Area of the rhombus bounded by the four lines | vedclass

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(B) The four lines are $ax + by + c = 0$,$ax + by - c = 0$,$ax - by + c = 0$,and $ax - by - c = 0$.The vertices of the rhombus are found by solving th

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$\frac{2c^2}{ab}$

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(B) The four lines are $ax + by + c = 0$,$ax + by - c = 0$,$ax - by + c = 0$,and $ax - by - c = 0$.The vertices of the rhombus are found by solving the intersection of these lines:For $y=0$,$ax = \pm c \implies x = \pm \frac{c}{a}$. So,vertices are $(\frac{c}{a}, 0)$ and $(-\frac{c}{a}, 0)$.For $x=0$,$by = \pm c \implies y = \pm \frac{c}{b}$. So,vertices are $(0, \frac{c}{b})$ and $(0, -\frac{c}{b})$.The lengths of the diagonals $d_1$ and $d_2$ are:$d_1 = \frac{c}{a} - (-\frac{c}{a}) = \frac{2c}{a}$$d_2 = \frac{c}{b} - (-\frac{c}{b}) = \frac{2c}{b}$The area of the rhombus is given by $\frac{1}{2} 	imes d_1 	imes d_2$:$	ext{Area} = \frac{1}{2} 	imes \frac{2c}{a} 	imes \frac{2c}{b} = \frac{2c^2}{ab}$.

Area of the rhombus bounded by the four lines,$ax \pm by \pm c = 0$ is :

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