Question

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

• A. 10
• B. 8
• C. 13
• D. 5

Hint:

For angle bisectors, always use unit vectors along the sides meeting at the vertex.

Answer 1

The Correct Option is B

Step 1: Finding direction vectors of $BA$ and $BC$.
\[ \vec{BA}=A-B=(1-2,\;0+1)=(-1,1) \] \[ \vec{BC}=C-B=\left(\frac{7}{3}-2,\;\frac{4}{3}+1\right)=\left(\frac{1}{3},\frac{7}{3}\right) \] Step 2: Finding unit vectors.
\[ |\vec{BA}|=\sqrt{(-1)^2+1^2}=\sqrt{2} \Rightarrow \hat{u}_1=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \] \[ |\vec{BC}|=\sqrt{\left(\frac{1}{3}\right)^2+\left(\frac{7}{3}\right)^2} =\frac{\sqrt{50}}{3} \Rightarrow \hat{u}_2=\left(\frac{1}{\sqrt{50}},\frac{7}{\sqrt{50}}\right) \] Step 3: Equation of angle bisector.
Direction ratios of the internal bisector are \[ \hat{u}_1+\hat{u}_2 \] Simplifying, the equation of the angle bisector through $B(2,-1)$ is \[ 2x+2y=5 \] Thus, \[ \alpha=2,\quad \beta=2 \] Step 4: Final calculation.
\[ \alpha^2+\beta^2=4+4=8 \]