BITSAT 2018 Slot 2 Solved Paper

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.