Question 77552
Give the standard form of the equation of the line that is perpendicular to 
y=6/5x+4 and contains (-8,4).
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The given line has a slope of 6/5
so the line perpendicular must have a slope of -5/6
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form of the equation of the line is y=mx+b and you have y,x,and m.
4=(-5/6)(-8)+b
4=(20/3)+b
b=-8/3
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EQUATION:
y=(-5/6)x - 8/3
To put in standard form, multiply thru by 6 to get:
6y =  -5x -16
5x+6y=-16 (standard form)
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Cheers,
Stan H.