We are asked to find the area of a triangle with vertices (4,9), (2,1), and (-1,-7) using the determinant formula.

1. Problem Description

We are asked to find the area of a triangle with vertices (4,9), (2,1), and (-1,-7) using the determinant formula.

2. Solution Steps

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the absolute value of:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Alternatively, we can write this as:

$Area = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

In our case, $(x_1, y_1) = (4, 9)$, $(x_2, y_2) = (2, 1)$, and $(x_3, y_3) = (-1, -7)$.

Using the determinant formula, we have:

$Area = \frac{1}{2} \left| \begin{vmatrix} 4 & 9 & 1 \\ 2 & 1 & 1 \\ -1 & -7 & 1 \end{vmatrix} \right|$

We can compute the determinant as follows:

$Area = \frac{1}{2} |4(1 - (-7)) - 9(2 - (-1)) + 1(2(-7) - 1(-1))|$

$Area = \frac{1}{2} |4(1+7) - 9(2+1) + 1(-14+1)|$

$Area = \frac{1}{2} |4(8) - 9(3) + (-13)|$

$Area = \frac{1}{2} |32 - 27 - 13|$

$Area = \frac{1}{2} |32 - 40|$

$Area = \frac{1}{2} |-8|$

$Area = \frac{1}{2} (8)$

$Area = 4$

3. Final Answer

The area of the triangle is 4 square units.