BITSAT 2013 Paper

4a2+b2+2c2 + 4ab – 6ac – 3bc = 0
= (2a+b)2 - 3 (2a + b) + 2c2 = 0
⇒ (2a + b - 2c) (2a + b - c) = 0
⇒ c= 2a + b
or c = a +

1

2

b
The equation of the family of lines is
a (x + 2) + b (y + 1) = 0 or a (x + 1) + b (y+

1

2

) = 0
giving the point of concurrence (–2, –1) or
(−1,

1

2

)