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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the determinant formula:$\Delta = \frac{1}{2} \left| \begin{

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$\frac{15}{2}$ square units

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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the determinant formula:$\Delta = \frac{1}{2} \left| \begin{array}{lll} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|$Substituting the given vertices $(1,0), (6,0), (4,3)$:$\Delta = \frac{1}{2} \left| \begin{array}{lll} 1 & 0 & 1 \ 6 & 0 & 1 \ 4 & 3 & 1 \end{array} ight|$Expanding along the second column (which contains two zeros):$\Delta = \frac{1}{2} \left| -0 \left| \begin{array}{ll} 6 & 1 \ 4 & 1 \end{array} ight| + 0 \left| \begin{array}{ll} 1 & 1 \ 4 & 1 \end{array} ight| - 3 \left| \begin{array}{ll} 1 & 1 \ 6 & 1 \end{array} ight| ight|$$\Delta = \frac{1}{2} | -3(1 - 6) |$$\Delta = \frac{1}{2} | -3(-5) | = \frac{1}{2} | 15 | = \frac{15}{2}$ square units.

Find the area of the triangle with vertices at the points $(1,0), (6,0), (4,3)$.

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