(2,-9)(3,2) and (3,-9) find the area of the triangle
Several ways:
1. Heron's Formula,
google it and apply
2.
- find the equation of one of the sides and write it in the form Ax + By + C = 0.
- find the length of the line used to find the equation.
- Use the method of finding the distance from a point to a line.
- You now have the base and the height of the triangle, easy from here.
3. Easiest way is to use the following algorithm
--- list the points in a column, repeat the first one you listed
2 -9
3 2
3 -9
2 -9
area = (1/2)|sum of downproducts - sum of uproducts|
= (1/2)|4 - 27- 27 -(-27 + 6 - 18)|
= (1/2)|-11|
= 11/2
= 5.5
Try plotting the points.
Note that (3,2) and (3,-9) form a vertical line of length 11
The point (2,-9) is one unit away, so you have a triangle with base 11 and height 1.
Now, is that so hard?
To find the area of a triangle with three given vertices, you can use the formula for the area of a triangle:
Area = 0.5 * |(x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2))|
Given the three vertices: (2, -9), (3, 2), and (3, -9), we can plug in the values into the formula to find the area.
Let's break down the formula step by step:
1. Calculate the differences between the y-coordinates:
y2 - y3 = 2 - (-9) = 11
y3 - y1 = -9 - (-9) = 0
y1 - y2 = -9 - 2 = -11
2. Calculate the differences between the x-coordinates, multiplied by the corresponding y-coordinate differences:
x1 * (y2 - y3) = 2 * 11 = 22
x2 * (y3 - y1) = 3 * 0 = 0
x3 * (y1 - y2) = 3 * (-11) = -33
3. Sum up the results from step 2:
(x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2)) = 22 + 0 + (-33) = -11
4. Take the absolute value of the sum:
|(x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2))| = |-11| = 11
5. Multiply the absolute value by 0.5:
0.5 * |(x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2))| = 0.5 * 11 = 5.5
Therefore, the area of the triangle with the given vertices is 5.5 square units.