Q55 / 125 · BITSAT 2011 · Mathematics

The family of straight lines $(2 a + 3 b) x + (a - b) y + 2 a - 4 b = 0$ is concurrent at the point

Correct answer
A
$(\frac{2}{5}, \frac{- 14}{5})$
B
$(\frac{- 2}{5}, \frac{- 14}{5})$
C
$(\frac{- 2}{5}, \frac{14}{5})$
D
$(\frac{2}{5}, \frac{14}{5})$

Solution

Rewriting the equation

(2 x + y + 2) a + (3 x - y - 4) b = 0

and for all

a

,

b

the straight lines pass through the intersection of

2 x + y + 2 = 0

and

3 x - y - 4 = 0

i.e., the point

(\frac{2}{5}, - \frac{14}{5})

.

Hence, the correct answer is option: 'A'.

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