Question 344636
Perpendicular lines have slopes that are negative reciprocals.
{{{m1*m2=-1}}}
{{{3*m2=-1}}}
{{{m2=-1/3}}}
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{{{y=-(1/3)x+b}}}
Use the point (2,2) to solve for {{{b}}}.
{{{2=-(1/3)(2)+b}}}
{{{b=6/3+2/3}}}
{{{b=8/3}}}
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{{{highlight(y=-(1/3)x+8/3)}}}
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{{{drawing(300,300,-6,6,-6,6,grid(1),circle(2,2,0.2),graph(300,300,-6,6,-6,6,3x+2,-(1/3)x+8/3))}}}