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:

Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

In a $\triangle ABC$,$2x+3y+1=0$ and $x+2y-2=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively. If $A=(3,2)$,then the equation of the side $BC$ is

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:

If $P(1, 2), Q(4, 6), R(5, 7)$ and $S(a, b)$ are the vertices of a parallelogram $PQRS$,then:

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: