Show that the straight-line through the points $x$ and $y$ in $\Bbb Q^2$ is given by the set How can I proof this statement? I have no clue what to do first. $x = (x_1, x_2), y = (y_1, y_2)$ in $\Bbb Q^2$ with $x \neq y$. Show that the straight-line through the points $x$ and $y$ in $\Bbb Q^2$ is given by the set: $$\left\lbrace(z_1,z_2)\in \Bbb Q^2,\det\left(\matrix{1& x_1&x_2\\1& y_1&y_2\\1& z_1&z_2\\}\right)=0\right\rbrace$$
PS: Could someone explain me how MathJax works. It does not seem to work for me.
 A: Hint: 
Observe that a line $(AB)$ is the set of all affine combinations (or barycentres) of the two points, for all possible weight systems. 
On the other hand, show that $\det\left(\matrix{1& x_1&x_2\\1& y_1&y_2\\1& z_1&z_2\\}\right)$ is $0$ if and only if the last row is a linear combination of the first two rows.
Finally, show that a linear combination as above gives you exactly $(z_1,z_2)$ as an affine combination of $(x_1,x_2)$ and $(y_1,y_2)$.
A: You want $$ (x_1,x_2),( y_1,y_2), ( z_1,z_2) $$ to be co-linear.
That is the vectors $V=( y_1-x_1, y_2-x_2)$ and $W=( z_1-x_1, z_2-x_2)$  to be linearly dependent.
That is $$V\times W  =\det\left(\matrix{i& j&k\\y_1-x_1& y_2-x_2&   0 \\ z_1-x_1 &   z_2-x_2 &0\\}\right)=(0,0,0)$$
Note that 
$ V\times W = 0 $ iff 
$$\det\left(\matrix{1& x_1&x_2\\1& y_1&y_2\\1& z_1&z_2\\}\right)=0$$
A: $\begin{vmatrix} 1 & x_1&x_2\\1& y_1&y_2\\1& z_1&z_2\\ \end{vmatrix} $ is the (signed) area of the triangle with vertices $(x_1, x_2), (y_1, y_2), (z_1, z_2)$ (see here).
Therefore, this determinant is zero iff the points are collinear.
