-1

I am going through one of the the scan conversion techniques and it mentions the implicit equation of line as follows:

$$F(x,y) = ax + by + c = 0 .$$

The text also mentions that $F(x,y)=0$ if the point is on the line without any proof. I am looking for a proof.

gebruiker
  • 6,154
sajis997
  • 105

1 Answers1

1

In learning coordinate geometry, we usually start with plugging some $x_{i}$ into $ax+by+c=0$ to find an unknown $y_{i}$ with known $a, b, c$. Then we observe that these ordered-pairs $(x_{i},y_{i})$ are always collinear in Cartesian plane. Here, the process are reversed. A point $(x',y')$ satisfying $F(x',y')=0$ where $F(x,y)=ax+by+c$. Usual text seldom provides a proof. If so, we need to define what is a straight line on Cartesian plane. Teacher may use dynamic software to illustrate the idea by varying a point on the graph and checking the relation.

Ng Chung Tak
  • 18,990
  • What software did you use to illustrate the image above ? It is very well demostrated. – sajis997 Jan 17 '16 at 17:51
  • 1
    Geogebra which is a freeware. – Ng Chung Tak Jan 17 '16 at 18:48
  • There are more issues here: The text also mentions that F(x,y) is positive for points below the line and negative for points above the line. Now if take the points P(-1,3) and P(-1,1), I am getting a something totally opposite. Lets take P(-1,3), I am getting F(-1,3) > 0 and for P(-1,1) , F(-1,1) < 0. Any explanations ? – sajis997 Jan 17 '16 at 18:48
  • Need to rearrange $F(x,y)=0$ into $y=-\frac{ax+c}{b}$, if $y'>-\frac{ax'+c}{b}$, then the point is above the graph. If $b=0$, check $x', -\frac{by'+c}{a}$ instead (right or left). – Ng Chung Tak Jan 17 '16 at 18:55
  • The text did not mention anything about the derivative here that you are suggesting. Should it not work with the just mapping the example coordinate points I mentioned ? Could you explain the purpose of taking derivative here ? – sajis997 Jan 17 '16 at 19:05
  • Lazy notation: $(x',y')$ is a point for testing. If you don't like it, use $(x_{1},y_{1})$. – Ng Chung Tak Jan 18 '16 at 03:47
  • Thanks, I do not understand why do we have re-arrange to proof F(x,y) > 0 or F(x,y) < 0. The example for F(x,y) = 0 with a sample line in the form ax + by + c map very well with the sample coordinate point. Should not the remainings work in the same manner. – sajis997 May 31 '16 at 07:00
  • @sajis997 Have you learnt slope-intercept form in your school curriculum? That's what I transformed to and it is convenient for graph sketching, illustrating inequalities and linear programming. – Ng Chung Tak May 31 '16 at 08:29
  • While proving the proof for F(x,y) = 0, you did the following : 1. ax + by + c = 0 as the standard equation of line in implicit form is F(x,y) = ax + by + c. 2. You defined some values for the constants a, b ,c and a sample point and then plot the point to proof the case, no re-arrangment was necessary. Why not follow the same steps to proof F(x,y) < 0 and F(x,y) > 0 – sajis997 May 31 '16 at 08:48
  • Look back to your very very earlier comments: $F(x,y)$ is positive for points below the line and negative for points above the line which is not sufficient coz you need to check the sign of $a$ and $b$. What I'd tried to express $y=g(x), b\neq 0$ for easy identification. – Ng Chung Tak May 31 '16 at 08:52
  • Exactly, F(x,y) > 0 for points below the line. Then I picked a point (-1,1) and plotted it (By the way - (-1,1) is below the line from your example line) . To my surprise I found F(x,y) = 3x + 2y + 1 < 0. This is where the confusion starts. Then you suggested to re-arrange, but I did not understand why do I have to re-arrange. – sajis997 May 31 '16 at 09:00
  • The graph $y=g(x)$, generally whatever well-defined function $g(x)$, divides the Cartesian plane in two regions one is $y>g(x)$ (the region above the graph) and $y<g(x)$ (the region below the graph) and of course $y=g(x)$ is the boundary among the two regions. – Ng Chung Tak May 31 '16 at 09:10