Question

Quincy draws this image of a triangle on a coordinate grid. Which of the following triangles has the same area as Quincy's triangle?
Quincy's triangle:
Point A: (2, 1)
Point B: (2, 7)
Point C: (7, 5)

Triangle A:
Point A: (2, 1)
Point B: (5, 9)
Point C: (2, 7)

Triangle B:
Point A: (2, 1)
Point B: (10, 2)
Point C: (2, 7)

Triangle C:
Point A: (2, 1)
Point B: (2, 7)
Point C: (4, 4)

Triangle D:
Point A: (2, 1)
Point B: (7, 2)
Point C: (2, 7)

Answer 1

To find which of the provided triangles has the same area as Quincy's triangle, we can calculate the area of Quincy's triangle using the formula for the area of a triangle given vertices at coordinates \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

Using the coordinates of Quincy's triangle:
- \( A(2, 1) \)
- \( B(2, 7) \)
- \( C(7, 5) \)

Substituting these coordinates into the formula:

\[
\text{Area} = \frac{1}{2} \left| 2(7-5) + 2(5-1) + 7(1-7) \right|
\]

Calculating each term:

1. \( 2(7-5) = 2 \times 2 = 4 \)
2. \( 2(5-1) = 2 \times 4 = 8 \)
3. \( 7(1-7) = 7 \times -6 = -42 \)

Summing these:
\[
4 + 8 - 42 = -30
\]

Taking the absolute value and multiplying by \(\frac{1}{2}\):
\[
\text{Area} = \frac{1}{2} \times | -30 | = \frac{1}{2} \times 30 = 15
\]

Now that we know the area of Quincy's triangle is 15, we will check each of the other triangles using the same area formula.

**Triangle A:**
- \( (2, 1) \)
- \( (5, 9) \)
- \( (2, 7) \)

Area:
\[
\text{Area} = \frac{1}{2} \left| 2(9-7) + 5(7-1) + 2(1-9) \right|
\]
Calculating:
1. \( 2(9-7) = 4 \)
2. \( 5(7-1) = 30 \)
3. \( 2(1-9) = -16 \)
Summing:
\[
4 + 30 - 16 = 18
\]
Area = \(\frac{1}{2} \times 18 = 9\) (not equal)

**Triangle B:**
- \( (2, 1) \)
- \( (10, 2) \)
- \( (2, 7) \)

Area:
\[
\text{Area} = \frac{1}{2} \left| 2(2-7) + 10(7-1) + 2(1-2) \right|
\]
Calculating:
1. \( 2(2-7) = -10 \)
2. \( 10(7-1) = 60 \)
3. \( 2(1-2) = -2 \)
Summing:
\[
-10 + 60 - 2 = 48
\]
Area = \(\frac{1}{2} \times 48 = 24\) (not equal)

**Triangle C:**
- \( (2, 1) \)
- \( (2, 7) \)
- \( (4, 4) \)

Area:
\[
\text{Area} = \frac{1}{2} \left| 2(7-4) + 2(4-1) + 4(1-7) \right|
\]
Calculating:
1. \( 2(7-4) = 6 \)
2. \( 2(4-1) = 6 \)
3. \( 4(1-7) = -24 \)
Summing:
\[
6 + 6 - 24 = -12
\]
Area = \(\frac{1}{2} \times 12 = 6\) (not equal)

**Triangle D:**
- \( (2, 1) \)
- \( (7, 2) \)
- \( (2, 7) \)

Area:
\[
\text{Area} = \frac{1}{2} \left| 2(2-7) + 7(7-1) + 2(1-2) \right|
\]
Calculating:
1. \( 2(2-7) = -10 \)
2. \( 7(7-1) = 42 \)
3. \( 2(1-2) = -2 \)
Summing:
\[
-10 + 42 - 2 = 30
\]
Area = \(\frac{1}{2} \times 30 = 15\) (equal)

Thus, the triangle that has the same area as Quincy's triangle is Triangle D.