Determinant of a matrix and points on a line in $\mathbb{R}^2$

Question

I have two points $p$, $q$ in $\mathbb{R}^2$ for which $L$ denotes the line connecting them. $L$ is supposedly given by $\{(x_1, x_2) \in \mathbb{R}^2 | \det A = 0\}$ with $$A = \begin{pmatrix} 1 & p_1 & p_2 \\ 1 & q_1 & q_2 \\ 1 & x_1 & x_2 \end{pmatrix}$$ I can't see why. I know that from $\det A = 0$ means that $A$ has rank smaller $3$, but how does this relate to points on a line? Can someone explain it, please?

Answer 1

Whenever one line of a matrix is a linear combination of other two, the determinant is zero.

If $(x_1,x_2)$ is such that this determinant is zero, then there exists $\alpha,\beta$ such that:

$$ (x_1,x_2) = \alpha(p_1,p_2)+\beta(q_1,q_2) = r_0 $$

Since adding a line times any scalar $t$ would still maintain the determnant at zero:

$$ (x_1,x_2) = (\alpha+t)(p_1,p_2)+\beta(q_1,q_2) = r_0+t(p_1,p_2) $$

And this is a form of line equation.

Answer 2

The determinant is,

$$\mbox{det}A = q_1x_2+p_1q_2+p_2x_1-p_2q_1-p_1x_2-q_2x_1$$

Since $x$ is on the line segment $L$ we have that $x=(1-\gamma)p+\gamma q$ for some $\gamma\in[0,1]$. Substituting this gives,

\begin{align} \mbox{det}A &= q_1((1-\gamma)p_2+\gamma q_2)+p_1q_2+p_2((1-\gamma)p_1+\gamma q_1)-p_2q_1-p_1((1-\gamma)p_2+\gamma q_2)-q_2((1-\gamma)p_1+\gamma q_1)\\ &=\color{blue}{q_1p_2}\color{red}{-\gamma q_1p_2}\color{green}{+\gamma q_2p_1}\color{orange}{+q_2p_1}\color{purple}{+p_1p_2}\color{brown}{-\gamma p_1p_2}\color{red}{+\gamma q_1p_2}\color{blue}{-q_1p_2}\color{purple}{-p_1p_2}\color{brown}{+\gamma p_1p_2}-\gamma q_2p_1\color{orange}{-q_2p_1} + \gamma q_2p_1 \color{green}{-\gamma q_2p_1}\\ &=0 \end{align}

Answer 3

Remember the determinant is an alternating multilinear form in the rows of a square matrix.

So this is a linear equation in $x_1$ and $x_2$, hence it is the equation of a straight line.

The determinant is $0$ when $(x_1, x_2)=(p_1, p_2)$ or $(q_1, q_2)$ because in each case, two rows are equal, and the determinant is an alternating form. This means the straight passes through each of $M_1=(p_1, p_2)$ and $M_2=(q_1, q_2)$.

This way of writing the equation of a line through a pair of points comes from projective geometry.

Answer 4

$\det A = 0$ if and only if there exists $\alpha, \beta \in \mathbb{R}$ such that $$\pmatrix{x_1 \\ x_2 \\ 1} = \alpha \pmatrix{p_1 \\ p_2 \\ 1} + \beta \pmatrix{q_1 \\ q_2 \\ 1}$$

This gives $\alpha + \beta = 1$ and $$(x_1, x_2) = \alpha p + \beta q = (1-\beta)p + \beta q = p + \beta(q-p)$$

for $\beta \in \mathbb{R}$, which is precisely the equation of a line.