-5 and passing through(3, -2) I do not know how to begin , please help

Question 179403: find the equation , in standard form, of the line perpendicular to 2x-3y=-5 and passing through(3, -2)
I do not know how to begin , please help
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
when lines are perpendicular they have slopes that are negative inverses of each other
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so lets find the slope of our equation:
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2x-3y=-5---->-3y=-2x-5--->y=(2/3)x+(5/3)
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so our slope is 2/3. Therefore the line perpedicular to this line will have a slope of -3/2
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so we have a slope=m=-3/2 and we have a given point (3,-2,). So lets use the point slope formula which is y-k=m(x-h) where m is the slope and (h,k) is any point on the ine
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y+2=(-3/2)(x-3)
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2y+4=-3(x-3).......multiplied all terms by 2 to eliminate fraction
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2y+4=-3x+9.....distributed right side
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......added 3x and subtracted 4 from both sides
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lets graph to see if these are perpendicular
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