The points $A\, (5,-5)$ and $B\, (-1,-1)$ are the endpoints of the hypotenuse of an isosceles right triangle $\triangle ABC$. What is the area of $ABC$?
To find the area of $\triangle ABC$, I need to know the length of one of the congruent legs. Since it is an isosceles right triangle, the legs are perpendicular to each other, and I can use the Pythagorean theorem to relate them to the hypotenuse. I can also use the distance formula to find the length of the hypotenuse, since it is the same as the distance between $A$ and $B$. The distance formula says that $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$. Plugging in the given coordinates, I get $AB = \sqrt{(-1 - 5)^2 + (-1 - (-5))^2} = \sqrt{36 + 16} = \sqrt{52}$. Now, by the Pythagorean theorem, I have $AB^2 = AC^2 + BC^2$, where $AC$ and $BC$ are the lengths of the legs. Since the triangle is isosceles, I know that $AC = BC$, so I can write $AB^2 = 2AC^2$. Substituting $AB = \sqrt{52}$, I get $\sqrt{52}^2 = 2AC^2$, which simplifies to $52 = 2AC^2$. Dividing both sides by 2, I get $AC^2 = 26$. Taking the positive square root, I get $AC = \sqrt{26}$. This is also the length of $BC$, so I have $AC = BC = \sqrt{26}$. The area of $\triangle ABC$ is half the product of the base and the height, which are the same as the lengths of the legs. Therefore, the area is $\frac{1}{2} \cdot \sqrt{26} \cdot \sqrt{26} = \frac{1}{2} \cdot 26 = 13$. 