The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is

In a $\triangle ABC$,$D$ and $E$ are points on $AB$ and $AC$ respectively such that $DE \parallel BC$. Suppose $BE$ and $CD$ intersect at $O$. If the areas of $\triangle ADE$ and $\triangle ODE$ are $3$ and $1$ respectively,find the area of $\triangle ABC$.

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:

Three vertices of a triangle are $A(4, 3)$,$B(1, -1)$,and $C(7, k)$. The value$(s)$ of $k$ for which the centroid,orthocentre,incentre,and circumcentre of the $\Delta ABC$ lie on the same straight line is/are:

If the points of intersection of the coordinate axes and $|x+y|=2$ form a rhombus,then its area is

The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,is: