The triangle formed by joining the points $P(2, 7)$,$Q(4, -1)$,and $R(-2, 6)$ is:

Let $ABC$ be a triangle and $M$ be a point on side $AC$ closer to vertex $C$ than $A$. Let $N$ be a point on side $AB$ such that $MN$ is parallel to $BC$ and let $P$ be a point on side $BC$ such that $MP$ is parallel to $AB$. If the area of the quadrilateral $BNMP$ is equal to $\frac{5}{18}$ of the area of $\triangle ABC$,then the ratio $AM/MC$ equals

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, \sqrt{3})$ are the vertices of a triangle,then this triangle is:

The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

For all $\alpha, \beta \in R$ and $\alpha \beta > 0$,the line $\alpha x + \beta y + \sqrt{\alpha \beta} = 0$ is such that it

Let $A(-2,-1)$,$B(1,0)$,$C(\alpha, \beta)$,and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to . . . . . . .