First slide

Pair of straight lines

Question

If $m_{1}$ and $m_{2}$ are the roots of the equation $x^{2} - a x - a - 1 = 0$ , then the area of the triangle formed by the three straight lines $y = m_{1} x, y = m_{2} x a n d y = a (a \neq - 1)$ is

Easy

A

$\frac{- a^{2} (a + 2)}{2 (a + 1)} i f a > - 1$
B

$\frac{+ a^{2} (a + 2)}{2 (a + 1)} i f a < - 1$
C

$\frac{- a^{2} (a + 2)}{2 (a + 1)} i f - 2 < a < - 1$
D

$\frac{a^{2} (a + 2)}{2 (a + 1)} i f a > - 2$

Solution

Since $m_{1}$ and $m_{2}$ are the roots of the equation $x^{2} - a x - a - 1 = 0$
so $m_{1} + m_{2} = a; m_{1} m_{2} = - (a + 1)$
$\Rightarrow {(m_{1} - m_{2})}^{2} = {(m_{1} + m_{2})}^{2} + 4 m_{1} m_{2}$
$= a^{2} + 4 (a + 1)$ $= {(a + 2)}^{2}$
$∴$ The required area is $Δ = \pm \frac{a^{2} (a + 2)}{2 (a + 1)}$
since area is positive quantity, so we get two answers

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