To find which of the given triangles has the same area as Quincy's triangle with coordinates A(2, 1), B(2, 7), and C(7, 5), we first need to calculate the area of Quincy's triangle using the formula for the area of a triangle given its vertices (x1, y1), (x2, y2), and (x3, y3):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
For Quincy's triangle:
- A(2, 1) → (x1, y1) = (2, 1)
- B(2, 7) → (x2, y2) = (2, 7)
- C(7, 5) → (x3, y3) = (7, 5)
Plugging the coordinates into the formula, we have:
\[ \text{Area} = \frac{1}{2} \left| 2(7-5) + 2(5-1) + 7(1-7) \right| \]
Now calculating step by step:
\[ = \frac{1}{2} \left| 2 \cdot 2 + 2 \cdot 4 + 7 \cdot (-6) \right| \] \[ = \frac{1}{2} \left| 4 + 8 - 42 \right| \] \[ = \frac{1}{2} \left| -30 \right| \] \[ = \frac{1}{2} \cdot 30 = 15 \]
So the area of Quincy's triangle is 15.
Now, we need to determine the area of each of the provided triangles using the same formula:
Triangle 1: C(2, 7), B(10, 2), A(2, 1)
\[ \text{Area} = \frac{1}{2} \left| 2(2-1) + 10(1-7) + 2(7-2) \right| \]
Calculating: \[ = \frac{1}{2} \left| 2(1) + 10(-6) + 2(5) \right| \] \[ = \frac{1}{2} \left| 2 - 60 + 10 \right| \] \[ = \frac{1}{2} \left| -48 \right| = 24 \quad \text{(not equal to 15)} \]
Triangle 2: C(2, 7), B(7, 2), A(2, 1)
\[ \text{Area} = \frac{1}{2} \left| 2(2-1) + 7(1-7) + 2(7-2) \right| \]
Calculating: \[ = \frac{1}{2} \left| 2(1) + 7(-6) + 2(5) \right| \] \[ = \frac{1}{2} \left| 2 - 42 + 10 \right| \] \[ = \frac{1}{2} \left| -30 \right| = 15 \quad \text{(equal to 15)} \]
Triangle 3: C(2, 7), B(4, 4), A(2, 1)
\[ \text{Area} = \frac{1}{2} \left| 2(4-1) + 4(1-7) + 2(7-4) \right| \]
Calculating: \[ = \frac{1}{2} \left| 2(3) + 4(-6) + 2(3) \right| \] \[ = \frac{1}{2} \left| 6 - 24 + 6 \right| \] \[ = \frac{1}{2} \left| -12 \right| = 6 \quad \text{(not equal to 15)} \]
Triangle 4: C(2, 7), B(5, 9), A(2, 1)
\[ \text{Area} = \frac{1}{2} \left| 2(9-1) + 5(1-7) + 2(7-9) \right| \]
Calculating: \[ = \frac{1}{2} \left| 2(8) + 5(-6) + 2(-2) \right| \] \[ = \frac{1}{2} \left| 16 - 30 - 4 \right| \] \[ = \frac{1}{2} \left| -18 \right| = 9 \quad \text{(not equal to 15)} \]
Based on the calculations, only Triangle 2 has the same area as Quincy’s triangle.
The correct answer is Triangle 2.