JEE Main 2023 — Mathematics Coordinate Geometry

On plotting the diagram of the given data we get,

Now given, $x=y$x=y and $y-2x=2$y−2x=2

On solving we get, $A(-2,-2)$A(−2,−2)

Now height $h$h of $\Delta ABC$ΔABC $=$= Distance of point $A$A to origin,

$$h=\sqrt{{(-2)}^{2}+(-2{)}^{2}}$$h=(−2)2+(−2)2​

$$\Rightarrow h=2\sqrt{2}$$⇒h=22​

Given $\Delta ABC$ΔABC is an equilateral triangle,

So, $\mathrm{sin}(B)=\frac{h}{AB}$sin(B)=ABh​

$$\Rightarrow \mathrm{sin}({60}^{^{\circ}})=\frac{h}{AB}(\because \angle A=\angle B=\angle C={60}^{^{\circ})}$$⇒sin(60∘)=ABh​(∵∠A=∠B=∠C=60∘)

Area of $\Delta =\frac{\sqrt{3}}{4}{(AB)}^{2}$Δ=43​​(AB)2

$\Rightarrow$⇒Area of $\Delta =\frac{\sqrt{3}}{4}\frac{{h}^{2}}{{\mathrm{sin}}^{2}{60}^{\circ }}=\frac{\sqrt{3}}{4}\times \frac{32}{3}$Δ=43​​sin260∘h2​=43​​×332​

So, Area of $\Delta ABC=\frac{8}{\sqrt{3}}$ΔABC=3​8​