The orthocentre of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ is

Let the area of the triangle with vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$ be $4 \text{ sq. units}$. If the points $(\alpha, -\alpha)$,$(-\alpha, \alpha)$,and $(\alpha^2, \beta)$ are collinear,then $\beta$ is equal to:

The incentre of the triangle formed by the lines $x = 0$,$y = 0$,and $3x + 4y = 12$ is

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

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis is:

The sides of a quadrilateral are all positive integers and three of them are $5, 10, 20$. How many possible values are there for the fourth side?