Question

Find a formula for the area of the triangle whose vertices are (0,0),(a₁₁,a₂₁), and (a₁₂,a₂₂). Hint: Think about how the area of the triangle is related to the area of the parallelogram with vertices (0,0),(a₁₁,a₂₁),(a₁₁+a₁₂,a 21+a₂₂), and (a₁₂,a₂₂) Put your answer here

Answer 1

The area of a triangle with given vertices can be found using the determinant of a matrix, which is half the absolute value of the determinant of a 2x2 matrix formed by the coordinates of the vertices excluding the origin.

Step-by-step explanation:

To find the area of a triangle with vertices at points (0,0),(a₁₁,a₂₁), and (a₁₂,a₂₂), we can use the determinant of a matrix. The area of the parallelogram formed by these points is base times height, and since the triangle is half of the parallelogram, we halve the area of the parallelogram to find the area of the triangle.

Consider the matrix:

| 1 1 1 |
| 0 a₁₁ a₁₂ |
| 0 a₂₁ a₂₂ |

The area of the parallelogram is the absolute value of the determinant of this matrix:

| a₁₁ a₁₂ |
| a₂₁ a₂₂ |

So, the area of the triangle is half of this value:

Area of the Triangle = ½ |a₁₁a₂₂ - a₁₂a₂₁|