Question

The area of the triangle whose vertices are A(2, 3), B(−1, 0) and C(2, −4) (in square units) is

• A. 11.5
• B. 10
• C. 10.5
• D. 11

Hint:

When calculating the area of a triangle using its vertices, use the formula involving the determinant of the matrix formed by the coordinates.

Answer 1

The Correct Option is C

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:
Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
The given vertices are A(2,3), B(-1,0), and C(2,-4).
Let $(x_1, y_1) = (2, 3)$.
Let $(x_2, y_2) = (-1, 0)$.
Let $(x_3, y_3) = (2, -4)$.
Substitute these coordinates into the area formula:
Area = $\frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$.
Simplify the expression inside the absolute value:
Area = $\frac{1}{2} |2(4) + (-1)(-7) + 2(3)|$.
Area = $\frac{1}{2} |8 + 7 + 6|$.
Area = $\frac{1}{2} |21|$.
Area = $\frac{21}{2}$.
Area = 10.5 square units.