I've been looking for an answer in the forums and Google but couldn't find one.
Given two different points in the plane, $(x_1,\,y_1)$ and $(x_2,\,y_2)$, we know that the equation of that line can be written as: $$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$$It's possible to rewrite this expression as a determinant equation: $$\begin{vmatrix} 1 & x & y \\ 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \end{vmatrix}=0$$
I do know how to reach this last equation starting from the first one.
However, I've been asked how to derive the determinant equation of a line given its slope ($m$) and a point $(x_1,\,y_1)$.
I know the solution is: $$\begin{vmatrix} 1 & x & y \\ 1 & x_1 & y_1 \\ 0 & 1 & m \end{vmatrix}=0$$ But I'm not very sure how to reach that determinant. What I do know is that the equation can be written as: $$y-y_1=m(x-x_1)$$ How do I transform this algebraic equation to the above determinant equation?
$\endgroup$2 Answers
$\begingroup$These determinants can be interpreted as expressing linear dependence, but of vectors in a $3$-dimensional vector space containing your affine place as a plane not going through the origin. Concretely let $w,x,y$ be coordinates of that space, then your affine plan is the one given by $w=1$.
Now flipping your determinants (because I like to stack coordinates vertically), points with coordinates $\binom{x_1}{y_1}$, $\binom{x_2}{y_2}$ and $\binom{x}{y}$ are collinear whenever the corresponding vector (with an addition coordinate $w=1$) are coplanar (the plane cuts $w=1$ in a line), and this is expressed by $$\left|\matrix{1&1&1\\x_1&x_2&x\\y_1&y_2&y}\right|=0.$$ Now if instead of two points you give a point and a true vector (the slope corresponds to a translation of your plane, with coordinates $\binom1m$) then that vector should get extra coordinate $0$ rather than $1$. The coplanarity condition then becomes $$\left|\matrix{1&0&1\\x_1&1&x\\y_1&m&y}\right|=0$$ which up to some minor transformations is what you wrote.
$\endgroup$ $\begingroup$Hint: Take your previous formula and subtract the second line from the third. Then divide the last line by $(x_2-x_1)\neq 0$ to obtain the final result.
Note that the determinant will not change, because the determinant is $0$.
Edit:
$$0=\begin{vmatrix} 1 & x & y \\ 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ 1 & x_1 & y_1 \\ 0 & x_2-x_1 & y_2-y_1 \end{vmatrix} = \begin{vmatrix} 1 & x & y \\ 1 & x_1 & y_1 \\ 0 & \frac{x_2-x_1}{x_2-x_1} & \frac{y_2-y_1}{x_2-x_1} \end{vmatrix}$$
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