What is the general equation for the arclength of a line?

The general equation of a line is y=mx+b
Recall the formula for arc length is $A={\int }_a^b\sqrt{1+(\frac{dy}{dx}{)}^2}dx$
The derivative of the linear function is y'=m
$A={\int }_a^b\sqrt{1+m^2}dx$
m is simply a constant, we can use the power rule to integrate.
$A=[\sqrt{1+m^2}x{]}_a^b$
$A=b\sqrt{1+m^2}-a\sqrt{1+m^2}$
$A=(b-a)\sqrt{1+m^2}$
Now let's verify to see if our formula is correct. Let y=2x+1 and the arc length we wish to find being on the x-interrval [2,6]
$A=(6-2)\sqrt{1+2^2}=4\sqrt{5}$
If we were to use pythagoras, by connecting a horizontal line to a vertical line, we would get the following"
y(2)=5
y(6)=13
$\mathrm{△}y=13-5=8$
$\mathrm{△}x=4$
Thus $A^2={\mathrm{△}}^{2}y+{\mathrm{△}}^{2}x=8^2+4^2$
$A=\sqrt{80}=\sqrt{16\ast 5}=4\sqrt{5}$
As obtained using our formula.

For the arc length of a linear function given its slope m and an interval [a,b], using the arc length formula:
$S={\int }_a^b\sqrt{1+(\frac{dy}{dx}{)}^2}dx$
Let y=mx+b
$\Rightarrow \frac{dy}{dx}=m$
$S={\int }_a^b\sqrt{1+m^2}dx$
This may look scary because of all of the variables, but m is technically just a constant: the slope of the line.
The antiderivative is $\sqrt{1-m^2}\ast x$, and substituting the limits of integration:
$S=\sqrt{1-m^2}\ast b-\sqrt{1-m^2}\ast a$
$S=(b-a)\sqrt{1-m^2}$