Let A(1,0),B(2,-1) and C((7)/(3),(4)/(3)) be three points.

Question

Let $A(1,0),B(2,-1)$ and $C(\frac{7}{3},\frac{4}{3})$ 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

solution : $A(1,0),B(2,-1)$ and $C(\frac{7}{3},\frac{4}{3})$

equation of line AB using two point form

$\begin{gathered} y-0=\frac{0-(-1)}{1-2}(x-1) \ y=-x+1 \ {L}_{1}:x+y-1=0 \end{gathered}$

similarly equation of line $BC$.

$\begin{gathered} y+1=\frac{\frac{4}{3}-(-1)}{\frac{7}{3}-2}(x-2) \ y+1=\frac{7/3}{1/3}(x-2) \ y=7x-14-1 \ {L}_{2}:7x-y-15=0 \end{gathered}$

from graph ( 2,0 ) lies in the region of $\angleABC$.

calculating ${L}_{1}(2,0)\cdot {L}_{2}(2,0)$

$(2+0-1)\cdot (7	imes 2-0-15)=1\cdot (-1)=-1$

equation of angular bisector of $\angleABC$ is given as

$\begin{gathered} \frac{x+y-1}{\sqrt{{1}^{2}+{1}^{2}}}=\pm \frac{7x-y-15}{\sqrt{{7}^{2}+(-1{)}^{2}}} \ \frac{x+y-1}{\sqrt{2}}=\pm \frac{7x-y-15}{\sqrt{50}} \ 5x+5y-5=\pm (7x-y-15) \end{gathered}$

since ${L}_{1}(2,0)\cdot {L}_{2}(2,0)$ is negative so equation of $\angleABC$ bisector in the region $(2,0)$ can be

find using negative sign from equation (1)

$\begin{gathered} 5x+5y-5=-(7x-y-15) \ 5x+5y-5+7x-y-15=0 \ 12x+4y-20=0 \end{gathered}$

$3x+y=5$ comparing with bisector $\alpha x+\beta y=5$.

$\alpha =3$ and $\beta =1$

${\alpha }^{2}+{\beta }^{2}=9+1=10$.

Solutions · 1