Does the slope of the line stay constant if we change $x_0$?

Question

Let $y$ be a linear function and let $x_0$ be an arbitrary number in its domain. So far I've proved that $\forall x \in \mathbb R \setminus{\{x_0}\}:\dfrac{y(x)-y(x_0)}{x-x_0}=const.=m_a$, allowing us to write the function's equation as $f(x)=f(x_0)+m_a(x-x_0)$.

If we change the arbitrary point we started with to some $x_1$, does the newly defined $m_b=\dfrac{y(x)-y(x_1)}{x-x_1}$ equal $m_a$? How to prove this?

Answer 1

The answer is yes. By definition of a linear function, $y(x)=ax+b$ for some $a,b\in\mathbb{R}$, which are in fact unique. Indeed, if $a',b'\in\mathbb R$ such that $y(x)=a'x+b$ for all $x\in\mathbb{R}$, then $$\forall x\in\mathbb{R},\,ax+b=a'x+b'\tag{1}$$ for all $x\in\mathbb{R}$. Choose some $x\in\mathbb{R}$ such that $(1)$ implies that $b=b'$, and then use $(1)$ again to show that $a=a'$.

Now, given $x_0,x_1\in\mathbb R$, and using the formula $y(x)=ax+b$ of $y$, what can you get?

Answer 2

$x_0$ is arbitrary. If a property $P$ over a domain $D$ holds for an arbitrary $x_0 \in D$, it holds for all $x \in D$