Determinant of matrix and equation of a line

Question

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two points in the plane. Show that the equation of the line through $P$ and $Q$ is given by $\det(A) = 0$, where $$ A = \left [ \begin{array}{ccc} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{array}\right ] $$

How do I show that the points are given by the matrix and how do I find the slope?

score 1 · Answer 1 · answered Dec 01 '13 at 23:24

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When the entries of a matrix are vectors, the determinant of a matrix is $0$ precisely when the vectors are linearly-dependent. If $(x_1,y_1), (x_2,y_2)$ are linearly-dependent, then...

answered Dec 01 '13 at 23:24

user99680

6,708

Do I multiply matrix A by P and Q? – antotony Dec 01 '13 at 23:28
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No; use the entries $a_{11}=x_1$ ; $a_{12}=y_1$; $a_{21}=x_2$ and $a_{22}=y_2$. If $Det=0$ , then $(x_1,y_1), (x_2,y_2)$ are linearly-dependent, which means they are on the same line. – user99680 Dec 01 '13 at 23:31
I don't understand. Isn't a11 = x? and a12 = y? – antotony Dec 01 '13 at 23:35
BEcause your two points are $(x_1,y_1)$ and $(x_2,y_2)$ – user99680 Dec 01 '13 at 23:36
Ah, sorry, it is a bit different after the edit. Let me think. – user99680 Dec 01 '13 at 23:38
Can this problem be solved by using the vandermonde determinant? – antotony Dec 02 '13 at 00:42
I don't think your matrix is a Vandemonde matrix; you don't have $x_i=y_i^{\alpha}$. But expanding the determinant should help give you the solution. – user99680 Dec 02 '13 at 00:44
This is what I have so far: (x-x2)y1 + (x2-x1)y + (x1-x)y2 = 0. That is what I get when computing the determinant of A. Where do I go from here? – antotony Dec 02 '13 at 01:07

Determinant of matrix and equation of a line

1 Answers1