A triangle has vertices at points A(2, 3), B(7, 8), and C(8, 2). Calculate the area of this triangle using the determinant method

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In this problem, we need to calculate the area of a triangle with vertices at points A(2, 3), B(7, 8), and C(8, 2). We'll use the determinant method, which is a powerful technique for finding the area of a polygon given its vertices. The determinant formula for calculating the area of a triangle is shown here. We can express it in terms of the coordinates of the three vertices. The simplified formula is one-half times the absolute value of this expression. Now, let's substitute the coordinates of our vertices into the formula. A is at (2, 3), B is at (7, 8), and C is at (8, 2). We get 2 times (8 minus 2), plus 7 times (2 minus 3), plus 8 times (3 minus 8). Simplifying, we get 2 times 6, plus 7 times negative 1, plus 8 times negative 5. This equals 12 minus 7 minus 40, which is negative 35. Taking the absolute value and multiplying by one-half, we get 17.5 square units as the area of the triangle. Therefore, the area of the triangle with vertices at A(2, 3), B(7, 8), and C(8, 2) is 17.5 square units. Let's explore an alternative approach to calculating the area using the determinant method. We can express the area as one-half times the absolute value of a 3 by 3 determinant. This determinant contains the x and y coordinates of the three vertices, with a column of ones. To evaluate this determinant, we can expand it using the cofactor method. We multiply each element in the first row by its cofactor, alternating signs. This gives us a more detailed expression that we can further simplify. Now, let's substitute the coordinates of our vertices into the determinant. For A(2, 3), B(7, 8), and C(8, 2), we get this 3 by 3 matrix. Expanding the determinant, we multiply 2 by the determinant of the submatrix, minus 3 times its cofactor, plus 1 times the last cofactor. This gives us 2 times 6, minus 3 times negative 1, plus 14 minus 64. Simplifying, we get 12 plus 3 minus 50, which equals negative 35. Taking the absolute value of negative 35 and multiplying by one-half, we get 17.5 square units as the area of the triangle. This confirms our previous calculation using the simplified determinant formula. Let's explore the geometric interpretation of the determinant method for calculating triangle area. This approach can be understood in terms of vectors and the cross product in three-dimensional space. If we consider the vectors AB and AC, starting from vertex A and going to vertices B and C respectively, we can use these vectors to calculate the area. The magnitude of the cross product of these two vectors, divided by two, gives us the area of the triangle. The cross product can be calculated using a determinant. Since we're working in a two-dimensional plane, the z-components of our vectors are zero, which simplifies the calculation. The resulting cross product is perpendicular to the plane, pointing in the k-direction, with magnitude equal to the determinant of the 2 by 2 matrix formed by the x and y components of our vectors. The magnitude of the cross product equals the area of the parallelogram formed by the two vectors. Since the area of the triangle is half the area of this parallelogram, we divide by two to get our final formula. This is equivalent to the determinant method we've been using. This vector-based approach is mathematically equivalent to the determinant formulas we used earlier. It's just a different way of expressing the same calculation, highlighting the geometric meaning behind the determinant method for finding triangle areas. Let's make our exploration more interactive by seeing how the area changes when we move one of the vertices. The determinant formula automatically adjusts to calculate the new area as the coordinates change. Remember the formula we've been using. As we move point C to different positions, the area of the triangle changes. Let's see what happens when we move C to a few different locations. Let's look at a special case. When the three vertices are collinear, meaning they lie on the same straight line, the area of the triangle becomes zero. This makes sense geometrically, as a triangle with all vertices on a line has no area. Now, let's return to our original triangle with vertices at A(2, 3), B(7, 8), and C(8, 2), which has an area of 17.5 square units. This confirms our earlier calculations using different methods. To summarize what we've learned: The determinant method provides an elegant way to calculate the area of a triangle given its vertices. For our specific triangle with vertices at A(2, 3), B(7, 8), and C(8, 2), we found the area to be 17.5 square units. The determinant formula automatically handles any triangle configuration and updates as vertices move. This method has a geometric interpretation as half the magnitude of the cross product of two vectors. And as a special case, when three points are collinear, the area becomes zero. The determinant approach is powerful because it works for any triangle in the coordinate plane without requiring additional measurements.