Let A(1,0), B(2,-1) and C(73,43) be three points.

Question

Let $A(1,0),\ B(2,-1)$ and $C\left(\dfrac73,\dfrac43\right)$ be three points. If the equation of the bisector of the angle $ABC$ is $\alpha x + \beta y = 5$, then the value of $\alpha^2+\beta^2$ is:

Accepted Answer

Line $AB$: through $(1,0),(2,-1)$: $y = -(x-1) \implies L_1: x+y-1=0$.

Line $BC$: through $(2,-1),\left(\frac73,\frac43\right)$: slope $=\dfrac{4/3+1}{7/3-2}=7 \implies L_2: 7x-y-15=0$.

At $B(2,0)$-test: $L_1(2,0)=1,\ L_2(2,0)=-1$, opposite signs, so the bisector of $\angle ABC$ (the region containing the interior) uses the negative sign:
$$\frac{x+y-1}{\sqrt2} = -\frac{7x-y-15}{\sqrt{50}}$$
$$5x+5y-5 = -(7x-y-15) \implies 12x+4y-20=0 \implies 3x+y=5$$

So $\alpha=3,\ \beta=1$, giving $\alpha^2+\beta^2 = 9+1=10$.

Solutions · 1