Question 1062846
First find the point of intersection,
{{{x-3=-x/3-1/3}}}
{{{x+x/3=3-1/3}}}
{{{(4/3)x=8/3}}}
{{{x=2}}}
So then,
{{{y=2-3}}}
{{{y=-1}}}
Now find the perpendicular bisector of the line, {{{y=-x/3-1/3}}}}, 
you know that the slope of the bisector is equal to {{{m=3}}} since perpendicular lines have slopes that are negative reciprocals.
So the line has the form {{{y=3x+b}}}
Choose a point on the line.
I'll choose (5,-2). 
So the equation of the perpendicular bisector using point slope form would be,
{{{y+2=3(x-5)}}}
{{{y=3x-15-2}}}
{{{y=3x-17}}}
The intersection point of the perpendicular bisector with the original line can be found,
{{{x-3=3x-17}}}
{{{-2x=-14}}}
{{{x=7}}}
and
{{{y=7-3}}}
{{{y=4}}}
(7,4)
So then using (5,-2) as the center point and (7,4) as the intersection point, find the corresponding intersection point with the symmetric line. 
To go from the center point to the intersection point, you move 2 units in x and 6 units in y. The intersection point of the symmetric line will have the same value but just different sign. 
So starting at (5,-2) you would move -2 in the x and -6 in the y. 
(5,-2)+(-2,-6)=((3,-8)
So now you know the symmetric line goes through (3,-8) and (2,-1). 
Find the slope,
{{{m=(-8-(-1))/(3-2)=-7}}}
Then use the point slope form,
{{{y-(-8)=-7(x-3)}}}
{{{y+8=7x+21}}}
{{{highlight(y=13-7x)}}}
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