A question regarding the point-slope formula : does the formula really hold for any point of the straight line?

Question

I can see only one way to derive the point slope formula, but this derivation also seems to bring a question.

Let $D$ be the straight line of slope $m$ passing through point $P=(a,b)$.

Let $Q=(x,y)$ be any arbitrary point belonging to $D$.

We must have :

$\large\frac {y-b} {x-a}=m \iff y-b = m(x-a) \iff y= m(x-a)+b$.

So, any point $Q=(x,y)$ belonging to $D$ must satisfy this equation.

Now, $P=(a,b)$ is one of these points.

So, isn't there a problem for point $P=(a,b)$ itself, inasmuch as , when $x=a$, the denominator $x-a$ equals $0$, implying that $\frac {y-b} {x-a}=m$ is meaningless.

I can see two options :

(1) either the derivation I consider is not the right one

(2) or the problem I see does not really exist.

Which one is correct?

Answer 1

Your derivation is generally fine, it just needs a little tweak: After you introduce $Q$, say "Either $Q$ is $P$, which trivially satisfies $y = m(x-a)+b$, or $Q$ is not $P$, in which case ....", and then give your argument as before.

And, as @lulu points out, to be 100% valid, you need to exclude vertical lines. Because, in the $Q \ne P$ branch, you'll need to argue that $Q \ne P$ implies $x \ne a$. You might want to use the fact that two different lines can intersect at at most one point, and apply it to your line and the vertical line through $(a,b)$.

Answer 2

The given equation is for finite straight line segment differences of projections.

Using quotient rule of differentiation for infinitesimal segment differences

$$\large\frac {y-b} {x-a}=m=\large\frac {dy/dx} {1}$$.

which makes sense at $(x,y)$ everywhere including at P.