$A(0,2)$ and $C(6,4)$ are opposite vertices of square $ABCD$. The sum of the slopes of the sides passing through vertex $A$ is:

Consider four triangles with side lengths $(5, 12, 9)$,$(5, 12, 11)$,$(5, 12, 13)$,and $(5, 12, 15)$. Among these,the triangle with the maximum area has sides:

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ 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

The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,is:

Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .... .