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Triangle on a grid

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A triangle in the plane has vertices $(1, 2)$ , $(6, 4)$ and $(3, 9)$ . What is its area? Give the answer as a fraction in lowest terms.

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Hint

Take the $2$ edge vectors leaving one vertex; the triangle is half of the parallelogram they span.

Solution

From $(1, 2)$ the edges go to $(6, 4)$ and $(3, 9)$ , which gives the vectors $(5, 2)$ and $(2, 7)$ .
The parallelogram on these vectors has area equal to the absolute value of $5 \times 7 - 2 \times 2 = 31$ .
The triangle is exactly half of that parallelogram, so its area is $\frac{31}{2}$ .

Answer: $\frac{31}{2}$