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

Suppose the vertices of a triangle are given by $A(0,3)$,$B(-2,0)$,and $C(6,1)$. For the point $(\alpha, \alpha+1)$ to lie inside the triangle,$\alpha$ must lie in the interval:

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

The equations of the sides $AB$,$BC$,and $CA$ of a triangle $ABC$ are $2x + y = 0$,$x + py = 21a$ $(a \neq 0)$,and $x - y = 3$ respectively. Let $P(2, a)$ be the centroid of $\triangle ABC$. Then $(BC)^2$ is equal to $........$

The area of the parallelogram formed by the lines $L_1 \equiv \lambda x+4 y+2=0$,$L_2 \equiv 3 x+4 y-3=0$,$L_3 \equiv 2 x+\mu y+6=0$,and $L_4 \equiv 2 x+y+3=0$,where $L_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$,is

If the equation of the base of an equilateral triangle is $x+y=6$ and the opposite vertex is the point $(-1, -1)$,then the area of the triangle is equal to $k$ sq. units. Find the value of $k$.