find the area of the triangle with the given vertices: A(12,4), B(1,1), C(0,8)

To find the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), we can use the formula:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substitute the given coordinates of the vertices:
- \( A(12,4) \)
- \( B(1,1) \)
- \( C(0,8) \)

The formula becomes:

\[
\text{Area} = \frac{1}{2} \left| 12(1 - 8) + 1(8 - 4) + 0(4 - 1) \right|
\]

Now, calculate each part:

\[
= \frac{1}{2} \left| 12(-7) + 1(4) + 0 \right|
\]
\[
= \frac{1}{2} \left| -84 + 4 \right|
\]
\[
= \frac{1}{2} \left| -80 \right|
\]
\[
= \frac{1}{2} \times 80 = 40
\]

Thus, the area of the triangle is \( \boxed{40} \) square units.

Would you like further clarification or more details on this? Here are some related questions:

1. How do you find the area of a triangle using the distance formula between points?
2. How does the shoelace (or determinant) formula relate to the area of polygons?
3. Can the area of a triangle be negative? If so, what does it represent?
4. How can this method be adapted to find the area of a quadrilateral?
5. What happens if two of the points are collinear?

**Tip**: When working with triangles, ensure the vertices are not collinear by checking the area. If it's zero, the points are on the same line!

Find the area of the triangle with the given vertices: A(12,4), B(1,1), C(0,8)

This problem involves finding the area of a triangle with vertices A(12,4), B(1,1), and C(0,8) using the coordinate geometry formula. It explains step-by-step how to apply the determinant method (or Shoelace Theorem). Suitable for students in Grades 9-10.

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