The area of the triangle (in $$\text{unit}^2$$​) whose vertices are A(4, 8), B(-6, 2) and C(5, 4) is:

Correct option is B

​$$\text{The area of a triangle with vertices } A(x_1, y_1), B(x_2, y_2), \text{ and } C(x_3, y_3) \text{ is given by:} \\\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \\\text{Given:} \\A(4, 8), \quad B(-6, 2), \quad C(5, 4) \\\text{Substitute these into the formula:} \\\text{Area} = \frac{1}{2} \left| 4(2 - 4) + (-6)(4 - 8) + 5(8 - 2) \right| \\\textbf{Simplify the Expression} \\\text{Calculate each term inside the absolute value:} \\4(2 - 4) = 4(-2) = -8 \\-6(4 - 8) = -6(-4) = 24 \\5(8 - 2) = 5(6) = 30 \\\text{Now, sum these results:} \\-8 + 24 + 30 = 46 \\\text{Take the absolute value and multiply by } \frac{1}{2}: \\\text{Area} = \frac{1}{2} \times 46 = 23$$​​