Let A(1,0), B(2,-1) and C ( 73, 43) be three points. If the equat… — Solution | JEE Mathematics

MCQMediumJEE 2026Straight Line Equations

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
$5$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The points are $A(1,0)$ , $B(2,-1)$ and $C\left(\dfrac{7}{3},\dfrac{4}{3}\right)$ .

Find: The value of $\alpha^2+\beta^2$ if the bisector of angle $ABC$ is $\alpha x+\beta y=5$ .

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

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

Therefore, the correct option is B.

Common mistakes

Using the vectors $\vec{AB}$ and $\vec{CB}$ instead of the vectors from the vertex $B$ . The angle bisector at $B$ must be formed using directions along $\vec{BA}$ and $\vec{BC}$ . Always take both vectors starting from the angle vertex.
Adding the raw direction vectors directly without converting them to unit vectors. For an angle bisector, the correct direction comes from the sum of the corresponding unit vectors, not the sum of arbitrary side vectors.
Making an error while finding $\vec{BC}$ from the coordinates of $C$ and $B$ . The subtraction must be coordinate-wise: $C-B=\left(\frac{7}{3}-2,\frac{4}{3}+1\right)$ . A sign error here changes the bisector completely.

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Answer

Step-by-step solution

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