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If we were given two points on a linear equation $(x_1,y_1),(x_2,y_2)$, it is quite easy to find the slope and use substitution to find the slope intercept form $y=mx+b$, to graph it.

Is it possible to solve for $b$ strictly in terms of $x_1,y_1,x_2$, and $y_2$?

Saketh Malyala
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  • Why not? $y_1=mx_1+b$ and $y_2=mx_2+b$ gives $b=y_1-mx_1=y_2-mx_2$, so $m=y_1-y_2/(x_1-x_2)$ (how is that for a surprise). Now plug this back in in either of the first two equations. – B. Pasternak Apr 08 '16 at 22:53
  • Substitute $y=y_1$, $x=x_1$, $m=\frac{y_2-y_1}{x_2-x_1}$ and solve for $b$. – molarmass Apr 08 '16 at 22:54

2 Answers2

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Yes it is. As you say, we can find the slope: $$m=\frac{y_2-y_1}{x_2-x_1}.$$ Thus, $$y_1=mx_1+b=\frac{y_2-y_1}{x_2-x_1}x_1+b,$$ so $$y_1-\frac{y_2-y_1}{x_2-x_1}x_1=\frac{x_2y_1-x_1y_2}{x_2-x_1}=b.$$

Plutoro
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Yes, by imposing that the three points$(0,b),(x_1 ,y_1 ),(x_2 ,y_2 )$ are aligned (determinant, or other way you know).

G Cab
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