Question

Let $ABC$ be a triangle with $A (-3,1)$ and $\angle ACB =\theta, 0<\theta<\frac{\pi}{2}$. 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 cqual to:

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $2$

Answer 1

Answer: (C)

$\therefore M \left(\frac{ a -3}{2}, \frac{ b +1}{2}\right)$ lies on $ 2 x + y -3=0 $
$\Rightarrow 2 a + b =11 \ldots \ldots .$(i)
$\because $ C lies on $7 x -4 y =1$
$\Rightarrow 7 a -4 b =1 \ldots $(ii)
$\therefore $ by (i) and (ii) :
$ a =3, b =5 $
$\Rightarrow C (3,5) $
$\therefore m _{ AC }=2 / 3$
Also, $m _{ CD }=7 / 4$
$\Rightarrow \tan \frac{\theta}{2}=\left|\frac{\frac{2}{3}-\frac{4}{4}}{1+\frac{14}{12}}\right|$
$ \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2}$
$\Rightarrow \tan \theta=\frac{2 \cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}$