Q41 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let $\mathrm{A}(1,0), \mathrm{B}(2,-1)$ and $\mathrm{C}\left(\frac{7}{3}, \frac{4}{3}\right)$ 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
- A13
- Correct answerB10
- C5
- D8
Solution
For angle bisector of angle ABC where A(1,0), B(2,-1), C($\frac{7}{3}$, $\frac{4}{3}$).
Direction BA: $(1-2, 0-(-1)) = (-1, 1)$, unit vector $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Direction BC: $\left(\frac{1}{3}, \frac{7}{3}\right)$, magnitude $\frac{5\sqrt{2}}{3}$, unit vector $\left(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}\right)$
Angle bisector direction (sum of unit vectors): $\left(\frac{-4}{5\sqrt{2}}, \frac{12}{5\sqrt{2}}\right)$ → direction $(-1, 3)$
Normal to bisector: $(3, 1)$. Bisector passes through B(2, -1):
$3(x-2) + 1(y+1) = 0$ → $3x + y - 5 = 0$
Therefore $\alpha = 3$, $\beta = 1$, so $\alpha^2 + \beta^2 = 9 + 1 = 10$
Direction BA: $(1-2, 0-(-1)) = (-1, 1)$, unit vector $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Direction BC: $\left(\frac{1}{3}, \frac{7}{3}\right)$, magnitude $\frac{5\sqrt{2}}{3}$, unit vector $\left(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}\right)$
Angle bisector direction (sum of unit vectors): $\left(\frac{-4}{5\sqrt{2}}, \frac{12}{5\sqrt{2}}\right)$ → direction $(-1, 3)$
Normal to bisector: $(3, 1)$. Bisector passes through B(2, -1):
$3(x-2) + 1(y+1) = 0$ → $3x + y - 5 = 0$
Therefore $\alpha = 3$, $\beta = 1$, so $\alpha^2 + \beta^2 = 9 + 1 = 10$
Hence, the correct answer is option: 'B'.