If the coordinates of points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively,and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is:

The points $(0, 8/3), (1, 3)$ and $(82, 30)$ are:

The points $(1,3)$ and $(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie on the line $y=2x+c$,then one of the vertices on the other diagonal is

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

Let $B$ and $C$ be two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on the line $y-2x=2$ such that $\triangle ABC$ is an equilateral triangle. Then,the area of the $\triangle ABC$ is