To find the area of triangle $\triangle ABC$ with vertices at $A(1, 8)$, $B(6, 6)$, and $C(-5, -7)$, we can use the formula for the area of a triangle when the coordinates of its vertices are known. The formula is:

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

Where:

• $(x_1, y_1)$ are the coordinates of point $A$,
• $(x_2, y_2)$ are the coordinates of point $B$,
• $(x_3, y_3)$ are the coordinates of point $C$.

Substituting the given coordinates $A(1, 8)$, $B(6, 6)$, and $C(-5, -7)$ into the formula:

$\text{Area} = \frac{1}{2} \left| 1(6 - (-7)) + 6((-7) - 8) + (-5)(8 - 6) \right|$

Let's compute this step by step. I'll calculate the value now.The area of triangle $\triangle ABC$, given the vertices $A(1, 8)$, $B(6, 6)$, and $C(-5, -7)$, is $43.5$ square units.

Would you like any further details or explanations on this? Here are five related questions you might find helpful:

1. How can you calculate the area of a triangle using other methods?
2. What is the significance of using absolute values in the area formula?
3. Can you generalize this formula to higher dimensions, such as for tetrahedrons in 3D?
4. How would you find the perimeter of this triangle using the distance formula?
5. What are the properties of a triangle that determine if it can be classified as right, obtuse, or acute?

Tip: When working with coordinates, it’s always good to sketch the graph and verify the triangle shape visually to check for potential errors.