If area of quadrilateral formed by lines $2 x^{2} + 3 x y - 2 y^{2} = 0$ and $2 x^{2} + 3 x y - 2 y^{2} + 3 x + y + 1 = 0$ is $A$ sq. units, then the value of $20 A$ is

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Solution

Given, $2 x^{2} + 3 x y - 2 y^{2} = 0$
$\Rightarrow 2 x^{2} + 4 x y - x y - 2 y^{2} = 0$
$\Rightarrow (2 x - y) (x + 2 y) = 0$
$2 x^{2} + 3 x y - 2 y^{2} + 3 x + y + 1$
$= (2 x - y + λ_{1}) (x + 2 y + λ_{2})$
By comparing coefficients,
$3 = 2 λ_{2} + λ_{1}$
$1 = - λ_{2} + 2 λ_{1}$
Solving, we get
$λ_{1} = 1, λ_{2} = 1$

So, the lines representing quadrilateral are
$2 x - y = 0 \dots (1) x + 2 y = 0 \dots (2) 2 x - y + 1 = 0 \dots (3) x + 2 y + 1 = 0 \dots (4)$
$∵ (1) and (3), (2) and (4)$ are parallel.
Also, $(1), (2)$ are $⊥^{r}$ , so the quadrilateral can be either be a square or rectangle.

Distance between $(1) and (3) =$ Distance between $(2) and (4) = \frac{1}{\sqrt{5}}$ units.
So, it represent square, whose area
$A = \frac{1}{5} ∴ 20 A = 4 sq.units$

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