Real Algebraic Curves

Jan 2010
594
5
Hobart, Tasmania, Australia
I am reading C. G. Gibson's book: Elementary Geometry of Algebraic Curves.

I need some help with aspects of Example 1.4

The relevant text from Gibson's book is as follows:



Gibson - Example 1.4.png



Question 1


In the above text, Gibson writes the following:

" ... ... Then a brief calculation verifies that any point \(\displaystyle p + t(q - p) = (1- t)p + tq\) also lies on the line ... ... "

I am unable to perform the brief calculation that Gibson refers to ... can someone please help me by showing the calculation and how it works ... ...



Question 2



" ... Since at least one of a,b is non-zero, the system has a non-trivial solution, By linear algebra the \(\displaystyle 3 x 3\) matrix of coefficients is singular, so the rows \(\displaystyle (p_1, p_2 , 1) , ( q_1, q_2, 1) , ( r_1, r_2, 1)\) are linearly independent. ... ... "

Can someone lease explain how we know that the \(\displaystyle 3 \times 3\) matrix of coefficients is singular?



Hope someone can help with the above two questions ...

Peter
 
Jul 2015
217
116
Ilford
Q1)

$ap_1+bp_2+c\ =\ 0\ \ldots\ \fbox1$

$aq_1+bq_2+c\ =\ 0\ \ldots \fbox2$​

$\fbox2-\fbox1:\quad a(q_1-p_1)+b(q_2-p_2)\ = 0\ \ldots\ \fbox3$

$\fbox1+t\times\fbox3:\quad a\left[p_1+t(q_1-p_1)\right]+b\left[p_2+t(q_2-p_2)\right]+c\ = 0$

$\implies\ \left(p_1+t(q_1-p_1),p_2+t(q_2-p_2)\right)$ lies on the line.


Q2) We have

$\begin{pmatrix} p_1 & p_2 & 1 \\ q_1 & q_2 & 1 \\ r_1 & r_2 & 1 \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix}\ =\ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$​

If the 3×3 matrix were not singular we could multiply by its inverse to get $a=b=c=0$. But we are given that “at least one of $a$, $b$ is non-zero”.

It follows that the points $p,q,r$ are linearly dependent. Hence there can only be at most two linearly independent points, $p$ and $q$, on a straight line; all other points on the line are linear combinations of these two points (and conversely all linear combinations of them lie on the line).
 
Last edited: