If one side of an isosceles triangle is given by the line $y=2$ and the base is provided by the points $(2,0)$ and $(0,2)$,then its area (in sq. units) is

The mid-points of the sides of a triangle are $(2, 1)$,$(-1, -3)$,and $(4, 5)$. The coordinates of its vertices are:

The perimeter of a triangle is $16 \text{ cm}$,one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$,then the triangle is:

The straight lines $x+y=0$,$5x+y=4$,and $x+5y=4$ form

Let the line $x+y=1$ meet the axes of $x$ and $y$ at $A$ and $B$,respectively. $A$ right-angled triangle $AMN$ is inscribed in the triangle $OAB$,where $O$ is the origin and the points $M$ and $N$ lie on the lines $OB$ and $AB$,respectively. If the area of the triangle $AMN$ is $\frac{4}{9}$ of the area of the triangle $OAB$ and $AN : NB = \lambda : 1$,then the sum of all possible values of $\lambda$ is:

The diagonals of the parallelogram whose sides are $lx + my + n = 0$,$lx + my + n' = 0$,$mx + ly + n = 0$,and $mx + ly + n' = 0$ include an angle of: