Coordinate Geometry – Straight Lines – JEE Main 24 Jan 2026 Shift 1

Solution:

Let the angle bisector of $\angle ABC$ meet $AC$ at point $D$.
By the Angle Bisector Theorem, the ratio in which $D$ divides $AC$ is equal to the ratio of the lengths of the sides $AB$ and $BC$.

Calculate lengths $AB$ and $BC$:
$$AB = \sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$
$$BC = \sqrt{(2-\frac{7}{3})^2 + (-1-\frac{4}{3})^2} = \sqrt{(-\frac{1}{3})^2 + (-\frac{7}{3})^2} = \sqrt{\frac{1}{9} + \frac{49}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$$

Ratio $AB : BC = \sqrt{2} : \frac{5\sqrt{2}}{3} = 1 : \frac{5}{3} = 3 : 5$.
So, $D$ divides $AC$ internally in the ratio $3:5$.

Coordinates of $A(1,0)$ and $C(\frac{7}{3}, \frac{4}{3})$.
Using the section formula for $D(x,y)$:
$$x_D = \frac{3(\frac{7}{3}) + 5(1)}{3+5} = \frac{7+5}{8} = \frac{12}{8} = \frac{3}{2}$$
$$y_D = \frac{3(\frac{4}{3}) + 5(0)}{3+5} = \frac{4}{8} = \frac{1}{2}$$
So, $D \equiv (\frac{3}{2}, \frac{1}{2})$.

The angle bisector passes through $B(2,-1)$ and $D(\frac{3}{2}, \frac{1}{2})$.
Equation of line $BD$:
$$y – (-1) = \frac{\frac{1}{2} – (-1)}{\frac{3}{2} – 2} (x – 2)$$
$$y + 1 = \frac{\frac{3}{2}}{-\frac{1}{2}} (x – 2)$$
$$y + 1 = -3(x – 2)$$
$$y + 1 = -3x + 6$$
$$3x + y = 5$$

Comparing with $\alpha x + \beta y = 5$:
$\alpha = 3$, $\beta = 1$.

We need $\alpha^2 + \beta^2$:
$$\alpha^2 + \beta^2 = 3^2 + 1^2 = 9 + 1 = 10$$

Ans. (4)