Straight Lines
Straight Line
star_batch_jee_advanced_2025
Grade 11
Question:
$ABC$ is a triangle, whose vertex $A$ is $(3,4)$, $L_1 = 0$, $L_2 = 0$ are the angle bisectors of angle $B$ and $C$ respectively where $L_1 = x + 2y - 5 = 0$, $L_2 = x - 2y - 3 = 0$ also $AB = KAI$ where $I$ is the incentre then $K$ is ____.
Step-by-Step Solution
Key Concept: The incenter divides angle bisectors in a ratio determined by the triangle's sides, which relates to the sine rule applied to the angle subtended.
Point $A(3,4)$ is the incenter of triangle $ABC$. Using the angle formula with $m_l = -1$ and $m_{Al} = -\frac{7}{2}$, we compute $\tan\alpha = \frac{-7/2 - (-1)}{1 + 7/4} = \frac{16}{3}$. Then $\beta = 90° - \frac{\beta}{2}$ gives $\tan\beta = \frac{16}{3}$ and $\sin\frac{\beta}{2} = \frac{3}{\sqrt{265}}$. By the sine rule $\frac{AI}{\sin(\beta/2)} = \frac{AB}{\sin\alpha}$, we obtain $AB = 2\sqrt{53}$ and $k = 4$ from $AB = k\cdot AI$.
Correct Answer: 4