Asked by eller
Fiona draws a triangle on a coordinate grid with vertices at (6, 2), (4, 60, and (1, 2). What is the area in square units of Fiona's triangle?
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Answered by
GPT-4o mini
To find the area of the triangle with vertices at \( (6, 2) \), \( (4, 6) \), and \( (1, 2) \) using the coordinates of the vertices, we can use the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Let the vertices be:
- \( (x_1, y_1) = (6, 2) \)
- \( (x_2, y_2) = (4, 6) \)
- \( (x_3, y_3) = (1, 2) \)
Plugging the coordinates into the formula:
\[
\text{Area} = \frac{1}{2} \left| 6(6 - 2) + 4(2 - 2) + 1(2 - 6) \right|
\]
Now we calculate each term inside the absolute value:
1. \( 6(6 - 2) = 6 \times 4 = 24 \)
2. \( 4(2 - 2) = 4 \times 0 = 0 \)
3. \( 1(2 - 6) = 1 \times (-4) = -4 \)
Adding these together:
\[
24 + 0 - 4 = 20
\]
Now we can find the area:
\[
\text{Area} = \frac{1}{2} \left| 20 \right| = \frac{1}{2} \times 20 = 10
\]
Therefore, the area of Fiona's triangle is
\[
\boxed{10} \text{ square units.}
\]
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Let the vertices be:
- \( (x_1, y_1) = (6, 2) \)
- \( (x_2, y_2) = (4, 6) \)
- \( (x_3, y_3) = (1, 2) \)
Plugging the coordinates into the formula:
\[
\text{Area} = \frac{1}{2} \left| 6(6 - 2) + 4(2 - 2) + 1(2 - 6) \right|
\]
Now we calculate each term inside the absolute value:
1. \( 6(6 - 2) = 6 \times 4 = 24 \)
2. \( 4(2 - 2) = 4 \times 0 = 0 \)
3. \( 1(2 - 6) = 1 \times (-4) = -4 \)
Adding these together:
\[
24 + 0 - 4 = 20
\]
Now we can find the area:
\[
\text{Area} = \frac{1}{2} \left| 20 \right| = \frac{1}{2} \times 20 = 10
\]
Therefore, the area of Fiona's triangle is
\[
\boxed{10} \text{ square units.}
\]
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