Why does this have to be $f(0)=g(0)$? For the problem, I am not given any solution so no idea

Prove that any two continuous maps $f,g; I \to X$ such that
$$f(0)=g(0) \in X$$
are homotopic where $I=[0,1]$ is the unit line.

I just thought, well

$I \in \mathbb{R}$ which is convex, so I can have $h(s,t)=(1-t)f(s)+tg(s)$ as my homotopy.

...No? how should

$$f(0)=g(0)$$

come into play to determine the solution? My reasoning would tell us in fact that any paths are homotopic which shouldn't be the case. I don't understand how I can solve this.

If, for example, $X$ is not path-connected and $f(0)$ is in a different path component from $g(0)$, then $f$ and $g$ can't be homotopic.  But if $f(0) = g(0) = x$, then you can "homotope" both $f$ and $g$ to be the constant map $I \to X$ with value $x$, and therefore $f$ and $g$ are homotopic.
As @JackLee points out, your construction of a homotopy does not make sense because in general $X$ there is no notion of multiplying points by numbers or adding points together.
