how do i find the area and desscribe the following figure?

Question

how do i find the area and desscribe the following figure
1,-1 3,2 -2,1

Answer 1

I assume these are ordered pairs
(1, -1)
(3, 2) and
(-2, 1)

That means it's a triangle.

A = (1, -1)
B = (3, 2)
C = (-2, 1)

Check to see if it's a right triangle.
slope of AB = (2 - -1) / (3 - 1) = 3/2
slope of AC = (1 - -1) / (-2 - 1) = -2/3

Since
(slope AB) * (slope AC) = -1, then we know those two line segments are perpendicular.
Which means the triangle is a right triangle with AB and AC as the legs (and BC as the hypotenuse.)

Area = (1/2)(AB)(AC)

Distance AB = sqrt 13
Distance AC = sqrt 13

So...
Area = (1/2)(sqrt 13)(sqrt 13) = 13/2 = 6.5

Answer 2

Three points and it is probably a triangle (unless the points are all on a straight line which they are not so it is a triangle).

If a right triangle then it is easy to find the area. It is just 1/2 the base times the height. If a side has slope m then a perpendicular side to it would have slope -1/m so let us check the slopes of the sides.

Let A = (1,-1), B = (3,2) and C = (-2,1)

Line AB. Slope (1,-1) - (3,2) = (2 - (-1))/(3 - 1) = 3/2
Line AC. Slope (1,-1) - (-2,1) = (1 - (-1))/(-2 - 1) = -2/3

Since 3/2 = -(1/(-2/3)) this is a right triangle

The area will then be (1/2)AB*AC (you can call either AB or AC the base and the height will be the other, it is not important which is which)

Length AB = SQRT(3^2 + 2^2) = SQRT(13)
Length AC = SQRT(2^2 + 3^2) = SQRT(13)

So area = (1/2)SQRT(13)*SQRT(13)
Area = 13/2 = 6.5

Answer 3

Well there are three points that do not make a line so by definition it would be desribed as a triangle. To find the area try drawing a line in somewhere so you can get right triangles and then just calculate 1/2 * B * H=A