ax+by+c =0

apple2357

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So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0

The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0

Why can't this be made to work?
Tell me if i am talking nonsense!
 

pka

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So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0

The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0

Why can't this be made to work?
Tell me if i am talking nonsense!
Your are talking nonsense!
Surely you know that if a line contains \(\displaystyle (x_1,y_1)~\&~(x_2,y_2)\) and if \(\displaystyle x_1\ne x_2\) then its slope is \(\displaystyle m=\dfrac{y_2-y_1}{x_2-x_1}\).
 

apple2357

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Yes I know all that. I just couldn’t explain why the method and approach above fails ?
 

pka

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Yes I know all that. I just couldn’t explain why the method and approach above fails ?
Two linear equations in three unknowns???
 

Dr.Peterson

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So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0

The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0

Why can't this be made to work?
Tell me if i am talking nonsense!
It works fine! Keep going ... (after fixing a typo).

The only problem is that the answer is not unique -- you can multiply an equation of the form ax+by+c=0 by any non-zero number and the resulting equation is equivalent, and has the same form.

So at some point in your work you will be picking an arbitrary value for either a, b, or c.
 

JeffM

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Sep 14, 2012
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So suppose you want the equation of the line that passes through (2,5) and (3, 10) in the form ax+by+c=0

The standard way to do this is to find the gradient and substitute to find the y intercept and tidy up, or you could use y-y_1= m(x-x_1) etc.
I don't understand why it is not possible to substitute the points directly into ax+by+c =0. Two points should be sufficient to find the equation of the line but if you take this approach we have only 2 equations and 3 unknowns?
2a+5b+c = 0
3a+10y+c= 0

Why can't this be made to work?
Tell me if i am talking nonsense!
A system of two equations with three unknowns may have an infinite number of valid solutions.

\(\displaystyle 2a + 5b + c = 0 \text { and } 3a + 10b + c = 0 \implies\)

\(\displaystyle c = -\ (2a + 5b) \implies 3a + 10b - (2a + 5b) = 0 \implies a = -\ 5b.\)

\(\displaystyle \text {Let } b = 1 \implies a = -\ 5 \implies c = -\ (-\ 10 + 5) = 5.\)

\(\displaystyle -\ 5x + y + 5 = 0 \implies y = 5x - 5.\)

\(\displaystyle \therefore x = 2 \implies 5x - 5 = 5 = y, \text { which checks, and}\)

\(\displaystyle x = 3 \implies 5x - 5 = 10 = y, \text { which also checks.}\)
 

JeffM

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Two linear equations in three unknowns???
But it does work, of course. The answer simply is not unique. The uniqueness and simplicity of the resulting equation is the advantage of the traditional methods of finding the equation of a line joining two distinct points. However, simply add b = 1 as a third equation to get the traditional, simple answer.

I think you missed the thrust of the OP. It was asking why a proposed method does NOT work, but it does work. The OP incorrectly assumed that a system of two equations with three unknowns has no valid solutions. The premise of the question was therefore confusing.
 
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apple2357

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Ok. So the method does completely work.
You just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?
 

JeffM

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Ok. So the method does completely work.
You just have to pick a value for b ( which can be anything) and the resulting equation will still be unique ( after simplifying) - so there is no problem with this approach at all?
There is no conceptual problem, but, in terms of communication, it is very common (and therefore readily comprehensible) to state a linear equation as

\(\displaystyle y = ux + v.\)

This results from setting b = 1. This form immediately gives the slope and the y-intercept and so is mathematically useful.
 
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HallsofIvy

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The problem is that while any straight line can be written as ax+ by+ c= 0 that form is not unique! Multiplying each term by d gives a'x+ b'y+ c'= 0, where a'= ad, b'= bd, and c'= cd, a different equation for the same line. If you (arbitrarily) take a= 1 then you have x+ by+ c= 0. The line passes through (2,5) and (3, 10) so you have the two equations, 2+ 5b+ c= 0 and 3+ 10b+ c= 0, to solve for the two unknowns, b and c.
 

apple2357

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That makes sense!
 

Jomo

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The line y=3x+4 and the line 2y=6x+8 both have the same exact points! The form y = 3x+4 or y = mx+b just insists that the coefficient of y is 1. In ax+by +c=0, b does not have to be 1!
For the record, the form is usually written as ax+by = c
 
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