Question 362513
First find the slope of the line,
{{{x-3y-5=0}}}
{{{3y=x-5}}}
{{{y=x/3-5/3}}}
{{{m[1]=1/3}}}
Perpendicular lines have slopes that are negative reciprocals.
{{{m[1]*m[2]=-1}}}
{{{(1/3)*m[2]=-1}}}
{{{m[2]=-3}}}
Now use the point slope form,
{{{y=-3x+b}}}
Use the point to solve for {{{b}}},
{{{-2=-3(2)+b}}}
{{{-2=-6+b}}}
{{{b=4}}}
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{{{highlight(y=-3x+4)}}}
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{{{highlight(3x+y=4)}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(2,-2,0.2),graph(300,300,-5,5,-5,5,x/3-5/3,-3x+4))}}}