A straight line 4 x + y - 1 = 0 through the point A ( 2 , - 7 ) meets the line B C whose equation is 3 x - 4…

The equation of line given as $4x+y-1=0, 3x-4y+1=0$.

The slopes of line are $m_1=-4$ and $m_2=\frac{3}{4}$.

Triangle $ABC$ is an isosceles; $AB=AC$.

$AB$ and $AC$ both pass through points $(2,-7)$.

The slope is,

$m_3=\left|\frac{-4-{\displaystyle \frac{3}{4}}}{1+(-4)\left({\displaystyle \frac{3}{4}}\right)}\right|=\frac{19}{8}$

The required equation of line $AC$ is,

$\frac{y+7}{x-2}=\frac{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{19}{8}}}{1+{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{19}{8}}}$

After simplification,

$89y+623=-52x+104$

$\Rightarrow 52x+89y+519=0$