Question 929360
A(-2, 5), B(2, -3), C(8, 0), and P(4, 3). 
{{{drawing(300,300,-10,10,-10,10,grid(1),
blue(line(-2,5,2,-3)),
blue(line(2,-3,8,0)),
circle(-2,5,0.2),circle(2,-3,0.2),circle(8,0,0.2),circle(4,3,0.2))}}}
Find the slope of AB and BC.
AB:
{{{m=(-3-5)/(2-(-2))=-8/4=-2}}}
BC:
{{{m=(0-(-3))/(8-2)=3/6=1/2}}}
AB and BC are perpendicular since their slopes are negative reciprocals.
{{{(-2)(1/2)=-1}}}
The angle between the two lines is then 90 so a bisector would split the angle into 2 45 degree segments.
The line BC makes an angle with the x-axis determined by the slope.
{{{tan(alpha)=1/2}}}
{{{alpha=26.6}}} {{{degrees}}}
Adding {{{45}}} to this and then taking the tangent to find the slope,
{{{m=tan(alpha+45)=tan(71.6)=3}}}
Using the point slope form of a line with point B,
{{{y-(-3)=3(x-2)}}}
{{{y+3=3x-6}}}
{{{highlight(y=3x-9)}}}
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{{{drawing(300,300,-10,10,-10,10,grid(1),
blue(line(-2,5,2,-3)),
blue(line(2,-3,8,0)),
circle(-2,5,0.2),circle(2,-3,0.2),circle(8,0,0.2),circle(4,3,0.2),graph(300,300,-10,10,-10,10,3x-9))}}}
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You could have also used point P and found the slope of BP.
{{{m=(3-(-3))/(4-2)=6/2=3}}}
Continuing the same way as shown previously.