Question 1079557
Slope of LM=-5/4, point slope of line is y-y1=m(x-x1), m slope, (x1,y1) point
y+2=(-5/4)x, and y=-(5/4)x-2
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MN needs a slope of 4/5, to be perpendicular to LM, and the slope is 4/5
y+2=(4/5)(x) and y=(4/5)x+2
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NO needs a slope of -5/4, and and it is
y-2=-(5/4)(x-5)
y=-(5/4)x+33/4
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OL needs a slope of 4/5, and it is
y-7=(4/5)(x-1)
y=(4/5)x+31/5
{{{graph(300,300,-10,10,-10,10,-(5/4)x-2,(4/5)x+2,-(5/4)x+(33/4),(4/5)x+(31/5))}}}