Show that the points A(5,-6), B(-3,0) and C(-1,2) form an isosceles triangle. 2) Show that the points A(-1,2), B(3,4) and C(2,-4) form a right angled triangle and find its area.

$$AB = \sqrt{(-3 - 5)^2 + (0 - (-6))^2}$$ $$AB = \sqrt{(-8)^2 + (6)^2}$$ $$AB = \sqrt{64 + 36}$$ $$AB = \sqrt{100}$$ $$AB = 10$$

$$BC = \sqrt{(-1 - (-3))^2 + (2 - 0)^2}$$ $$BC = \sqrt{(2)^2 + (2)^2}$$ $$BC = \sqrt{4 + 4}$$ $$BC = \sqrt{8}$$ $$BC = 2\sqrt{2}$$

$$AC = \sqrt{(-1 - 5)^2 + (2 - (-6))^2}$$ $$AC = \sqrt{(-6)^2 + (8)^2}$$ $$AC = \sqrt{36 + 64}$$ $$AC = \sqrt{100}$$ $$AC = 10$$

Since $AB = 10$ and $AC = 10$, two sides of the triangle (AB and AC) have equal lengths.
Therefore, the points A, B, and C form an isosceles triangle.

: Show that the points A(-1, 2), B(3, 4), and C(2, -4) form a right-angled triangle and find its area.
To show it's a right-angled triangle, we will use the Pythagorean theorem by calculating the square of the lengths of all sides. The area will be calculated using the formula for a right-angled triangle.

$$AB^2 = (3 - (-1))^2 + (4 - 2)^2$$ $$AB^2 = (4)^2 + (2)^2$$ $$AB^2 = 16 + 4$$ $$AB^2 = 20$$

$$BC^2 = (2 - 3)^2 + (-4 - 4)^2$$ $$BC^2 = (-1)^2 + (-8)^2$$ $$BC^2 = 1 + 64$$ $$BC^2 = 65$$

$$AC^2 = (2 - (-1))^2 + (-4 - 2)^2$$ $$AC^2 = (3)^2 + (-6)^2$$ $$AC^2 = 9 + 36$$ $$AC^2 = 45$$

The longest side is BC, so we check if $AB^2 + AC^2 = BC^2$.

$$20 + 45 = 65$$ $$65 = 65$$

Since $AB^2 + AC^2 = BC^2$, the triangle ABC is a right-angled triangle with the right angle at A.

For a right-angled triangle, the area is $\frac{1}{2} \times \text{base} \times \text{height}$. The base and height are the two sides forming the right angle, which are AB and AC.

$$Area = \frac{1}{2} \times AB \times AC$$ $$Area = \frac{1}{2} \times \sqrt{20} \times \sqrt{45}$$ $$Area = \frac{1}{2} \times \sqrt{20 \times 45}$$ $$Area = \frac{1}{2} \times \sqrt{900}$$ $$Area = \frac{1}{2} \times 30$$ $$Area = 15$$

The area of the triangle is $\boxed{\text{15 square units}}$.