Given family of lines can be written as
3x + 4y – 7 +l(2x + 3y – 5) = 0
The point of intersection is (1, 1). Let Q ≡ (2, 3)
L1 will be perpendicular to PQ through P and L2 will be passing through Q. Therefore,
L1 ≡ 2y + x = 3 L2 ≡ 2x − y = 1 As angle between the line is same for L1 & L2. Therefore,
m1 = (1+ 2)/(1 - 2) = -3
Other slope will be perpendicular to m1, that is,
m2 = 1/3
So, equation of lines passing through (1, 2) are
3x + y = 5
x –3y = –5