Question

What is the area of a triangle whose vertices are J(−2, 1), K(0, 3), and L(3, −4)?



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Answer 1

To find the area of a triangle given its vertices at \( J(-2, 1) \), \( K(0, 3) \), and \( L(3, -4) \), we can use the formula for the area of a triangle defined by its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

Let's assign \( J(-2, 1) \) as \((x_1, y_1)\), \( K(0, 3) \) as \((x_2, y_2)\), and \( L(3, -4) \) as \((x_3, y_3)\):

- \( x_1 = -2, y_1 = 1 \)
- \( x_2 = 0, y_2 = 3 \)
- \( x_3 = 3, y_3 = -4 \)

Substituting these coordinates into the area formula:

\[
\text{Area} = \frac{1}{2} \left| -2(3 - (-4)) + 0(-4 - 1) + 3(1 - 3) \right|
\]

Calculating each term inside the absolute value:

1. For the first term:
\[
-2(3 + 4) = -2(7) = -14
\]

2. For the second term:
\[
0(-4 - 1) = 0
\]

3. For the third term:
\[
3(1 - 3) = 3(-2) = -6
\]

Putting it all together inside the absolute value:

\[
\text{Area} = \frac{1}{2} \left| -14 + 0 - 6 \right| = \frac{1}{2} \left| -20 \right| = \frac{1}{2} \times 20 = 10
\]

Thus, the area of the triangle is \( \boxed{10} \).