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 equal to

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

Answer 1

The correct option is C $\frac{4}{3}$

$E(\frac{x_3-3}{2},\frac{y_3+1}{2})$ lies on $2x+y-3=0$
$\Rightarrow 2\left(\frac{x_3-3}{2}\right)+\frac{y_3+1}{2}-3=0$


$\Rightarrow 2x_3+y_3-11=0\cdots \left(i\right)$
$C(x_3,y_3)$ lies on $7x-4y-1=0$
$\therefore 7x_3-4y_3-1=0\cdots \left(ii\right)$
From equation $\text{(i) \& (ii)},$ we get
$C\equiv (x_3,y_3)\equiv (3,5)$

$A^{\prime }(x_2,y_2)$ is image of $A$ and is given by
$\frac{x_2+3}{7}=\frac{y_2-1}{-4}=-2\left(\frac{7(-3)-4\left(1\right)-1}{7^2+(-4{)}^2}\right)=\frac{4}{5}$
$\Rightarrow A^{\prime }\equiv (x_2,y_2)\equiv (\frac{13}{5},\frac{-11}{5})$
$m_{AC}=m_1=\frac{5-1}{3-(-3)}=\frac{2}{3}$
$m_{BC}=m_2=\frac{5+\frac{11}{5}}{3-\frac{13}{5}}=18$
$\therefore \tan \theta =\frac{18-\frac{2}{3}}{1+18\times \frac{2}{3}}=\frac{4}{3}$