Let A B C be a triangle with A ( - 3 , 1 ) and ∠ A C B = θ , 0 < θ < π 2 . If…

Let, $x-$coordinate of the point $C$ is $\alpha$, So coordinate of the point   $C\left(\alpha ,\frac{7\alpha -1}{4}\right)$  $\because (\mathrm{C}\text{ lies on }7x-4y-1=0)$

 Hence, mid point of $AC$ is $m\left(\frac{\alpha -3}{2},\frac{7\alpha +3}{8}\right)$

As, point $m$ lies on the line $2x+y-3=0$, Substituting, we get $\alpha =3$.

So. $C(3,5)$

Hence, slope of AC is $\frac{2}{3}$ and slope of $7x-4y-1=0$ is$\frac{7}{4}$

$\therefore \tan \frac{\theta }{2}=\frac{\frac{7}{4}-\frac{2}{3}}{1+\frac{7}{4}\times \frac{2}{3}}=\frac{1}{2}$

$\Rightarrow \tan \theta =\frac{2·\frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}\Rightarrow \tan \theta =\frac{4}{3}$