<s>[INST] 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 $m = \frac{f(b) - f(a)}{b - a}$
 [/INST] 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.
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