F4 ADM geometry

Let me be the slope of the angle bisector, then by formula:
|(m - 1)/(1 + m)| = |(m + 1/7)/(1 - m/7)|
|(m - 1)/(1 + m)| = |(7m + 1)/(7 - m)|
|m - 1| |7 - m| = |7m + 1| |1 + m|
|-m2 + 8m - 7| = |7m2 + 8m + 1|
7m2 + 8m + 1 = ±(-m2 + 8m - 7)
6m2 + 16m - 6 = 0 or 8m2 + 8 = 0 (no real solutions)
3m2 + 8m - 3 = 0
(3m - 1) (m + 3) = 0
m = 1/3 or -3
So the pair of angle bisectors have their slopes equal to 1/3 and -3.

Given two lines,
L1: ax+by+c=0, and L2: a1x+b1y+c1=0
The angle bisectors are given by:
(ax+by+c)/sqrt(a^2+b^2) = +/- (a1x+b1y+c1)/sqrt(a1^2+b1^2)
If we simplify the lines to be
L1: y=mx+c, and L2: y=m1x+c1
Then the bisectors are given by:
(y-mx-c)/sqrt(1+m^2) = +/- (y-m1x-c1)/sqrt(1+m1^2)
on simplifying, taking only the + sign,
y=(m1/sqrt(m1^2+1)+m/sqrt(m^2+1))/(1/sqrt(m1^2+1)+1/sqrt(m^2+1)) x + ....
or the slope of one of the bisectors is:
mb=(m1/sqrt(m1^2+1)+m/sqrt(m^2+1))/(1/sqrt(m1^2+1)+1/sqrt(m^2+1))
Setting m=1, m1=-1/7
and substitute into the above, we obtain the slope of one of the angle bisectors,
mb=1/3
For the slope of the other angle bisector, we get
mb'=-(1/mb)=-3