Position of a point$\ \ C(x|y)\ \ $ on a line between $\ \ A(1|1)\ \ $ and $\ \ B(5|2)\ \ $ when $x (i.e. 1)$ away from $A$

Question

How to calculate the position of a point on a line when the start end endpoint of the line and the distance of the point from the start are given$?$

Example:

$A(1|1)$

$B(5|2)$

Distance of point $C$ from $A = 1$

Answer 1

Apologies for the incorrect answer I hastily wrote before.

Since the line passes through (1, 1) and (5, 2) and $C = (x, y)$ also lies on that line, AC has the same slope as AB.

So, $$\frac{y - 1}{x - 1} = \frac{2 - 1}{5 - 1} = \frac{1}{4}$$ $$x-1 = 4y - 4$$ $$x = 4y - 3$$

Now, $$(x-1)^2 + (y-1)^2 = 1$$ Since distance of C from A is 1.

Substituting x = 4y - 3 $$(4y-4)^2 + (y-1)^2 = 1$$ $$=17y^2 - 34y + 16 = 0$$

This has solutions, $y = 1.2425$ which gives $x = 1.97$ and another solution which does not lie between A and B.

So $C = (1.97, 1.2425)$