Equation of a straight line

jcf

y = mx + c

but the other form y-y1 = m(x-x1)


how is it derived from y = mx + c

y - y1 = m(x -x1)
=> y = mx - mx1 +y1
=> y = mx + (y1-mx1)
=> y = mx + c

c = (y1 - mx1)

jcf

Deleted User wrote: »

y - y1 = m(x -x1)
=> y = mx - mx1 +y1
=> y = mx + (y1-mx1)
=> y = mx + c

c = (y1 - mx1)

Thanks, from y = mx + c , c = y - mx, but is this the same as

c = (y1 - mx1) ?

yup,

if (y1 - mx1) = c = y - mx

=> y1- mx1 = y - mx
=> y - y1 = mx - mx1
=> y - y1 = m(x-x1)

and we're back to the start.

jcf

but is there a derivation from y= mx + c as a starting point ?

sponsoredwalk

If you think of the the slope formula

[latex]m = \frac{y-y_0}{x - x_0}[/latex]

you can derive everything from this.

From

[latex]m = \frac{y-y_0}{x - x_0}[/latex]

we get

[latex]y -y_0 = m(x - x_0)[/latex].
Set [latex] x_0 = 0[/latex] & then you have [latex]y -y_0 = m(x - 0)[/latex].
So [latex]y -y_0 = mx [/latex], or [latex]y = mx + y_0[/latex]
This is also written as [latex]y = mx + c[/latex], where [latex] c = y_0[/latex] is the y-intercept.
Thinking about it this way we see

[latex]y = mx + c[/latex]

is just a special case of the more general equation.

If you want to derive it backwards then:

[latex]y = mx + c[/latex]
[latex]y = m(x - 0) + c[/latex]
[latex]y - c= m(x - 0)[/latex]


This is the point-slope equation of the line through [latex](0,c)[/latex].
For the point-slope equation of the line through [latex](x_0,y_0)[/latex]
we have [latex]y -y_0 = m(x - x_0)[/latex].