Test Index

BITSAT 2013 Paper

Section: Mathematics

Question : 94 of 150

Marks: +1, -0

4a^2 + b^2 + 2c^2

Solution:

4a^2 + b^2 + 2c^2

+ 4ab – 6ac – 3bc = 0

(2a+b)^2

- 3 (2a + b) +

2c^{2}

= 0

⇒ (2a + b - 2c) (2a + b - c) = 0

⇒ c= 2a + b

or c = a +

\frac{1}{2}

The equation of the family of lines is

a (x + 2) + b (y + 1) = 0 or a (x + 1) + b

\left(y+\frac{1}{2}\right)

= 0

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

\left(-1,\frac{1}{2}\right)

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