The two vertices of a triangle are $(2, -1)$ and $(3, 2)$,and the third vertex lies on the line $x + y = 5$. If the area of the triangle is $4$ square units,then the third vertex is:

Suppose the hypotenuse and its opposite vertex of an isosceles right-angled triangle are $3x + 4y - 4 = 0$ and $(2, 2)$ respectively. Then,which of the following is another side of the triangle?

If the vertices of a triangle are $(0, 4)$,$(4, 1)$,and $(7, 5)$,find its perimeter.

If the points $(a, b)$,$(a', b')$ and $(a - a', b - b')$ are collinear,then:

Let $\left(5, \frac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2\right)$. Let $\alpha$ denote the circumradius,$\beta$ denote the area,and $\gamma$ denote the perimeter of the triangle. Then $\alpha + \beta + \gamma$ is

All points lying inside the triangle formed by the points $(1, 3)$,$(5, 0)$,and $(-1, 2)$ satisfy which of the following inequalities?