Math Problem Statement

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

Solution

To find the area of triangle JKLJKL, given its vertices J(7,8)J(-7, -8), K(4,4)K(-4, -4), and L(8,2)L(-8, -2), we can use the following formula for the area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3):

Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\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)J(-7, -8), K(4,4)K(-4, -4), and L(8,2)L(-8, -2) into the formula, we get:

Area=12(7)((4)(2))+(4)((2)(8))+(8)((8)(4))\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 JKLJKL 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.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Triangle Area

Formulas

Area of a triangle = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Theorems

Triangle area in coordinate plane

Suitable Grade Level

Grades 9-12