Let ABC be a triangle with $\mathrm{A}(-3,1)$ and $\angle \mathrm{ACB}=\theta, 0<\theta<\frac{\pi}{2}$. If the equation of the median through $B$ is $2 x+y-3=0$ and the equation of angle bisector of $C$ is $7 x-4 y-1=0$, then $\tan \theta$ is equal to:

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Straight Lines

Let ABC be a triangle with

A (- 3, 1)

and

∠ ACB = θ, 0 < θ < \frac{π}{2}

. If the equation of the median through

B

2 x + y - 3 = 0

and the equation of angle bisector of

C

7 x - 4 y - 1 = 0

, then

tan θ

is equal to:

Answer: (A)

Let

C = (c, \frac{7 c - 1}{4})

on the bisector

7 x - 4 y - 1 = 0

. The median through

B

passes through the midpoint of

A C

, which is

(\frac{c - 3}{2}, \frac{7 c + 3}{8})

, and this midpoint lies on

2 x + y - 3 = 0

(c - 3) + \frac{7 c + 3}{8} = 3 \Rightarrow 8 c - 24 + 7 c + 3 = 24 \Rightarrow c = 3,

C = (3, 5)

.
The slope of

C A

\frac{1 - 5}{- 3 - 3} = \frac{2}{3}

and the slope of the bisector is

\frac{7}{4}

. The bisector makes an angle

\frac{θ}{2}

with

C A

, so

tan \frac{θ}{2} = \frac{\frac{7}{4} - \frac{2}{3}}{1 + \frac{7}{4} \cdot \frac{2}{3}} = \frac{13/12}{26/12} = \frac{1}{2} .

Hence

tan θ = \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{4}{3} .

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option