If the line $3x + 4y - 24 = 0$ intersects $X$ and $Y$ axes at points $A$ and $B$ respectively,then the incenter of the triangle $OAB$,where $O$ is the origin,is:

$A$ triangle is formed by the $Y$-axis,the straight line $L$ passing through the points $(3,0)$ and $(1, 4/3)$,and the straight line perpendicular to the line $L$ and passing through the point $(8,1)$. The area of the triangle (in sq units) is

Let $a, b, c$ be the side-lengths of a triangle and $l, m, n$ be the lengths of its medians. Put $K = \frac{l+m+n}{a+b+c}$. Then,as $a, b, c$ vary,$K$ can assume every value in the interval

If the line $3x + 4y - 24 = 0$ intersects the $x-$axis at the point $A$ and the $y-$axis at the point $B$,then the incentre of the triangle $OAB$,where $O$ is the origin,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:

$A$ straight line $4x+y-1=0$ passing through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. Then the equation of the line $AC$ such that $AB=AC$,is