Rewrite the line equations:
y=x/5+3 and y=3x/10+2, from which we get two slopes or gradients: 1/5=0.2 and 3/10=0.3. These are the tangents of the angles of the slopes, θ1 and θ2 respectively.
Since θ1 is the smaller (acute) angle the difference between them is θ2-θ1 so the bisector measures (θ2-θ1)/2. This gives us the slope (measured anticlockwise from the x-axis) of one of the bisectors:
θ1+(θ2-θ1)/2=(θ2+θ1)/2, that is, the average of the two angles, as you might expect. If we take the obtuse angle between them it measures 180-(θ2-θ1), half of which is 90-(θ2-θ1)/2. The gradient of the other bisector is:
θ2+90-(θ2-θ1)/2=90+(θ2+θ1)/2. This means that the bisector lines will be perpendicular to one another. Their gradients m1 and m2 will therefore be related: m1m2=-1.
We should now work out p=tan(θ2+θ1)=(tanθ2+tanθ1)/(1-tanθ2tanθ1)=(0.3+0.2)(1-0.06)=0.5/0.94=25/47.
We need tan((θ2+θ1)/2), so, if we let θ=(θ2+θ1)/2, then p=tan(2θ) and m1=tanθ, m2=-cotθ:
p=2tanθ/(1-tan2θ), 1-tan2θ=2tanθ/p, tan2θ+2tanθ/p=1,
tan2θ+2tanθ/p+1/p2=1+1/p2,
(tanθ+1/p)2=1+1/p2, m1=tanθ=-1/p+√(1+1/p2)=(-1+√(p2+1))/p and m2=(-1-√(p2+1))/p. (m1m2=(1-(p2+1))/p=-1)
We also need to find out where the given lines intersect:
x/5+3=3x/10+2, 2x+30=3x+20, x=10, so y=5.
y-5=m1(x-10) and y-5=m2(x-10) are the bisector equations.
These can be written:
y-5=m1x-10m1, -m1x+y+10m1-5=0 and -m2x+y+10m2-5=0.
m1=0.2494 and m2=-4.0094 approx.
So: -0.2494x+y-2.5059=0 and 4.0094x+y-45.0941=0 are the actual equations of the bisectors.