Quincy draws this image of a triangle on a coordinate grid. Which of the following triangles has the same area as Quincy's triangle?

Question

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.