Hi all,
I am so lost with my algebra class; I am taking this online and would appreciate any help, suggestions, feedback.
The problem I am to do is this: Find an area of a triangle with vertices that have coordinates of (0,9), (0, -4), and (5, -4). Thanks if you can help.
Given the rectangular coordinates of the three vertices, the area of a triangle whose vertices are P1(x1,y1), P2(x2,y2), P3(x3,y3) is given by Area triangle P1P2P3 = 1/2[(x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)].
Method:
First step-Write down the vertices in two columns, abscissas in one, ordinates in the other, repeating the coordinates of the first vertex. They must be in order around the triangle, clockwise preferably.
Second step-Multiply each abscissa by the ordinate of the next row, and add the results. This gives x1y2+x2y3+x3y1.
Third step-Multiply each ordinate by the abscissa of the next row, and add the results. This gives y1x2+y2x3+y3x1.
Fourth step-Subtract the result of the third step from that of the second step, and divide by 2. This gives the required area. (If you list the vertices in clockwise order the result will be positive. If you list them is counterclockwise order the result will be negative.)
For example, consider the coordinates (0, 4), (4, 2), (-4, -6) from which our table looks like:
x1 y1 0 +4
x2 y2 +4 +2
x3 y3 -4 -6
x1 y2 0 +4
For the second step--0x3 + 4x(-6) + (-4)x4 = 0-24-16 = -40
For the third step------4x4 + 2x(-4) + (-6)x0 = 16-8+0 = 8
For the fourth step--[8-(-40)]/2 = 24 sq units. That was easy.
This method applies to any multiple sided figure that a set of vertices produces.
