# How to find 'b' in straight line equation?

In a straight line equation ' y = mx + b ' if i have 'm' and 'x' how can i get 'b' (where the line cross the Y axis)? I search on the internet they said to get it by drawing it and see where it cross the Y axis, and this not possible since the numbers am dealing with is very large.

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How is that calculus? – user824294 Oct 30 '12 at 7:40
Will you please share more information on the set-up of the problem? Typically $x$ is not a constant, so you do not "have" it. It is a variable. You may be given a particular pair $(x,y)$ that satisfies the equation in addition to $m$. Is that what you mean? – Jonas Meyer Oct 30 '12 at 7:41
@Swaiss: Could you please give an example of what information you are given? – Jonas Meyer Oct 30 '12 at 7:46
this is two points on the line am working on it: point 1 = (419445.4712 , 1683997.1179) point 2 = (419438.9394 , 1683999.5773) – IBRA Oct 30 '12 at 7:55
@Swaiss: Your problem with two points is not what you described previously. Just knowing $m$ and a value of $x$ would not tell you much: Every value of $x$ will be part of a solution to an equation with a given $m$. But with two points, we're getting somewhere. If you know how to find $m$, and you also know a particular solution $(x,y)$, which can be given by either of your two points, plug in those three numbers into the equation $y=mx+b$ to leave $b$ as the only unknown, then solve. See also purplemath.com/modules/strtlneq.htm – Jonas Meyer Oct 30 '12 at 7:58

It looks like you have two points $(x_1,y_1)$ and $(x_2,y_2)$ so you can set up two simultaneous equations
$$y_1=mx_1+b$$$$y_2=mx_2+b$$Multiply the first by $x_2$ and the second by $x_1$
$$y_1x_2=mx_1x_2+bx_2$$$$y_2x_1=mx_1x_2+bx_1$$subtract:$$y_1x_2-y_2x_1=b(x_2-x_1)$$ and go from there.
You might also like to investigate the form:$$y=y_1\frac{(x-x_2)}{(x_1-x_2)}+y_2\frac{(x-x_1)}{(x_2-x_1)}$$ which is a direct way of writing the equation of a line through two points. The form can be developed to give the equation of a quadratic through three points, and generally the lowest degree curve through $n$ points.
You could also get the slope $m$ from $m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$, and then use either equation to get $b$. – littleO Oct 30 '12 at 8:41