One of the vertices of a square is the origin,and the adjacent sides of the square lie along the positive $x$ and $y$ axes. If the side length is $5$,which of the following is $NOT$ a vertex of the square?

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

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:

Without using the Pythagoras theorem,show that the points $(4,4), (3,5),$ and $(-1,-1)$ are vertices of a right-angled triangle.

The sides of a triangle are $3x + 4y$,$4x + 3y$,and $5x + 5y$ units,where $x, y > 0$. The triangle 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: