MHB Real Algebraic Curves: Solving Example 1.4 from C.G. Gibson's Book

[Math Amateur]

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:
View attachment 4562
Question 1In the above text, Gibson writes the following:

" ... ... Then a brief calculation verifies that any point $$p + t(q - p) = (1- t)p + tq$$ 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 $$3 \times 3$$ matrix of coefficients is singular, so the rows $$(p_1, p_2 , 1) , ( q_1, q_2, 1) , ( r_1, r_2, 1)$$ are linearly independent. ... ... "

Can someone lease explain how we know that the $$3 \times 3$$ matrix of coefficients is singular?
Hope someone can help with the above two questions ...

Peter

 

[GJA]

Hi Peter,

I will try to address each of your questions individually below.

Question 1

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 ... ...

Before showing the computation I thought I would try to give a hint that may help you to see what the author is intimating at. The key is to determine what $a$ and $b$ are given that $p$ and $q$ are distinct points on the line. Think of what they would be if we were to write the equation of the line through $p$ and $q$. Once you have that, it may be easier to see what needs to happen.

Question 2

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

This matrix is singular (i.e. not invertible) because the linear transformation to which the matrix corresponds is not one-to-one: it sends $(0,0,0)^{T}$ and the non-zero vector $(a,b,c)^{T}$ to $(0,0,0)^{T}.$

Let me know if anything is unclear/not quite right. Good luck!

 

[Math Amateur]

Hi Peter,

I will try to address each of your questions individually below.
Before showing the computation I thought I would try to give a hint that may help you to see what the author is intimating at. The key is to determine what $a$ and $b$ are given that $p$ and $q$ are distinct points on the line. Think of what they would be if we were to write the equation of the line through $p$ and $q$. Once you have that, it may be easier to see what needs to happen.
This matrix is singular (i.e. not invertible) because the linear transformation to which the matrix corresponds is not one-to-one: it sends $(0,0,0)^{T}$ and the non-zero vector $(a,b,c)^{T}$ to $(0,0,0)^{T}.$

Let me know if anything is unclear/not quite right. Good luck!

Hello GJA,

In answer to my Question 1, you write the following:

"Before showing the computation I thought I would try to give a hint that may help you to see what the author is intimating at. The key is to determine what [FONT=MathJax_Math]a and [FONT=MathJax_Math]b are given that [FONT=MathJax_Math]p and [FONT=MathJax_Math]q are distinct points on the line. Think of what they would be if we were to write the equation of the line through [FONT=MathJax_Math]p and [FONT=MathJax_Math]q. Once you have that, it may be easier to see what needs to happen."

Following what you suggest, I proceeded as follows:https://www.physicsforums.com/attachments/4567As you can see, at this point I have come to a bit of a dead end ...

Are you able to help further ...

Peter

 

[Math Amateur]

Hi Peter,

I will try to address each of your questions individually below.
Before showing the computation I thought I would try to give a hint that may help you to see what the author is intimating at. The key is to determine what $a$ and $b$ are given that $p$ and $q$ are distinct points on the line. Think of what they would be if we were to write the equation of the line through $p$ and $q$. Once you have that, it may be easier to see what needs to happen.
This matrix is singular (i.e. not invertible) because the linear transformation to which the matrix corresponds is not one-to-one: it sends $(0,0,0)^{T}$ and the non-zero vector $(a,b,c)^{T}$ to $(0,0,0)^{T}.$

Let me know if anything is unclear/not quite right. Good luck!

Hello again GJA ... thought I would try a vector calculus approach ... seems that this gets around Gibson's "brief calculation" ... but note that I would very much like to be able to understand your suggested approach ...But anyway, to articulate the vector analysis approach to derive

$$p + t(q-p) = (1-t)p +tq $$

for a line through $$p = (p_1, p_2)$$ and $$q = (q_1, q_2)$$

using the approach from Susan Colley's book, Vector Calculus ... we proceed as follows:Following Colley (page 13) ... ... but using Gibson's symbols/notation we have the following ... In general, given two arbitrary points

$$ P = p = (p_1, p_2) $$ and $$ Q = q = (q_1, q_2) $$

the line joining them has a vector parametric equation as follows: (see Figure 1 below)

$$r(t) = OP + t PQ $$

That is

$$r = p + t(q-p) = (1-t)p + t q$$ where $$r = (x,y) $$
Figure 1 visually "explains" the derivation ...
View attachment 4568Note that I would still want to see your approach GJA as that really answers what Gibson's "brief calculation" consists of ...

Thanks again for all your help GJA ...

Peter

 

[GJA]

Hi Peter,

I was thinking that if we use the slope-intercept form of a line
$$y=mx+b$$
then we know
$$m=\frac{q_{2}-p_{2}}{q_{1}-q_{2}},$$
so that after clearing the fraction
$$a = -(q_{2}-p_{2})=p_{2}-q_{2}$$
and
$$b = q_{1}-p_{1}$$
Does that seem to point things in the right direction? Let me know if anything is unclear/not quite right.

 

[Math Amateur]

Hi Peter,

I was thinking that if we use the slope-intercept form of a line
$$y=mx+b$$
then we know
$$m=\frac{q_{2}-p_{2}}{q_{1}-q_{2}},$$
so that after clearing the fraction
$$a = q_{2}-p_{2}$$
and
$$b = q_{1}-p_{1}$$
Does that seem to point things in the right direction? Let me know if anything is unclear/not quite right.

Hi GJA,

Thanks again ...

I follow that

$$m=\frac{q_{2}-p_{2}}{q_{1}-q_{2}},$$

but then, how do you get

$$a = q_{2}-p_{2}$$
and
$$b = q_{1}-p_{1}$$ ... ... ?Can you be more explicit ...

Sorry ... probably being a bit slow ...:(*** EDIT ***

Just noticed what I think is a typo ...

... should it be $$m=\frac{q_{2}-p_{2}}{q_{1}-p_{1}},$$ ??

Peter

 

[Math Amateur]

Hi GJA,

Thanks again ...

I follow that

$$m=\frac{q_{2}-p_{2}}{q_{1}-q_{2}},$$

but then, how do you get

$$a = q_{2}-p_{2}$$
and
$$b = q_{1}-p_{1}$$ ... ... ?Can you be more explicit ...

Sorry ... probably being a bit slow ...:(*** EDIT ***

Just noticed what I think is a typo ...

... should it be $$m=\frac{q_{2}-p_{2}}{q_{1}-p_{1}},$$ ??

Peter

Hi again GJA,

OK now ... just derived it from your post ... can see it now ... indeed, I was being slow ...

thanks for for all your help ...

Peter