Straight Line Equation: Intercept & Slope

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We all know that at any instant of the line:

m = dy/dx

integrating on both sides:

y = mx + c

So, how do we know that "c" is the intercept cut on the Y-axis?
 
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You are mistaking the general problem of finding the constant of integration, with the much simpler one you have presented. In general, the constant of integration is not known unless there exist boundary conditions. When integrating m=dy/dx, the answer given is the most general class, or set of functions, which will satisfy your given ODE. The solution set may be something like (mx+c, where c is a real number) in which case there are uncountably many solutions to the ODE. Now, it just so happens that functions of that form are lines. In general, given a function of the form y=f(x), if we wish to inquire at what point (x,y) the y-intercept is, we make the observation that when the function crosses the y-axis, its x value will be zero. In which case we arrive after substitution to the equation y=f(0), which may be solved for y,to find the y-value, of the y-intercept. In your case f(x)=mx+b, therefore we solve y=m(0)+b=b. Implying the y-value at which the line mx+b will cross the y-axis is b.
 
shanu_bhaiya said:
We all know that at any instant of the line:
m = dy/dx
integrating on both sides:
y = mx + c
So, how do we know that "c" is the intercept cut on the Y-axis?

The y-intercept is, by definition, the y value when x= 0. Putting x= 0 in y= mx+ c gives y= c. Therefore, the "c" in the equation is the y-intercept.
 
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