The area (in square unit) of the triangle formed by $x + y + 1 = 0$ and the pair of straight lines $x^{2} - 3xy + 2y^{2} = 0$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $x^{2} - 2xy - xy + 2y^{2} = 0$
$\Rightarrow \ (x - 2y)(x - y) = 0$
$\Rightarrow \ x = 2y$ …(i)
$x = y$ …(ii)
Also, $x + y + 1 = 0$ …(iii)
On solving Eq. (i) and (ii) and (iii), we get
$A\left( - \frac{2}{3},\  - \frac{1}{3} ight),\ B\left( - \frac{1}{2},\  - \frac{1}{2} ight),\ C(0,\ 0)$
$	herefore Area\ of\ \mathrm{\Delta}ABC = \frac{1}{2}\left| \begin{matrix}
 - \frac{2}{3} & - \frac{1}{3} & 1 \
 - \frac{1}{2} & - \frac{1}{2} & 1 \
0 & 0 & 1
\end{matrix} ight|$
$= \frac{1}{2}\left\lbrack \frac{1}{3} - \frac{1}{6} ightbrack$
$= \frac{1}{2}\left\lbrack \frac{1}{6} ightbrack = \frac{1}{12}$

Q. The area (in square unit) of the triangle formed by $x + y + 1 = 0$ and the pair of straight lines $x^{2} - 3xy + 2y^{2} = 0$ is

Explanation

Reviewed by Dr. Abubacker

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