Difference between revisions of "2024 AMC 12B Problems/Problem 15"
(→Solution 2 (Determinant)) |
(→Solution 2 (Determinant)) |
||
Line 49: | Line 49: | ||
Thus, the area is:<math>\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|</math> = <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math> | Thus, the area is:<math>\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|</math> = <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math> | ||
− | ~Athmyx | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] |
Revision as of 05:32, 14 November 2024
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 (Determinant)
To calculate the area of a triangle formed by three points
The coordinates are:
,
,
Taking a numerical value into account:
Simplify:
Thus, the area is:
=