Question

Prove that the diagonals of the parallelogram formed by the four straight lines
$\sqrt{3}x+y=0,\sqrt{3}y+x=0$, $\sqrt{3}x+y=1$, and $\sqrt{3}y+x=1$ are at right angles to one another.

Answer 1

$\sqrt{3}y+x=\mathrm{0........}\left(i\right)\sqrt{3}x+y=\mathrm{1........}\left(ii\right)\sqrt{3}y+x=\mathrm{1......}\left(iii\right)\sqrt{3}x+y=\mathrm{0........}\left(iv\right)$

Solving $\left(iv\right)$ and $\left(i\right)$

$\Rightarrow A(0,0)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow B(\frac{\sqrt{3}}{2},-\frac{1}{2})$

solving $\left(ii\right)$ and $\left(iii\right)$

$\Rightarrow C(\frac{\sqrt{3}-1}{2},\frac{\sqrt{3}-1}{2})$

solving $\left(iii\right)$ and $\left(iv\right)$

$\Rightarrow D(-\frac{1}{2},\frac{\sqrt{3}}{2})$

Slope of $AC$ $=m_{AC}=\frac{\frac{\sqrt{3}-1}{2}-0}{\frac{\sqrt{3}-1}{2}-0}=1$

Slope of $BD$ $=m_{BD}=\frac{\frac{\sqrt{3}}{2}-(-\frac{1}{2})}{-\frac{1}{2}-\frac{\sqrt{3}}{2}}=-1$

$m_{AC}\times m_{BD}=1\times -1=-1$

Hence the lines are perpendicular, so the diagonals of a parallelogram are at right angles.