If the line l_1:3y-2x=3 is the angular bisector of the lines l_2:x-y+1=0 and l_3: x+ y+17=…

Question

If the line ${l}_{1}:3y-2x=3$ is the angular bisector of the lines ${l}_{2}:x-y+1=0$ and ${l}_{3}:\alpha x+\beta y+17=0$, then ${\alpha }^{2}+{\beta }^{2}-\alpha -\beta $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 348

Accepted Answer

$l_1$ and $l_2$ intersect at $P$: solving $3y-2x=3$ and $x-y+1=0$: $y=x+1 \implies 3(x+1)-2x=3\implies x=0,\ y=1$. So $P=(0,1)$.

Since $P$ lies on $l_3$: $\beta(1)+17=0 \implies \beta=-17$.

Take a point $Q(-1,0)$ on $l_2$. Its reflection in $l_1$ (the bisector) must lie on $l_3$. Reflecting $Q(-1,0)$ in $3y-2x-3=0$:
$$\frac{x+1}{-2}=\frac{y-0}3=\frac{-2(-2(-1)+3(0)-3)}{13} = \frac{-2(-1)}{13}=\frac2{13}$$
$$x=-1-\frac4{13}=-\frac{17}{13}, \quad y=\frac6{13}$$

So $Q'=\left(-\dfrac{17}{13},\dfrac6{13}\right)$ lies on $l_3: \alpha x-17y+17=0$:
$$\alpha\left(-\frac{17}{13}\right)-17\left(\frac6{13}\right)+17=0 \implies -\frac{17\alpha}{13}-\frac{102}{13}+17=0$$
$$-17\alpha-102+221=0 \implies -17\alpha=-119 \implies \alpha=7$$

$$\alpha^2+\beta^2-\alpha-\beta = 49+289-7+17 = 348$$

Solutions · 1