What is the slope of a line half way between two lines of slope m1 and m2?

Question

I have two lines of slope m1 and m2 respectively. What is the slope of the line bisecting these two lines? Clearly there are 2 different lines which can be constructed, one perpendicular to the other. I am looking for the one which bisects the acute angle between the 2 lines.

Answer 1

We have:

$$m_1=\frac{y_2-y_1}{x_2-x_1}=\tan{(\alpha)}$$

And:

$$m_2=\frac{y_3-y_1}{x_3-x_1}=\tan{(\beta)}$$

Therefore,

$$m_3=\tan\left(\beta+\frac{\alpha-\beta}{2}\right)=\tan\left(\frac{\alpha+\beta}{2}\right)$$

Where $m_3$ is the slope of the bisector.

Now, we will use this result to find an expression in terms of $m_1$ and $m_2$.

One may show that:

$$\tan\left(\frac{\alpha+\beta}{2}\right)\equiv \frac{1-\cos(\alpha+\beta)}{\sin(\alpha+\beta)}$$

Now, you can use the double angle formulas for $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$, then apply the substitutions $\sin(x)=\frac{m}{\sqrt{1+m^2}}$ and $\cos(x)=\frac{1}{\sqrt{1+m^2}}$ to obtain an expression for $m_3=f(m_1,m_2)$.

Answer 2

First of all, there are two angle bisectors, so you’ll somehow have to decide which one you wanted.

You might use formulas for sums and half-angles of tangents to calculate this, but it’s pretty easy to find the slope of the bisector directly. In an isosceles triangle, the angle bisector between the two same-length sides passes through the midpoint of the opposite side. So, we move the same distance along each line, find the midpoint, and compute the resulting slope.

Specifically, we can take unit vectors parallel to the lines: $$\begin{align}u_1&=\left({1\over\sqrt{1+m_1^2}},{m_1\over\sqrt{1+m_1^2}}\right) \\ u_2&=\left({1\over\sqrt{1+m_2^2}},{m_2\over\sqrt{1+m_2^2}}\right)\end{align}$$ which you can derive from the identity $\tan^2\theta + 1=\sec^2\theta$, find their midpoint $(u_1+u_2)/2$, divide the $y$-coordinate by the $x$-coordinate and simplify: $$m_3={{\frac12\left({m_1\over\sqrt{1+m_1^2}}+{m_2\over\sqrt{1+m_2^2}}\right)}\over\frac12\left({1\over\sqrt{1+m_1^2}}+{1\over\sqrt{1+m_2^2}}\right)} = {m_1m_2+\sqrt{1+m_1^2}\sqrt{1+m_2^2}-1\over m_1+m_2}.$$ The slope of the other bisector is found via $(u_1-u_2)/2$, which gives $$m_3'={m_1m_2-\sqrt{1+m_1^2}\sqrt{1+m_2^2}-1\over m_1+m_2}$$ instead. (This amounts to choosing either the same sign or opposite signs for the two square roots, which are the cosines of the line angles.) You can verify that $m_3m_3'=-1$, which means that the two bisectors are perpendicular to each other, as we’d expect.

This formula is of course invalid when either line is parallel to the $y$-axis or $m_1+m_2=0$, but those cases are easily dealt with by the bisector method, too. If $m_1=-m_2$, the two midpoints are of the form $(a,0)$ and $(0,b)$ so the bisectors are the coordinate axes. When only one of the lines is vertical, the midpoints are $((0,1)\pm m)/2$, which yields $m_3=m\pm\sqrt{1+m}$ for the slopes. It should be obvious what the bisectors are when both lines are vertical.