The sides of a triangle are $3x + 4y$,$4x + 3y$,and $5x + 5y$ units,where $x, y > 0$. The triangle is:

$A$ 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

If two opposite vertices of a square are $(5, -4)$ and $(-3, 2)$,find its area.

The area of the parallelogram formed by the lines $y = mx, y = mx + 1, y = nx, y = nx + 1$ is:

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a^2$ is equal to

If the points $(2k, k), (k, 2k)$ and $(k, k)$ with $k > 0$ enclose a triangle of area $18$ square units,then the centroid of the triangle is equal to