UPSC NDA II 2015 Mathematics Paper

Concept:The area of a parallelogram equals half the product of its diagonals, and the intercept form of a line $\frac{x}{a} + \frac{y}{b} = 1$ gives the x-intercept and y-intercept directly.Explanation:Rewrite each line in intercept form to find the intercepts on the axes.For $ax + by + c = 0$: $ax + by = -c$ implies $\frac{x}{-\frac{c}{a}} + \frac{y}{-\frac{c}{b}} = 1$, so intercepts are $\left(-\frac{c}{a}, 0\right)$ and $\left(0, -\frac{c}{b}\right)$.For $ax - by + c = 0$: $ax - by = -c$ implies $\frac{x}{-\frac{c}{a}} + \frac{y}{\frac{c}{b}} = 1$, so intercepts are $\left(-\frac{c}{a}, 0\right)$ and $\left(0, \frac{c}{b}\right)$.For $ax + by - c = 0$: $ax + by = c$ implies $\frac{x}{\frac{c}{a}} + \frac{y}{\frac{c}{b}} = 1$, so intercepts are $\left(\frac{c}{a}, 0\right)$ and $\left(0, \frac{c}{b}\right)$.For $ax - by - c = 0$: $ax - by = c$ implies $\frac{x}{\frac{c}{a}} + \frac{y}{-\frac{c}{b}} = 1$, so intercepts are $\left(\frac{c}{a}, 0\right)$ and $\left(0, -\frac{c}{b}\right)$.These four lines form a parallelogram with vertices at the intersection points of the intercepts.The diagonals of this parallelogram lie along the x-axis and y-axis.The length of the diagonal along the x-axis is the distance between $\left(-\frac{c}{a}, 0\right)$ and $\left(\frac{c}{a}, 0\right)$, which is $\frac{2c}{a}$.The length of the diagonal along the y-axis is the distance between $\left(0, -\frac{c}{b}\right)$ and $\left(0, \frac{c}{b}\right)$, which is $\frac{2c}{b}$.Area of parallelogram = $\frac{1}{2} \times \left(\frac{2c}{a}\right) \times \left(\frac{2c}{b}\right) = \frac{2c^2}{ab}$.The figure formed by the lines $ax + by + c = 0$, $ax - by + c = 0$, $ax + by - c = 0$, and $ax - by - c = 0$ is shown belowAnswer:$\frac{2c^2}{ab}$