Solve Ax+By=C Equation for Line through (2,-1) & (4,6)

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To solve the equation of the line through the points (2, -1) and (4, 6) in standard form ax + by = c, first calculate the slope, which is 3.5. The y-intercept can be found by substituting one of the points into the slope-intercept form y = mx + b, leading to the equation y = 3.5x - 8. This can be rearranged to standard form as 2y - 7x = -16. The discussion highlights the difference between the standard form and slope-intercept form, noting that the former allows for multiple representations of the same line.
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I know this sounds simple, but my friend insists this is a problem in his textbook, and I can't solve it:

"Give the equation in standard form ax+by=c of the line through the points (2, -1) and (4,6)"

Don't you need one point for every variable? Isn't the standard form y = mx + b?
 
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[y-(-1)]/(x-2)=[6-(-1)]/(6-2)
0=7x-2y-16
 
Can you explain a bit more clearly, please?

Are there infinite possible values for A, B, and C? That's the reason I couldn't solve it.
 
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Point (x,y) is on the line. LHS of the equation find the slope of the line in term of the variable. RHS finds the slope by the 2 given points.

I can't answer your question about the variable a,b and c. the equation comes naturally when it is simplified to the form wanted.
 
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I get it now, I'm not actually solving for 3 variables, because it can be compressed to the y = mx + c form. We're just using integer values to avoid that nasty 3.5
 
If you rearrange the equation you can get to y=(c/b)-(a/b)x which is in the forme y=mx+b

(I hope I'm right)
 
The ax+by=c form is a little more general, as you can describe a vertical line this way (when b=0, a<>0). The drawback is your choice of coefficients is no longer unique, x+y=1 describes the same line as 2x+2y=2 and so on. This was your problem, you have infinitely many acceptable solutions, that are all multiples of one another.

You correct this lack of uniqueness in the y=mx+b form, you're essentially forcing the y coefficient to be 1, which is why you can't get vertical lines (m=infinity is not allowed). Which one you call standard form isn't terribly important (y=mx+b is probably the more commonly used one), as long as you know which form you want your line in.
 
Alkatran said:
I know this sounds simple, but my friend insists this is a problem in his textbook, and I can't solve it:

"Give the equation in standard form ax+by=c of the line through the points (2, -1) and (4,6)"

Don't you need one point for every variable? Isn't the standard form y = mx + b?

y = mx + b is called the slope-intercept form, and is probably more "standard" than the standard form ax + by = c.

To solve the problem, find the slope: m=\frac{6-(-1)}{4-2}=3.5 and y-intercept by substituting into y=mx+b: -1=3.5\cdot2+b\rightarrowb=-1-7=-8, giving y=3.5x-8 or 2y-7x=-16
 
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