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To find the area of triangle \( JKL \), given its vertices \( J(-7, -8) \), \( K(-4, -4) \), and \( L(-8, -2) \), we can use the following formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

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

Substituting the coordinates of \( J(-7, -8) \), \( K(-4, -4) \), and \( L(-8, -2) \) into the formula, we get:

\[
\text{Area} = \frac{1}{2} \left| (-7)((-4) - (-2)) + (-4)((-2) - (-8)) + (-8)((-8) - (-4)) \right|
\]

I'll compute this step by step.The area of triangle \( JKL \) is 11 square units.

Would you like further details or explanations? Here are some related questions for deeper understanding:

1. How does the area formula for a triangle work in terms of coordinate geometry?
2. Can the same formula be used for triangles in 3D space?
3. How would the area change if one vertex was moved?
4. What are some other methods to calculate the area of a triangle?
5. How does the area relate to the concept of the determinant of a matrix?

**Tip:** When working with coordinate geometry, always ensure that you substitute the correct coordinates into formulas to avoid errors.

What is the area, in square units, of triangle JKL with vertices J(-7, -8), K(-4, -4), and L(-8, -2)?

Learn how to calculate the area of triangle JKL in a coordinate plane, using the formula for the area of a triangle given its vertices. This problem demonstrates key concepts in coordinate geometry and is suitable for students in Grades 9-12.

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