Question

Show that the four lines $ax\pm by\pm c=0$ enclose a rhombus whose area is $\frac{2c^2}{ab}$.

Answer 1

The four sides of the rhombus are
$a+xby+c=0$
$ax+by-c=0$
$ax-by+c=0$
$ax-by-c=0$
Solving $\left(1\right)$ with $\left(3\right)$ and $\left(2\right)$ with $\left(4\right)$, we get the vertices $(c/a,0),(c/a,0)$.
Solving $\left(2\right)$ with $\left(3\right)$ and $\left(1\right)$ with $\left(4\right)$,we get the vertices as $(0,c/b)(0,-c/b)$
Above shown that one diagonal is of length $2c/a$ and is along $x-$axis whereas the other is along $y-$axis and is length $2c/b$. Since the diagonals being along the axes, they are perpendicular and hence a rhombus.
The area is $\frac{1}{2}(d_1.d_2)=\frac{1}{2}(\frac{2c}{a}.\frac{2c}{b})=\frac{2c^2}{ab}$
Note : You may do it as in Q.25.
Area$=\frac{p_1{p}_2}{\sin \theta }=\left|\frac{(c_1-d_1)(c_2-d_2)}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\right|$
$=\left|\frac{\left(2c\right)\left(2c\right)}{\left|\begin{array}{cc}a& b\\ a& -b\end{array}\right|}\right|=\frac{4c^2}{2ab}=\frac{2c^2}{ab}$