Question 322839
{{{5x+3y=-13}}}
{{{3y=-5x-13}}}
{{{y=-(5/3)x-13/3}}}
Perpendicular lines have slopes that are negative reciprocals,
{{{m1*m2=-1}}}
{{{-(5/3)*m2=-1}}}
{{{m2=3/5}}}
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{{{y=mx+b=(3/5)x+b}}}
Use the point (-8,1) to solve for {{{b}}}.
{{{1=(3/5)(-8)+b}}}
{{{b=5/5+24/5}}}
{{{b=29/5}}}
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{{{highlight(y=(3/5)x+29/5)}}}
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{{{drawing(300,300,-10,2,-2,10,grid(1),circle(-8,1,.3),graph(300,300,-10,2,-2,10,(3/5)x+29/5,-(5/3)x-13/3))}}}