Two mutually perpendicular straight lines through the origin form an isosceles triangle with the line $2x + y = 5$. Then the area of the triangle is:

Let $A(1, 1)$,$B(-4, 3)$,and $C(-2, -5)$ be the vertices of a triangle $ABC$. Let $P$ be a point on the side $BC$,and let $\Delta_{1}$ and $\Delta_{2}$ be the areas of triangle $APB$ and triangle $ABC$,respectively. If $\Delta_{1} : \Delta_{2} = 4 : 7$,then find the area enclosed by the lines $AP$,$AC$,and the $x$-axis.

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:

In an isosceles $\triangle ABC$,the coordinates of vertices $B$ and $C$ of the base $BC$ are $(3, 2)$ and $(2, 3)$ respectively. If the equation of the line $AB$ is $3y = 2x$,then the equation of the line $AC$ is

The equation of the base $BC$ of an equilateral triangle is $3x + 4y = 1$ and the vertex $A$ is $(-3, 2)$. Then the area of the triangle is-

$A$ pair of straight lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. The coordinates of the centre of the circle inscribed in the square are