Find the area of triangle with vertices at $A(1,13)$, $B(-4,1)$, $C(4,-5)$.
Step-by-Step Solution
Key Concept: The area of a triangle can be computed using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, where the height is the perpendicular distance from a vertex to the opposite side.
First, find the slopes of the sides: slope of $AB = \frac{13-1}{1-(-4)} = \frac{12}{5}$, slope of $BI = \frac{1-(-5)}{-4-4} = -\frac{3}{4}$, slope of $AI = \frac{13-(-5)}{1-4} = -6$. Using the condition that $m_1 \cdot m_2 = -1$, determine if any sides are perpendicular. After solving the system of equations for the lines containing $BC$ and $AC$, we get the equations $y = 1$ and $y - 13 = -4(x-1)$. Calculate the area using the distance formula and the perpendicular height from vertex to opposite side, yielding an area of $8$ square units.
Correct Answer: 8