Question 179403
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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{{{highlight(3x-2y=5)}}}......added 3x and subtracted 4 from both sides
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lets graph to see if these are perpendicular
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{{{graph(300,300,-10,10,-10,10,(2/3)x+(5/3),(-3/2)x+(5/2))}}}