Let ABC be a triangle with A(-3, 1) and angle | vedclass

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(C) Let $C = (a, b)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7a - 4b = 1 \quad \dots(i)$.Let $M$ be the midpoint of $AC$. Sin

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$\frac{4}{3}$

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(C) Let $C = (a, b)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7a - 4b = 1 \quad \dots(i)$.Let $M$ be the midpoint of $AC$. Since $A = (-3, 1)$,$M = (\frac{a-3}{2}, \frac{b+1}{2})$.Since $M$ lies on the median $2x + y - 3 = 0$,we have $2(\frac{a-3}{2}) + (\frac{b+1}{2}) - 3 = 0$,which simplifies to $2a - 6 + b + 1 - 6 = 0$,or $2a + b = 11 \quad \dots(ii)$.Solving $(i)$ and $(ii)$,we multiply $(ii)$ by $4$: $8a + 4b = 44$. Adding to $(i)$: $15a = 45 \Rightarrow a = 3$. Substituting into $(ii)$: $6 + b = 11 \Rightarrow b = 5$. Thus,$C = (3, 5)$.The slope of $AC$ is $m_{AC} = \frac{5-1}{3-(-3)} = \frac{4}{6} = \frac{2}{3}$.The slope of the angle bisector $7x - 4y - 1 = 0$ is $m_{bisector} = \frac{7}{4}$.Let $\alpha$ be the angle of $AC$ and $\beta$ be the angle of the bisector. The angle between them is $\frac{	heta}{2}$.$	an(\frac{	heta}{2}) = |\frac{m_{bisector} - m_{AC}}{1 + m_{bisector} \cdot m_{AC}}| = |\frac{7/4 - 2/3}{1 + (7/4)(2/3)}| = |\frac{(21-8)/12}{(12+14)/12}| = \frac{13}{26} = \frac{1}{2}$.Finally,$	an 	heta = \frac{2 	an(	heta/2)}{1 - 	an^2(	heta/2)} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.

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

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