In a triangle $ABC$,if $BC=5, CA=6, AB=7$,then the length of the median drawn from $B$ onto $AC$ 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 incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ is

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

The vertex of the right angle of a right-angled triangle lies on the straight line $2x + y - 10 = 0$. If the two other vertices are at points $(2, -3)$ and $(4, 1)$,then the area of the triangle in sq. units is:

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are