How to find slope of angle bisector of two lines with slope M1 and M2?

Question

Given two slopes(M1 and M2) of two distinct lines, is there any way to find the slope(M3) of angle bisectors of those two lines?

Answer 1

If $k_1$ and $k_2$ are two given slopes, then vectors of these two lines are $(1,k_1)$ and $(1,k_2)$, or in normalized form $u_1=(\frac{1}{\sqrt{1+k_1^2}},\frac{k_1}{\sqrt{1+k_1^2}})$ and $u_2=(\frac{1}{\sqrt{1+k_2^2}},\frac{k_2}{\sqrt{1+k_2^2}})$. Note that the vectors of bisectors are given by $u_1+u_2$ and $u_1-u_2$ as these are the diagonals of the rhombi determined by $u_1$, and $u_2$, and by $u_1$ and $-u_2$. So the vectors of the bisectors are: $$(\frac{1}{\sqrt{1+k_1^2}}+\frac{1}{\sqrt{1+k_2^2}},\frac{k_1}{\sqrt{1+k_1^2}}+\frac{k_2}{\sqrt{1+k_2^2}})$$ and $$(\frac{1}{\sqrt{1+k_1^2}}-\frac{1}{\sqrt{1+k_2^2}},\frac{k_1}{\sqrt{1+k_1^2}}-\frac{k_2}{\sqrt{1+k_2^2}}).$$ By scaling them we get vectors: $$(1,\frac{k_1\sqrt{1+k_2^2}+k_2\sqrt{1+k_1^2}}{\sqrt{1+k_2^2}+\sqrt{1+k_1^2}})\mbox{ and }(1,\frac{k_1\sqrt{1+k_2^2}-k_2\sqrt{1+k_1^2}}{\sqrt{1+k_2^2}-\sqrt{1+k_1^2}}),$$ so the slopes of the bisectors are given by: $$\frac{k_1\sqrt{1+k_2^2}+k_2\sqrt{1+k_1^2}}{\sqrt{1+k_2^2}+\sqrt{1+k_1^2}}\mbox{ and }\frac{k_1\sqrt{1+k_2^2}-k_2\sqrt{1+k_1^2}}{\sqrt{1+k_2^2}-\sqrt{1+k_1^2}}.$$