Question

Show that the lines $x^2-4xy+y^2=0$ and $x+y=\sqrt{6}$ form an equilateral triangle. Find its area of perimeter.

Answer 1

Let equilateral triangle be $ABC$

$x^2-4xy+y^2=0$ .........$\left(1\right)$

Above equation represents pair of straight lines passing through origin, hence $A(0,0)$ be one of the vertices.

$x+y=\sqrt{6}$

$\Rightarrow x=\sqrt{6}-y$

Put the value of $x$ in $\left(1\right)$ we get

${(\sqrt{6}-y)}^2-4(\sqrt{6}-y)y+y^2=0$

$\Rightarrow 6+y^2-2y\sqrt{6}-4y\sqrt{6}+4y^2+y^2=0$

$\Rightarrow 6y^2-6y\sqrt{6}+6=0$

$\Rightarrow y^2-y\sqrt{6}+1=0$

$\Rightarrow y=\frac{\sqrt{6}\pm \sqrt{6-4\times 1\times 1}}{2}$

$=\frac{\sqrt{6}\pm \sqrt{2}}{2}=\frac{\sqrt{2}(\sqrt{3}\pm 1)}{2}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2(\sqrt{3}\pm 1)}{2\sqrt{2}}=\frac{\sqrt{3}\pm 1}{\sqrt{2}}$

When $y=\frac{\sqrt{3}+1}{\sqrt{2}}$ we have

$x=\sqrt{6}-y=\sqrt{6}-\frac{\sqrt{3}+1}{\sqrt{2}}$

$=\frac{\sqrt{12}-\sqrt{3}-1}{\sqrt{2}}$

$=\frac{2\sqrt{3}-\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}$

When $y=\frac{\sqrt{3}-1}{\sqrt{2}}$ we have

$x=\sqrt{6}-y=\sqrt{6}-\frac{\sqrt{3}-1}{\sqrt{2}}$

$=\frac{\sqrt{12}-\sqrt{3}+1}{\sqrt{2}}$

$=\frac{2\sqrt{3}-\sqrt{3}+1}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}$

Hence coordinates of $B$ will be $B(\frac{\sqrt{3}-1}{\sqrt{2}},\frac{\sqrt{3}+1}{\sqrt{2}})$

and that of $C$ will be $C(\frac{\sqrt{3}+1}{\sqrt{2}},\frac{\sqrt{3}-1}{\sqrt{2}})$

Side of equilateral triangle$=AB=\sqrt{{(\frac{\sqrt{3}-1}{\sqrt{2}}-0)}^2+{(\frac{\sqrt{3}+1}{\sqrt{2}}-0)}^2}$

$=\sqrt{\frac{3+1-2\sqrt{3}}{2}+\frac{3+1+2\sqrt{3}}{2}}$

$=\sqrt{\frac{8}{2}}=\sqrt{4}=2$

Side of equilateral triangle$=AC=\sqrt{{(\frac{\sqrt{3}+1}{\sqrt{2}}-0)}^2+{(\frac{\sqrt{3}-1}{\sqrt{2}}-0)}^2}$

$=\sqrt{\frac{3+1+2\sqrt{3}}{2}+\frac{3+1-2\sqrt{3}}{2}}$

$=\sqrt{\frac{8}{2}}=\sqrt{4}=2$

Side of equilateral triangle$=BC=\sqrt{{(\frac{\sqrt{3}+1}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}})}^2+{(\frac{\sqrt{3}-1}{\sqrt{2}}-\frac{\sqrt{3}+1}{\sqrt{2}})}^2}$

$=\sqrt{{\left(\frac{2}{\sqrt{2}}\right)}^2+{\left(\frac{-2}{\sqrt{2}}\right)}^2}$

$=\sqrt{\frac{8}{2}}=\sqrt{4}=2$

We can see that,

$AB=AC=BC⟹$ given lines forms an equilateral triangle.

Area of an equilateral triangle$=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times 2^2=\sqrt{3}$sq.units

Perimeter of the equilateral triangle$=3\times side=3\times 2=6$ units.