UPSC NDA 2 2021 Mathematics Solved Paper

Concept:For an equilateral triangle, all sides are equal. The distance formula is used to find side lengths.Explanation:Let vertices be $A(0,0)$, $B(2,2)$, and $C(x,y)$.Calculate side $AB = \sqrt{(2-0)^2 + (2-0)^2} = 2\sqrt{2}$.Since $triangle ABC$ is equilateral, $AC = BC = AB = 2\sqrt{2}$.From $AC = BC$: $\sqrt{x^2 + y^2} = \sqrt{(x-2)^2 + (y-2)^2}$.Square both sides: $x^2 + y^2 = x^2 -4x +4 + y^2 -4y +4$, simplifying to $0 = 8 -4x -4y$, so $y = 2 - x$.Now use $AC = 2\sqrt{2}$: $x^2 + y^2 = 8$. Substitute $y$: $x^2 + (2-x)^2 = 8$.Expand: $x^2 + 4 -4x + x^2 = 8$, giving $2x^2 -4x -4 = 0$, or $x^2 -2x -2 = 0$.Solve the quadratic: $x = 1 \pm \sqrt{3}$. Choose $x = 1 + \sqrt{3}$, then $y = 2 - (1+\sqrt{3}) = 1 - \sqrt{3}$.The difference of coordinates of the third vertex is $x - y = (1+\sqrt{3}) - (1-\sqrt{3}) = 2\sqrt{3}$.Answer:$2\sqrt{3}$