Flächeninhalt eines Dreiecks mit der Schnürsenkelformel

Problem

Find the area of triangle $ABC$ with vertices $A(2,-3)$, $B(4,5)$, and $C(-5,1)$ using the shoelace formula.

Step 1: Set up the shoelace products

Using the coordinates in order, the three vertex contributions are

$$2\cdot 5 - (-3)\cdot 4,\quad 4\cdot 1 - 5\cdot (-5),\quad (-5)\cdot (-3) - 1\cdot 2.$$

Step 2: Simplify each bracket

This gives

$$10 - (-12) = 22,\quad 4 - (-25) = 29,\quad 15 - 2 = 13.$$

Adding the three results gives

$$22 + 29 + 13 = 64.$$

Step 3: Take half the total

The shoelace formula uses one-half of that sum, so the area is

$$\frac{1}{2}\cdot 64 = 32.$$

Answer

The area of triangle $ABC$ is $32$ square units.