Question

The angle between the diagonals of a quadrilateral formed by the lines

$\frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1,\frac{x}{b}+\frac{y}{a}=2,\frac{x}{b}+\frac{y}{a}=2$ is

Answer 1

Let $L_1:\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow$ $bx+ay=ab$

$L_2:\frac{x}{a}+\frac{y}{b}=2$

$\Rightarrow$ $bx+ay=2ab$

$L_3:\frac{x}{b}+\frac{y}{a}=1$

$ax+by=ab$

$L_4:\frac{x}{b}+\frac{y}{a}=2$

$\Rightarrow$ $ax+by=2ab$

From $L_1$ and$L_3$ we get

$bx+ay=ab....(\times a)$

$ax+by=ab.....(\times b)$

$abx+a^{2}y=a^{2}b$

$abx+b^{2}y=a{b}^2$

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$y=\frac{ab(a-b)}{a^2-b^2}=\frac{ab}{a+b}$

$bx+a(\frac{ab}{a}+b)=ab$

$x=a-\left(\frac{a^2}{a+b}\right)=\frac{a^2+ab-a^2}{a+b}=\frac{ab}{a+b}$

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

Similarly by $L_2$ and $L_4$

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

For point $Q$, $L_2,L_3$ we get following the same steps

$x=\frac{ab(a-2b)}{a^2-b^2}$

For $S$, by $L_1,L_4$, $x=\frac{ab(2a-b)}{a^2-b^2}$

MPR$=\frac{\frac{2ab}{a+b}-\frac{ab}{a+b}}{\frac{2ab}{a+b}-\frac{ab}{a+b}}=1$

MQS$=\frac{\frac{ab(a-2b)}{a^2-b^2}-\frac{ab(a-2b)}{a^2-b^2}}{\frac{ab(2a-b)}{a^2-b^2}-\frac{ab(2a-b)}{a^2-b^2}}=\frac{-a^{2}b-a{b}^2}{a^{2}b-a{b}^2}=-1$

$MPR\times MQS=1\times (-1)=-1$

Diagonals are perpendicular, so angle is $\frac{\pi }{2}\left({90}^o\right)$