Straight lines 3 x + 4 y = 5 and 4 x - 3 y = 15 intersect at the point A . If point B and C are chosen on…

The given straight lines are $3x+4y=5$ and $4x-3y=15$. Clearly, these straight lines are perpendicular to each other $\left(m_1{m}_2=-1\right)$ and intersect at $A$. Now, $B$ and $C$ are points on these lines such that $AB=AC$ and $BC$ passes through $\left(1,2\right)$.

From figure it is clear that $\angle B=\angle C=45°$

Let slope of $BC$ be $m$. Then,

$\tan 45°=\left|\frac{m+\frac{3}{4}}{1-\frac{3}{4}m}\right|$

$\Rightarrow \pm 1=\frac{4m+3}{4-3m}$

$4m+3=\pm \left(4-3m\right)$

$4m+3=4-3m$

or $4m+3=-4+3m$

$m=\frac{1}{7}$

or $m=-7$

Hence, equation of $BC$ is

$y-2=\frac{1}{7}\left(x-1\right)$

or $y-2=-7\left(x-1\right)$

$\Rightarrow 7y-14=x-1$

or $y-2=-7x+7$

$\Rightarrow x-7y+13=0$

or $7x+y-9=0$

Hence, option (c) is correct.