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smc

geometry senior

Problem

The vertices of have coordinates as follows: , where and are positive. The origin and point lie on opposite sides of . The area of may be found from the expression:
(A)
(B)
(C)
(D)
Solution
To solve this problem, we could use the distance formula to find the lengths of the sides and then Heron's Formula to find the area of the triangle, but that solution seems messy and prone to mistakes with lots of square roots and polynomial expansions. Therefore, we look for a simpler, easier solution. Suppose and . This makes half of a rectangle with side lengths and , so it has an area of . Plugging in and into all of the answer choices, the only one which returns is .
Final answer
B