The given straight lines are 3x+4y=5 and 4x−3y=15. Clearly, these straight lines are perpendicular to each other (m1m2=−1) and intersect at A. Now, B and C are points on these lines such that AB=AC and BC passes through (1,2) From figure it is clear that ∠B=∠C=45°
Let slope of BC be m. Then, tan‌45°=|
m+
3
4
1−
3
4
m
| ⇒ ±1=
4m+3
4−3m
4m+3=±(4−3m) 4m+3=4−3m or 4m+3=−4+3m m=
1
7
or m=−7 Hence, equation of BC is y−2=
1
7
(x−1) or y−2=−7(x−1) ⇒7y−14=x−1 or y−2=−7x+7 ⇒ x−7y+13=0 or 7x+y−9=0 Hence, option (c) is correct.