A line \(4 x+y=1\) passes through the point \(\mathrm{A}(2,-7)\) meets the line BC whose equation is \(3 x-4 y+1=0\) at the point B . The equation of the line \(A C\) so that \(A B=A C\) is

Slopes of \(A B\) and \(B C\) are -4 and \(\frac{3}{4}\) respectively.

Let \(\alpha\) be the angle between AB and BC .
Then, \(\tan \alpha=\left|\frac{-4-\frac{3}{4}}{1-4\left(\frac{3}{4}\right)}\right|=\frac{19}{8}\)...(i)
Since \(A B=A C\)
\(\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{ACB}=\alpha\)
\(\therefore \quad\) the line \(A C\) also makes an angle \(\alpha\) with \(B C\).
If \(m\) is the slope of the line \(A C\), then its equation is \(y+7=\mathrm{m}(x-2)\)...(ii)
Now, \(\tan \alpha= \pm\left[\frac{\mathrm{m}-\frac{3}{4}}{1+\mathrm{m} \cdot \frac{3}{4}}\right]\)
\(\Rightarrow \frac{19}{8}= \pm \frac{4 \mathrm{~m}-3}{4+3 \mathrm{~m}}\)
...[From (i)]
\(\Rightarrow \mathrm{m}=-4\) or \(-\frac{52}{89}\)
But slope of \(A B\) is -4 , so slope of AC is \(-\frac{52}{89}\).
Therefore, the equation of line AC given by (ii) is \(52 x+89 y+519=0\).