The four points whose coordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form:

$A$ vertex of an equilateral triangle is $(2, 3)$ and the equation of the opposite side is $x + y = 2$. Then,the equation of one of the other two sides is:

When are the points with coordinates $(2a, 3a)$,$(3b, 2b)$,and $(c, c)$ collinear?

In a $\triangle ABC$,$D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE \parallel BC$. Suppose $BE$ and $CD$ intersect at $O$. If the areas of $\triangle ADE$ and $\triangle ODE$ are $3$ and $1$ respectively,find the area of $\triangle ABC$.

The area of the triangle formed by the lines $x = 0$,$y = 0$,and $x/a + y/b = 1$ is:

Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side $a$ such that $a^{2}+11 a+3(m_{1}^{2}+m_{2}^{2})=220$. If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$,where $\alpha \in(0, \frac{\pi}{2})$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x +(\sin \alpha+\cos \alpha) y =10$,then $72(\sin ^{4} \alpha+\cos ^{4} \alpha)+a^{2}-3 a+13$ is equal to.