Equation of a line... This is a really embarrassing question, but could someone remind me why for any (continuous, to make this simple) function defined on $[a,b]$, the lines 
$$L_1 = m(x - a) + f(a)$$
and $$L_2 = m(x - b) + f(b)$$
are equivalent? Where $m = \frac{f(b) - f(a)}{b - a}$
 A: You can verify this straightforward. We assume $m\neq 0$, then we have
\begin{align*}
m(x-a)+f(a) & = m(x-b)+f(b) \\
\Leftrightarrow x-a+\frac{1}{m}f(a) & = x-b+\frac{1}{m}f(b)\\
\Leftrightarrow b-a & =\frac{1}{m}(f(b)-f(a)) \\
\Leftrightarrow 1 & = \frac{1}{m}\underbrace{\frac{f(b)-f(a)}{b-a}}_{=m}\\
& = 1 
\end{align*}
and since $1=1$ is true, the lines are equal. Hope this helps you.
By the way I don't think your question is embarassing, I find it important remind oneself from time to time why certain things we know are true are indeed true.
A: Another way to see this:


*

*Since their slope is the same, they are parallel. Thus we only need to find one common point.

*Clearly, substituting for example the point $b$, yields $L_2=f(b)$ and $L_2=f(b)-f(a)+f(a)=f(b)$

A: Or for the brute force approach, just put them both in $mx + b$ form:
$$L_1 = m \cdot (x - a) + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a} \cdot (x - a) + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x -  \frac {f(b) - f(a)}{b - a}  a + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x +  \frac {-f(b)a + f(a)a + f(a)b - f(a)a}{b - a}$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x +  \frac {f(a)b-f(b)a}{b - a}$$

$$L_2 = m \cdot (x - b) + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a} \cdot (x - b) + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x -  \frac {f(b) - f(a)}{b - a}  b + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x +  \frac {-f(b)b + f(a)b + f(b)b - f(b)a}{b - a}$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x +  \frac {f(a)b - f(b)a}{b - a}$$

$$L_1 = L_2$$

For the less brute force approach, a function $y = f(x - k) + j$ is the function $f$ shift right by $k$ and up by $j$ (this is very important for understanding functions).
Your function is $y = mx$.
So in case $L_1$ we shifted right by $a$ and up by $f(a)$.  In case $L_2$ we shifted right by $b$ and up by $f(b)$.  
$L_2$ is $(b - a)$ more to the right than $L_1$.  $L_2$ is $f(b) - f(a)$ more up than $L_1$.  But that ratio is your slope, so you have the same line.
