Let $$\Gamma$$ be the curve $$\mathrm{y}=\mathrm{be}^{-x / a}$$ & $$\mathrm{L}$$ be the straight line $$\frac{x}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$$ where $$\mathrm{a}, \mathrm{b} \in \mathbb{R}$$. Then
A square with each side equal to '$$a$$' above the $$x$$-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle $$\alpha$$ $$\left(0<\alpha< \frac{\pi}{4}\right)$$ with the positive direction of the axis. Equation of the diagonals of the square
If $$\mathrm{ABC}$$ is an isosceles triangle and the coordinates of the base points are $$B(1,3)$$ and $$C(-2,7)$$. The coordinates of $$A$$ can be
A rectangle ABCD has its side parallel to the line y = 2x and vertices A, B, D are on lines y = 1, x = 1 and x = $$-$$1 respectively. The coordinate of C can be