Question 324720
Parallel lines have identical slopes.
Your new line will have the form,
{{{y=-(3/2)x+b}}} 
Use the point (2,6) to solve for {{{b}}}.
{{{6=-(3/2)2+b}}}
{{{b-3=6}}}
{{{b=9}}}
{{{highlight(y=-(3/2)x+9)}}}
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Perpendicular lines have slopes that are negative reciprocals.
{{{m1*m2=-1}}}
{{{-(3/2)*m2=-1}}}
{{{m2=2/3}}}
{{{y=(2/3)x+b}}}
If the point is the origin then,
{{{0=(2/3)(0)+b
{{{b=0}}}
{{{highlight(y=(2/3)x)}}}
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{{{drawing(300,300,-5,9,-5,9,circle(0,0,.3),circle(2,6,.3),grid(1),graph(300,300,-5,9,-5,9,-(3/2)x+7/2,-(3/2)x+9,(2/3)x))}}}