equation of line as a determinant

Question

The question:

Prove that the equation of a line through the distinct points $(a_1,b_1)$ and $(a_2,b_2)$ can be written as $$\det \begin{bmatrix} x & y & 1 \\ a_1 & b_1 & 1 \\ a_2 & b_2 & 1 \end{bmatrix} = 0.$$

I started by writing the two points in the slop-intercept form $y=\frac{a_1-a_2}{b_1-b_2}x + b$ and then writing in standard form as $(a_1-a_2)y-(b_1-b_2)x - b(a_1-a_2) = 0$, and then I take the determinant of the matrix given and I get $(b_1-b_2)x - (a_1-a_2)y + a_1b_2-a_2b_1$. I expected the two to match up, and they are close, but I'm not sure what I'm missing here.

Answer 1

The thing you're missing is to express the $b$ in the equation of the line in terms of the two points $(a_1,b_1)$ and $(a_2,b_2)$.

And notice also that you have the slope backwards, it should be $\frac{b_1-b_2}{a_1-a_2}$ (but I think you got the next part right though). You might want to check this, but the equation of the line will turn out to be

$$y = \frac{b_1-b_2}{a_1-a_2}x+\frac{a_1b_2-b_1a_2}{a_1-a_2},$$

where the last term is equal to the $b$ in your expression.

Using this as your $b$, your rewritten equation $(a_1-a_2)y-(b_1-b_2)x - b(a_1-a_2) = 0$ looks like it becomes exactly the same as the determinant equation.

Answer 2

You already have a good answer and your idea is good, but I just thought I would add this alternative proof. Since the line through two points is the set of $\{t(P-Q)+Q :t\in\mathbb{R}\}$, we note that $t(P-Q)+Q=tP+(1-t)Q$, so the line is also the set of $\alpha P+\beta Q$ where $\alpha+\beta=1$. I.e. the set of points $R$ such that $(R,1)=\alpha(P,1)+\beta(Q,1)$. But this is just the set of $R$ such that $(R,1)$ is linearly dependent on $(P,1)$ and $(Q,1)$. Which is precisely the set of $R$ that make the given determinant 0.

I'm assuming your field is $\mathbb{R}$, but it doesn't really matter.

Answer 3

As the determinant is linear with respect to column vectors, this one is linear in $x$ and $y$, hence is the equation of a straight line. Thus it is enough to check the equation is satisfied by $(a_1,b_1)$ and $(a_2,b_2)$, which is trivial since in such cases two rows are equal.

Answer 4

A line in $\mathbb{R}^2$ is given by an equation of the form $$\tag{1}\xi x+\eta y+\zeta=0,$$ where either $\xi$ and/or $\eta$ are nonzero. This, together with equations $$ \xi a_1+\eta b_1+\zeta=0, \quad \xi a_2+\eta b_2+\zeta=0, $$ gives a (homogeneous) system of three equations for three unknowns $\xi$, $\eta$, and $\zeta$ in the matrix form $$\tag{2} M(x,y)\,c:=\pmatrix{x&y&1\\a_1&b_1&1\\a_2&b_2&1} \pmatrix{\xi\\\eta\\\zeta}=0. $$ A point $(x,y)$ hence lies on a line containing points $(a_1,b_1)$ and $(a_2,b_2)$ if and only if (2) has a nontrivial solution, that is, iff $M(x,y)$ is singular iff $\det M(x,y)=0$.

Remark It is easy to see that a singularity of $M(x,y)$ implies we have a well defined line with $\xi\neq 0$ and/or $\eta\neq 0$. A solution with $\xi=\eta=0$ would necessarily be trivial.

Remark For a unique line, $\xi$, $\eta$, and $\zeta$ are defined up to a constant nonzero multiple. The equation (2) says that the vector of these coefficients is an eigenvector of $M(x,y)$ associated with the zero eigenvalue. So for a unique line, we need that the corresponding eigenvector space is one-dimensional. But since $(a_1,b_1)\neq(a_2,b_2)$, one can check that the rank of $M(x,y)$ is at least two (and equal to two if $M(x,y)$ is singular). Hence the singular $M(x,y)$ has a one-dimensional and the line (1) is unique.