How was the area of this triangle found using the area of a right triangle?

Question

This type of question came up in my exam a month back and its troubling me why I wasn't able to understand it. I vaguely recall my teacher explaining you could find the area using a right triangle but I don't think I understood it then or in the exam. I am aware of the formula, $$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$ But I believe it won't be awarded marks if used because it's not part of the syllabus. So in summary,

How was the area found using the area of a right triangle? Is there a visual way to see why it could be done?

Answer 1

There is a right triangle with the origin, $(x_2, y_2)$ and $(x_3, y_3)$ as vertices, and another right triangle with vertices at the origin, $(x_1, y_1)$ and $(x_3, y_3)$.

Therefore the total area is big triangle - small triangle:

(a picture shows a thousand words)

Answer 2

The area of the obtuse triangle can be represented as the difference between the area of the right triangle with vertices $(0;0)$, $(x_2;y_2)$ and $(x_3;y_3)$, and the area of the right triangle with vertices $(0;0)$, $(x_1;y_1)$ and $(x_3;y_3)$.

However, I think it's easier to just calculate the product of the lengths of both green and red segments, then divide it by $2$ in order to find the area of the obtuse triangle.