Question

Prove that the lines $\sqrt{3}x+y=0,\sqrt{3}y+x=0,\sqrt{3}x+y=1$ and $\sqrt{3}y+x=1$ form a rhombus.

Answer 1

The given lines are as follows:

$\sqrt{3}x+y=0...\left(i\right)$

$\sqrt{3}y+x=0...\left(ii\right)$

$\sqrt{3}x+y=1...\left(iii\right)$

$\sqrt{3}y+x=1...\left(iv\right)$

In quadrilateral ABCD, let equations (i), (ii), (iii) and (iv) represent the sides AB, BC, CD and DA, respectively.

Lines (i) and (iii) and parallel and lines (ii) and (iv) are parallel.

Solving (i) and (ii):

$x=0,y=0$

Thus, AB and BC intersect at $B(0,0)$

Solving (i) and (iv):

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

Thus, AB and DA intersect at $A(-\frac{1}{2},\frac{\sqrt{3}}{2})$

Solving (iii) and (ii):

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

Thus, BC and CD intersect at $C(\frac{\sqrt{3}}{2},-\frac{1}{2})$

Solving (iii) and (iv):

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

Thus, DA and CD intersect at $D(\frac{\sqrt{3}-1}{2},\frac{\sqrt{3}-1}{\sqrt{3}})$

Let us find the lengths of sides AB, BC and CD and DA.

$AB=\sqrt{{(0-\frac{1}{2})}^2={(0-\frac{\sqrt{3}}{2})}^2}=1$

$BC=\sqrt{{(\frac{\sqrt{3}}{2}-0)}^2+{(-\frac{1}{2}-0)}^2}=1$

$CD=\sqrt{{(\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}}{2})}^2={(\frac{\sqrt{3}-1}{2}+\frac{1}{2})}^2}=1$

$DA=\sqrt{{(\frac{\sqrt{3}-1}{2}+\frac{1}{2})}^2={(\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}}{2})}^2}=1$

Hence, the given lines form a rhombus.