Question

equation bisector solution to pairs of angles formed by lines -3x + 10y - 20 = 0 and -x + 5y -15 = 0

Answer 1

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.