Straight lines 3x + 4y = 5 and 4x - 3y = 15 ~intersect at the point...

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$passes through \left(1, 2\right).</p><p>From figure it is clear that$∠ B = ∠ C = 45 °$</p><p><img src="/api/proxy-image?url=https$BC$be$m$. Then,</p><p>$\tan 45 ° = \left|\dfrac{m + \dfrac{3}{4}}{1 - \dfrac{3}{4}m}\right|$</p><p>$⇒ \pm 1 = \dfrac{4m + 3}{4 - 3m}$</p><p>$4m + 3 = \pm \left(4 - 3m\right)

$4m + 3 = 4 - 3m$

or $4m + 3 = - 4 + 3m$

$m =$\dfrac{1}{7}$</p><p>or$m = - 7$</p><p>Hence, equation of$BC$is</p><p>$y - 2 = \dfrac{1}{7}\left(x - 1\right)

or $y - 2 = - 7 \left(x - 1\right)$

$⇒ 7y - 14 = x - 1$

or $y - 2 = - 7x + 7$

$⇒ x - 7y + 13 = 0$

or $7x + y - 9 = 0$

Hence, option (c) is correct.