Question

Find the area of the triangle whose vertices are \( A(2,1) \), \( B(4,5) \) and \( C(0,3) \).

Hint:

Area of Triangle: Use determinant formula \( A = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \).

Answer 1

Using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] \[ = \frac{1}{2} \left| 2(5-3) + 4(3-1) + 0(1-5) \right| \] \[ = \frac{1}{2} \left| 2(2) + 4(2) + 0 \right| \] \[ = \frac{1}{2} \left| 4 + 8 \right| = \frac{1}{2} \times 12 = 6 \] Thus, the area is \( \mathbf{6} \) square units.
Correct Answer: \( 6 \)