Equating coefficients of linearly dependent vectors Consider a triangle in $\mathbb{R}^2$ with vertices A, B and C, which have position vectors $\underline a$, $\underline b$ and $\underline c$ respectively. The vector equations of two of the lines joining vertices of the triangle to the midpoints of opposite sides are as follows:
$$
\underline r_1 = \underline a+\lambda(\frac 12 \underline b + \frac 12 \underline c - \underline a)\\\underline r_2 = \underline b+\mu(\frac 12 \underline a + \frac 12 \underline c - \underline b)
$$
To find the position vector of the point of intersection of these two lines (ie the centroid of the triangle), you can equate $\underline r_1 = \underline r_2$ and attempt to extract a value for $\lambda$ and $\mu$. But it is impossible for all three of $\underline a$, $\underline b$ and $\underline c$ to be linearly independent, as in two dimensions every vector can be written as a linear combination of any two linearly independent vectors.
Why, then, does ignoring this issue and equating coefficients without considering whether it is valid to do so give the correct answer of $\lambda$, $\mu$ = $\frac 23$ ?
 A: This isn't a perfect answer, but hopefully it's still illuminating: note that the entire problem can be shifted by any fixed vector (moving that vector to the origin) without altering the geometry of the situation. In particular, let us subtract $\underline c$ from all vectors to move the third vertex $\underline c$ to the origin. If we set $\underline \rho_i = \underline r_i-\underline c$ and $\underline \alpha=\underline a-\underline c$ and $\underline \beta=\underline b-\underline c$, then the equations become
$$
\underline \rho_1 = \underline \alpha+\lambda(\frac 12 \underline \beta - \underline \alpha)\\
\underline \rho_2 = \underline \beta+\mu(\frac 12 \underline \alpha - \underline \beta).
$$
Now it is obvious that equating coefficients yields the unique correct values of $\lambda$ and $\mu$. Further, that "obviousness" relies upon the fact that $\underline\alpha$ and $\underline\beta$ are themselves linearly independent—but this is exactly the condition that the triangle is nondegenerate.
