let the triangle be ABC with A,B, and C matching the order of coordinate points the way they are written above.
Step 1: find equation for BC
slope = 5/9
equation is 5x-9y = 21
length of BC = √(9^2 + 5^2) = √106, I will use that as the base of the triangle.
distance for A to line BC (my height of triangle)
=│5(2) - 9(-5) - 21│/√106
= 34/√106
area of triangle = 1/2 base*height
= 1/2(√106)(34/√106) = 17
Find the area of a triangle whose vertices have coordinates (2,-5), (6,1), and (-3,-4)
2 answers
In addition to several methods for calculating the area of a triangle given the coordinates of its vertices, the area can be determined using only the coodinates themselves.
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.
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.
3 answers