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I was reading about Triangle Interiors on Wolfram Alpha:

and they have a simple equation: $$\mathbf{v} = \mathbf{v}_0 + a\mathbf{v}_1 + b\mathbf{v}_2,$$ and then they solve for $a$ and $b$, but I wasn't sure about the steps of how they went about doing that.

My algebra is rusty...

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Note that $\mathbf{v}$, $\mathbf{v}_0$, $\mathbf{v}_1$, and $\mathbf{v}_2$ are vectors, not numbers; you really need some analytic geometry to understand the expressions, not merely algebra at the precalculus level. – Arturo Magidin Jan 11 '12 at 21:44

The page defines a binary function $\times$ which maps vectors in $\mathbb{R}^2$ to real numbers, $\mathbb{R}$. This function is given as $\mathbf{v} \times \mathbf{w} = v_xw_y - w_xv_y$. It can be shown that it is a bilinear function, since

$$(\mathbf{a} + \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \times \mathbf{c}) + (\mathbf{b} \times \mathbf{c})$$ $$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = (\mathbf{a} \times \mathbf{b}) + (\mathbf{a} \times \mathbf{c})$$ $$(\lambda\mathbf{a})\times\mathbf{b}=\mathbf{a}\times(\lambda\mathbf{b})=\lambda(\mathbf{a}\times\mathbf{b})$$

$\forall\mathbf{a},\mathbf{b},\mathbf{c}\in\mathbb{R^2},\forall\lambda\in\mathbb{R}$ (to see this, set $\mathbf{a}=(a_x,a_y),$ $\mathbf{b}=(b_x,b_y),$ and $\mathbf{c}=(c_x,c_y),$ then substitute into the above). Using the above equations $$\mathbf{v}\times\mathbf{v_1}=(\mathbf{v_0} + a\mathbf{v_1} + b\mathbf{v_2})\times\mathbf{v_1}$$


$$\mathbf{v}\times\mathbf{v_2}=(\mathbf{v_0} + a\mathbf{v_1} + b\mathbf{v_2})\times\mathbf{v_2}$$

can be manipulated to obtain the desired results, so long as $\mathbf{v_1}\times\mathbf{v_2}\ne0$.

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