Question

Find the bisectors of the interior angles of the triangle whose sides are the line 7x+y-7=0, x+y+1=0 and x+7y-4=0.

Answer 1

7x+y-7=0 intersects x+y+1=0 at (4/3,-7/3).

7x+y-7=0 makes an angle of 98.13°
x+y+1=0 makes an angle of 135°
with the x-axis.

So, the bisector makes an angle of 116.56°, with slope -2

So, the equation of the line is

y + 7/3 = -2(x - 4/3)
3y+7 = -2(3x-4)
6x+3y+15=0

Now work the other pairs of lines.