Let $\rm A, B$ be distinct points on a line $L$, and let $e$ be a positive constant. By parametrizing $L$ show that when $e \neq 1$ there exist two distinct points $\rm P$ for which ${\rm P A} = e{\rm P B}$, and that when $e = 1$ there is just one.
Let $\rm A =(x_1, y_1)\, B = (x_2, y_2)$, Then $$\mathrm{L}(t) = \big((1-t)x_1 + tx_2, (1-t)y_1 + ty_2 \big)$$
Now if I let ${\rm P} = (a, b)$ and ${\rm P}' = (c, d)$,
Then $$\mathrm {\dfrac{PA}{PB}} = \dfrac{\sqrt{(x_1 - a)^2 + (y_1 - b)^2}}{\sqrt{(x_2 - a)^2 + (y_2 - b)^2}} = e \tag 1$$
I have to prove that $\mathrm{\dfrac{P^\prime A}{P' B}} = e$ for $e \ne 1$.
I tried brute force method by expanding $\mathrm{\dfrac{P' A}{P' B}}$ in distance formula form and equating it to $(1)$ but it does not work, I just end up in a mess.
Also I don't have any idea how to use the parametric form of line here. I think $\rm P^\prime$ would be the reflection of $\rm P$ by the line $L$, but I don't know how to prove it.
Any hints ?
You are making heavy weather of this. Write $\vec a$ for the position vector of $A$ and write $\vec u$ for the vector $\vec {AB}$. Then a typical point on the line joining $A$ and $B$ has position vector $\vec a+t\vec u$. The $|AP|=|t|\,|AB|$ and $|BP|=|1-t|\,|AB|$. So $|AP| =e|BP|$ iff $|t|=e|1-t|$ iff $t=\pm e(1-t)$ etc.