<PRE><B>Question 341221</B>
Write an equation that is perpendicular to 2x+3y=6 with point (3,5).
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The given line: y = (-2/3)x+2
slope is -2/3
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A perpendicular line will have slope = 3/2 and pass thru (3,5)
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y = mx+b 
5 = (3/2)3+b
5 = (9/2)+b
(10/2) = (9/2) + b
b = 1/2
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Equation:
y = (3/2)x + 1/2
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Graph:
{{{graph(400,300,-10,10,-10,10,(3/2)x+(1/2),(-2/3)x+2)}}}
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Cheers,
Stan H.</PRE>