To find the area of triangle ABC with given vertices, we can use the formula for the area of a triangle based on its vertices in a coordinate plane. The formula is:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
where \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are the coordinates of the vertices A, B, and C respectively.
### (i) A(1, 2), B(-2, 3), C(-3, -4)
1. Assign the coordinates:
- \( A(1, 2) \) → \( x_1 = 1, y_1 = 2 \)
- \( B(-2, 3) \) → \( x_2 = -2, y_2 = 3 \)
- \( C(-3, -4) \) → \( x_3 = -3, y_3 = -4 \)
2. Substitute into the area formula:
\[
\text{Area} = \frac{1}{2} \left| 1(3 - (-4)) + (-2)(-4 - 2) + (-3)(2 - 3) \right|
\]
3. Calculate each term:
- \( 1(3 + 4) = 1 \times 7 = 7 \)
- \( -2(-4 - 2) = -2 \times -6 = 12 \)
- \( -3(2 - 3) = -3 \times -1 = 3 \)
4. Combine the terms:
\[
\text{Area} = \frac{1}{2} \left| 7 + 12 + 3 \right| = \frac{1}{2} \left| 22 \right| = \frac{22}{2} = 11
\]
5. Final area:
\[
\text{Area} = 11 \text{ square units}
\]
### (ii) A(-5, 7), B(-4, -5), C(4, 5)
1. Assign the coordinates:
- \( A(-5, 7) \) → \( x_1 = -5, y_1 = 7 \)
- \( B(-4, -5) \) → \( x_2 = -4, y_2 = -5 \)
- \( C(4, 5) \) → \( x_3 = 4, y_3 = 5 \)
2. Substitute into the area formula:
\[
\text{Area} = \frac{1}{2} \left| -5(-5 - 5) + (-4)(5 - 7) + 4(7 - (-5)) \right|
\]
3. Calculate each term:
- \( -5(-10) = 50 \)
- \( -4(-2) = 8 \)
- \( 4(12) = 48 \)
4. Combine the terms:
\[
\text{Area} = \frac{1}{2} \left| 50 + 8 + 48 \right| = \frac{1}{2} \left| 106 \right| = \frac{106}{2} = 53
\]
5. Final area:
\[
\text{Area} = 53 \text{ square units}
\]
### (iii) A(3, 8), B(-4, 2), C(5, -1)
1. Assign the coordinates:
- \( A(3, 8) \) → \( x_1 = 3, y_1 = 8 \)
- \( B(-4, 2) \) → \( x_2 = -4, y_2 = 2 \)
- \( C(5, -1) \) → \( x_3 = 5, y_3 = -1 \)
2. Substitute into the area formula:
\[
\text{Area} = \frac{1}{2} \left| 3(2 - (-1)) + (-4)(-1 - 8) + 5(8 - 2) \right|
\]
3. Calculate each term:
- \( 3(3) = 9 \)
- \( -4(-9) = 36 \)
- \( 5(6) = 30 \)
4. Combine the terms:
\[
\text{Area} = \frac{1}{2} \left| 9 + 36 + 30 \right| = \frac{1}{2} \left| 75 \right| = \frac{75}{2} = 37.5
\]
5. Final area:
\[
\text{Area} = 37.5 \text{ square units}
\]
### (iv) A(10, -6), B(2, 5), C(-1, 3)
1. Assign the coordinates:
- \( A(10, -6) \) → \( x_1 = 10, y_1 = -6 \)
- \( B(2, 5) \) → \( x_2 = 2, y_2 = 5 \)
- \( C(-1, 3) \) → \( x_3 = -1, y_3 = 3 \)
2. Substitute into the area formula:
\[
\text{Area} = \frac{1}{2} \left| 10(5 - 3) + 2(3 - (-6)) + (-1)(-6 - 5) \right|
\]
3. Calculate each term:
- \( 10(2) = 20 \)
- \( 2(9) = 18 \)
- \( -1(-11) = 11 \)
4. Combine the terms:
\[
\text{Area} = \frac{1}{2} \left| 20 + 18 + 11 \right| = \frac{1}{2} \left| 49 \right| = \frac{49}{2} = 24.5
\]
5. Final area:
\[
\text{Area} = 24.5 \text{ square units}
\]
