2024 AMC 12B Problems/Problem 15
Problem
A triangle in the coordinate plane has vertices 
, 
, and 
. What is the area of 
?
Solution 1 (Shoelace Theorem)
We rewrite:
.
From here we setup Shoelace Theorem and obtain:
.
Following log properties and simplifying gives (B).
~MendenhallIsBald
Solution 2
To calculate the area of a triangle formed by three points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) on a Cartesian coordinate plane, you can use the following formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
The coordinates are:
- \( A(0, 1) \) - \( B(\log_2 3, 2) \) - \( C(\log_2 7, 3) \)
take a numerical value into account
\[ \text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right| \]
Simplify:
\[ = \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right| \]
\[ = \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right| \]
\[ = \frac{1}{2} \left| \log_2 \frac{9}{7} \right| \]
Thus, the exact area is:
\[ \text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right| \]
\[ \boxed{\text{B. \log_2 \frac{3}{\sqrt{7}}}} \]
