Question

Prove that the diagonals of a parallelogram whose sides are
$\frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1,\frac{x}{a}+\frac{y}{b}=2$ and $\frac{x}{b}+\frac{y}{a}=2$ are at right angle.

Answer 1

$\frac{x}{a}+\frac{y}{b}=\mathrm{1.......}\left(i\right)\frac{x}{b}+\frac{y}{a}=\mathrm{1......}\left(ii\right)\frac{x}{a}+\frac{y}{b}=\mathrm{2......}\left(iii\right)\frac{x}{b}+\frac{y}{a}=\mathrm{2.......}\left(iv\right)$

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

$\Rightarrow A(\frac{ab(2a-b)}{a^2-b^2},\frac{ab(a-2b)}{a^2-b^2})$

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

$\Rightarrow B(\frac{ab}{a+b},\frac{ab}{a+b})$

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

$\Rightarrow C(\frac{ab(a-2b)}{a^2-b^2},\frac{ab(2a-b)}{a^2-b^2})$

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

$\Rightarrow D(\frac{2ab}{a+b},\frac{2ab}{a+b})$

Slope of $AC=m_1$

$m_1=\frac{\{\frac{ab(2a-b)}{a^2-b^2}-\frac{ab(a-2b)}{a^2-b^2}\}}{\{\frac{ab(a-2b)}{a^2-b^2}-\frac{ab(2a-b)}{a^2-b^2}\}}{m}_1=\frac{ab(2a-b-a+2b)}{ab(a-2b-2a+b)}=-1$

Slope of $BD$

$m_2=\frac{\{\frac{2ab}{a+b}-\frac{ab}{a+b}\}}{\{\frac{2ab}{a+b}-\frac{ab}{a+b}\}}{m}_2=1m_1{m}_2=(-1)\left(1\right)m_1{m}_2=-1$

Product o slope is $-1$ so $AC$ and $BD$ are perpendicular.

Hence, the diagonals are at right angle.