Question 366453
write an equation in slope intercept form for th line that satisfies the following condition. passes through (10,16) perpendicular to the graph of 9x+12y=15 
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9x+12y=15
12y=-9x+15
/12
y=-9/12 x +15/9
y = -3/4 *x +5/3
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slope = -3/4
a line perpendicular to the given line will be 4/3
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point( 10,16)
(y-16)=4/3(x-10)
y-16 = 4/3 *x -40/3
y= 4/3*x-40/3+16
y=4x/3 +8/3
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{{{graph(300,300,-5,5,-5,5,((4x/3)+(8/3)),((-3x/4)+(5/3)))}}}
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