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:

Let the vertices be $$A(-3,1),\;B(x_B,y_B),\;C(x_C,y_C)$$.

The given data translate to the following two straight-line conditions:

• Median through $$B$$: $$2x+y-3=0$$ $$\;-(1)$$ (passes through $$B$$ and the midpoint of $$AC$$).
• Angle bisector at $$C$$: $$7x-4y-1=0$$ $$\;-(2)$$ (passes through $$C$$).

Step 1: Find the coordinates of $$C$$

The midpoint of $$AC$$ is $$M\Bigl(\dfrac{-3+x_C}{2},\,\dfrac{1+y_C}{2}\Bigr)$$. Because the median through $$B$$ also contains $$M$$, substitute $$M$$ in $$(1)$$:

$$2\left(\dfrac{-3+x_C}{2}\right)+\dfrac{1+y_C}{2}-3=0$$ $$\Longrightarrow\;(-3+x_C)+\dfrac{1+y_C}{2}-3=0$$ $$\Longrightarrow\;x_C-6+\dfrac{1+y_C}{2}=0$$ $$\Longrightarrow\;2x_C+y_C-11=0$$ $$\;-(3)$$

The point $$C$$ also lies on the bisector $$(2)$$: $$7x_C-4y_C-1=0$$ $$\;-(4)$$.

Solve $$(3)$$ and $$(4)$$ simultaneously:
From $$(3)$$, $$y_C=11-2x_C$$.
Insert in $$(4)$$: $$7x_C-4(11-2x_C)-1=0$$ $$\Rightarrow 7x_C-44+8x_C-1=0$$ $$\Rightarrow 15x_C=45\;\Longrightarrow\;x_C=3$$ $$\Rightarrow y_C=11-2(3)=5$$.

Hence $$C(3,5)$$.

Step 2: Slope of side $$CA$$

$$m_{CA}=\dfrac{5-1}{\,3-(-3)\,}= \dfrac{4}{6}= \dfrac{2}{3}.$$

Step 3: Slope of the angle-bisector at $$C$$

Rewrite $$(2)$$: $$4y=7x-1 \;\Longrightarrow\; y=\dfrac{7}{4}x-\dfrac{1}{4}$$, so $$m_{\text{bis}}=\dfrac{7}{4}.$$

Step 4: Angle between $$CA$$ and the bisector

For two lines with slopes $$m_1,\,m_2$$ the acute angle $$\phi$$ between them is $$\tan\phi=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|.$$ Here $$\tan\phi=\left|\dfrac{\dfrac{7}{4}-\dfrac{2}{3}}{1+\dfrac{2}{3}\cdot\dfrac{7}{4}}\right| =\dfrac{\dfrac{13}{12}}{\dfrac{13}{6}}=\dfrac{1}{2}.$$

Thus $$\phi=\arctan\!\left(\dfrac{1}{2}\right).$$

Step 5: Obtain $$\theta$$ from the angle bisector property

The line $$(2)$$ is the internal angle bisector at $$C$$; hence it divides the angle $$\angle ACB$$ into two equal parts. Therefore $$\theta=2\phi$$.

Using the double-angle identity $$\tan(2\phi)=\dfrac{2\tan\phi}{1-\tan^2\phi}$$, take $$\tan\phi=\dfrac{1}{2}$$:

$$\tan\theta=\dfrac{2\cdot\dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2} =\dfrac{1}{\,1-\dfrac{1}{4}\,}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}.$$

Answer: $$\displaystyle \tan\theta=\dfrac{4}{3}$$ (Option B).