How two line segments are not equal if function of one line is ($f(x)=x$) and of another ($g(x)=cx))$

Question

Line $c=g(x)=cx$;
Line $b=f(x)=x$

If for every point $(x,0)$ on line $(b)$ ,there exists a point $(x,cx)$ on line $(c)$ and vice versa,then how two given line segments are not equal (considering this reason).

Answer 1

Consider points $A(x_A, y_A) $ and $B(x_B, y_B)$:

If these points are on line b then:

b: $y=x$; $m=\frac{y_B-y_A}{x_B-x_A}=1$

And the length of AB is:

$AB=\sqrt{(x_B-x_A)^2+(y_B+y_A)^2}=(x_B-x_A)\sqrt{1+1=2}$

The gradient of line c; (y=cx) is $c$, so it can have one intersection with line b; (y=x), let this intersection is point A. Then point B will have different coordinates to when it was located on line b.Let mark this point as B', we have, $x_B=x_B'$ due to statement also:

$\frac{y_B'-y_A}{x_B-x_A}=c$

Then the length of AB' is:

$AB'=\sqrt{(x_B-x_A)^2+(y_B'+y_A)^2}=(x_B-x_A)\sqrt{1+c}$

Answer 2

They are not congruent!

In geometry "congruent" means that they can be overlapped and clearly $b=3$ and $c=5$ can't.

The bijection you found is all another concept. It proves that any point of $b$ is related to one and only one point of $c$ and viceversa. The set of points of segment $b$ and the set of points of $c$ have the same cardinality.

Shocking fact: it can be proved that $b$ has the same cardinality of $\mathbb{R}^3$.