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 $2x+y-3=0$ and the equation of angle bisector of $C$ is $7x-4y-1=0$ 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 $2x+y-3=0$
$\Rightarrow 2a+b=11\dots \dots .$(i)
$\because$ C lies on $7x-4y=1$
$\Rightarrow 7a-4b=1\dots$(ii)
$\therefore$ by (i) and (ii) :
$a=3,b=5$
$\Rightarrow C(3,5)$
$\therefore m_{AC}=2/3$
Also, $m_{CD}=7/4$
$\Rightarrow \tan \frac{\theta }{2}=\left|\frac{\frac{2}{3}-\frac{4}{4}}{1+\frac{14}{12}}\right|$
$\Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}$
$\Rightarrow \tan \theta =\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}$