UPSC NDA II 2018 Mathematics Solved Paper

Concept:Area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is half the absolute value of the determinant: $\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$.Explanation:Step 1: Substitute the given vertices into the area formula.$Area = \frac{1}{2} \begin{vmatrix} x_1 & \frac{1}{x_1} & 1 \\ x_2 & \frac{1}{x_2} & 1 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix}$.Step 2: Simplify using row operations: $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$.$Area = \frac{1}{2} \begin{vmatrix} x_1-x_2 & \frac{1}{x_1}-\frac{1}{x_2} & 0 \\ x_2-x_3 & \frac{1}{x_2}-\frac{1}{x_3} & 0 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix}$.Step 3: Simplify the fractions: $\frac{1}{x_1}-\frac{1}{x_2} = \frac{x_2-x_1}{x_1x_2} = -\frac{x_1-x_2}{x_1x_2}$, and $\frac{1}{x_2}-\frac{1}{x_3} = -\frac{x_2-x_3}{x_2x_3}$.$Area = \frac{1}{2} \begin{vmatrix} x_1-x_2 & -\frac{x_1-x_2}{x_1x_2} & 0 \\ x_2-x_3 & -\frac{x_2-x_3}{x_2x_3} & 0 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix}$.Step 4: Factor out common factors from rows: $(x_1-x_2)$ from row 1 and $(x_2-x_3)$ from row 2.$Area = \frac{1}{2} (x_1-x_2)(x_2-x_3) \begin{vmatrix} 1 & -\frac{1}{x_1x_2} & 0 \\ 1 & -\frac{1}{x_2x_3} & 0 \\ x_3 & \frac{1}{x_3} & 1 \end{vmatrix}$.Step 5: Expand the determinant along the third column (column 3 has entries 0,0,1). Only the 3×3 minor remains: $1 \times \left( 1 \cdot (-\frac{1}{x_2x_3}) - 1 \cdot (-\frac{1}{x_1x_2}) \right)$.$Area = \frac{1}{2} (x_1-x_2)(x_2-x_3) \times \left( -\frac{1}{x_2x_3} + \frac{1}{x_1x_2} \right)$.Step 6: Combine the fractions: $\frac{1}{x_1x_2} - \frac{1}{x_2x_3} = \frac{x_3 - x_1}{x_1x_2x_3}$.$Area = \frac{1}{2} (x_1-x_2)(x_2-x_3) \cdot \frac{x_3-x_1}{x_1x_2x_3}$.Step 7: Area is taken as absolute value (magnitude).Therefore, $Area = \left| \frac{(x_1-x_2)(x_2-x_3)(x_3-x_1)}{2 x_1 x_2 x_3} \right|$.Answer:$\left| \frac{(x_{1}-x_{2})(x_{2}-x_{3})(x_{3}-x_{1})}{2x_{1}x_{2}x_{3}} \right|$ (Option D).