Question 1031262
Let the points be P(0,4), Q(-4,-4) and R(6,1).

Then line PQ will have slope {{{m[PQ] = 2}}}, and the equation would be 2x - y + 4 = 0.
The line QR will have slope {{{m[QR] = 1/2}}}, and the equation would be x - 2y - 4 = 0.
Then line PR will have slope {{{m[PR] = -1/2}}}, and the equation would be x + 2y -8 = 0.

For the angle bisector at angle Q, let (x,y) be any point on it.  Then the distance of (x,y) from line PQ is {{{(2x-y+4)/sqrt(5)}}}, while its distance from the line QR is {{{-(x-2y-4)/sqrt(5)}}}.  (The negative sign denotes the fact that points (x,y) above the line x - 2y - 4 = 0 will give a net sign of negative for the expression x-2y-4 upon substitution.)

==> {{{(2x-y+4)/sqrt(5) = -(x-2y-4)/sqrt(5)}}}
==> 2x-y+4 = -x+2y+4  ==> 3x = 3y, or {{{highlight(x=y)}}}.

For the angle bisector at angle P, let (x,y) be any point on it.  Then the distance of (x,y) from line PQ is {{{(2x-y+4)/sqrt(5)}}}, while its distance from the line PR is {{{-(x+2y-8)/sqrt(5)}}}.  (The negative sign denotes the fact that points (x,y) below the line x + 2y - 8 = 0 will give a net sign of negative for the expression x+2y-8 upon substitution.)

==> {{{(2x-y+4)/sqrt(5) = -(x+2y-8)/sqrt(5)}}}
==> 2x -y +4 = -x-2y+8 ==> {{{highlight(3x + y - 4 = 0)}}}.

By a similar procedure, it easily found that the angle bisector for angle R is simply {{{highlight(y = 1)}}}.

a.) x-y=0 and x=0.
Let (x,y) be in the angle bisector with vertex at (0,0).  
The distance of (x,y) from the line x=0 (the y-axis) is x, while the distance of (x,y) from the line x - y=0 is {{{-(x-y)/sqrt(2)}}}.  The negative sign is for the fact that upon substitution of the coordinates of (x,y) into x - y, the sign of the expression is negative.

==> {{{x = -(x-y)/sqrt(2)}}}, or {{{highlight((sqrt(2)+1)x = y)}}}, after simplification.

b.) I leave up to you.