How do you derive Standard form from Slope form?

[Adrian_B]

Can someone show me, and fellow passers-by, how you go from Slope form to Standard form without using specific values, just the variables?
[MATH]m=\frac{y_2-y_1}{x_2-x_1}\rightarrow Ax+By=C[/MATH]What I do know is that from standard form we get the slope-intercept form [MATH]y=mx+b[/MATH] by solving for [MATH]y[/MATH] which means [MATH]m=-\frac{A}{B}[/MATH] and [MATH]b=\frac{C}{B}[/MATH] but I don't know how you deduce the reverse; standard form from slope-intercept form and slope form-- the change in notation confuses me.

Thanks, and I hope this will help a lot of other people too.

 

[lev888]

Can someone show me, and fellow passers-by, how you go from Slope form to Standard form without using specific values, just the variables?
[MATH]m=\frac{y_2-y_1}{x_2-x_1}\rightarrow Ax+By=C[/MATH]What I do know is that from standard form we get the slope-intercept form [MATH]y=mx+b[/MATH] by solving for [MATH]y[/MATH] which means [MATH]m=-\frac{A}{B}[/MATH] and [MATH]b=\frac{C}{B}[/MATH] but I don't know how you deduce the reverse; standard form from slope-intercept form and slope form-- the change in notation confuses me.

Thanks, and I hope this will help a lot of other people too.

You have: y=mx+b
You want: Ax+By=C
Can you move the terms from one side of the equation to the other, so that x and y terms are on the left and the constant term is on the right?

 

[Adrian_B]

So then it's -mx + y = b. Kind of feels like Standard form, if thats what you mean, but what about the variable infront of y and the negative?

 

[Dr.Peterson]

Can someone show me, and fellow passers-by, how you go from Slope form to Standard form without using specific values, just the variables?
[MATH]m=\frac{y_2-y_1}{x_2-x_1}\rightarrow Ax+By=C[/MATH]What I do know is that from standard form we get the slope-intercept form [MATH]y=mx+b[/MATH] by solving for [MATH]y[/MATH] which means [MATH]m=-\frac{A}{B}[/MATH] and [MATH]b=\frac{C}{B}[/MATH] but I don't know how you deduce the reverse; standard form from slope-intercept form and slope form-- the change in notation confuses me.

One important thing to realize is that the "standard form" is not unique. For example x + 2y = 3 and 5x + 10y = 15 represent the same line. (This is related to the fact that m and b correspond to ratios.) So you are really not looking for THE standard form, but for A standard form of the given line.

I commonly clear fractions first, just to make things look nicer, and then move variables to one side and constant to the other.

If you're asking not about what to do with a specific line such as y = -1/2 x + 3/2, but for a general formula, just move the x term in y = mx + b to the other side and you're done!

 

[Adrian_B]

One important thing to realize is that the "standard form" is not unique. For example x + 2y = 3 and 5x + 10y = 15 represent the same line. (This is related to the fact that m and b correspond to ratios.) So you are really not looking for THE standard form, but for A standard form of the given line.

I commonly clear fractions first, just to make things look nicer, and then move variables to one side and constant to the other.

If you're asking not about what to do with a specific line such as y = -1/2 x + 3/2, but for a general formula, just move the x term in y = mx + b to the other side and you're done!

Thanks Dr.P and lev888, this is what I had in mind, in general form or whatever you'd call it, which someone showed me:
[MATH]y=\frac{(y_2-y_1)}{(x_2-x_1)}x + b[/MATH]
[MATH]-\frac{(y_2-y_1)}{(x_2-x_1)}x + y =b[/MATH]
Or

[MATH]((y_2-y_1)x + (x_1-x_2)y) = b(x_1-x_2)[/MATH][MATH]Ax+ By =C [/MATH]
What I need someone to tell me now is why this would be wrong: [MATH]((y_1-y_2)x + (x_2-x_1)y) = b(x_2-x_1)[/MATH] Why do you get rid of the negative by producing [MATH](x_1-x_2)[/MATH]

 

[Dr.Peterson]

Thanks Dr.P and lev888, this is what I had in mind, in general form or whatever you'd call it, which someone showed me:
[MATH]y=\frac{(y_2-y_1)}{(x_2-x_1)}x + b[/MATH]
[MATH]-\frac{(y_2-y_1)}{(x_2-x_1)}x + y =b[/MATH]
Or

[MATH]((y_2-y_1)x + (x_1-x_2)y) = b(x_1-x_2)[/MATH][MATH]Ax+ By =C [/MATH]
What I need someone to tell me now is why this would be wrong: [MATH]((y_1-y_2)x + (x_2-x_1)y) = b(x_2-x_1)[/MATH] Why do you get rid of the negative by producing [MATH](x_1-x_2)[/MATH]

What you're starting with here is what I would call two-point form, written in slope-intercept form. You have done as I suggested and just moved the x term to the left, then cleared fractions to make it look nicer.

The final form is not wrong, but equivalent; that is because \(-(a-b) = -a + b = b-a\). That's a useful fact to be aware of. And the fact that you can swap the two points in the formula reflects the fact that the order of the points doesn't affect the line, or its slope. The line through points A and B is also the line through points B and A.

EDIT: I just looked again and realized your starting form is not very useful, since you use not only two points, but also a separate intercept. So the line will not necessarily pass through those points, but will be some line parallel to that line. The proper two-point form combines the slope from the two points with point-slope form with one of the points. Is there a reason you used what you did?

 

[HallsofIvy]

An important point about "standard form", Ax+ By= C- it is not "unique". We can multiply or divide the coefficients by any non-zero number and get a different equation for the same line.

For example 2x+ 4y= 6, x+ 2y= 3, and 3x+ 6y= 9 all designate the line that crosses the x-axis at (3, 0) and the y-axis at (0, 3/2). The "slope-intercept" form of that is y= (-1/2)x+ 3/2.