Because the origin is defined as a point. A line is a point moving. In this case arbitrarily at one unit per unit time and the axis goes on forever. A short line is a point moved, given its velocity, \(v\)
\(L=\int^{t}_0{v}dt=\sqrt{\left(\int^{t}_{0}{v(x,\,t)}dt\right)^2+\left(\int^{t}_{0}{v(y,\,t)}dt\right)^2}\)
where \(v(y,\,t)\) is the \(y\) component of \(v\) and \(v(x,\,t)\) is the \(x\) component of \(v\). \(v\) is a vector integrated over time. With,
\(\cfrac{d y}{d t}=a\)
\(\cfrac{d x}{d t}=b\)
and so,
\(\cfrac{d y}{d x}=\cfrac{a}{b}\)
with
\((x_1,\,y_1)\) as the starting point,
\(L=\sqrt{\left(\int^{x_2}_{x_1}{v(x,\,t)\cfrac{1}{b}}dx\right)^2+\left(\int^{y_2}_{y_1}{v(y,\,t)\cfrac{1}{a}}dy\right)^2}\)
\((x_2,\,y_2)\) is the end point after time \(t\), and \(y_2=\cfrac{a}{b}(x_2-x_1)+y_1\) and
\(x_1=b.t\)
Over one unit time,
\(x_2=b + x_1\)
and
\(y_2=a + y_1\)
but,
\(v(y,\,t)=\cfrac{d y}{d t}=a\)
\(v(x,\,t)=\cfrac{d x}{d t}=b\)
so,
\(L=\sqrt{\left(\int^{x_2}_{x_1}{b\cfrac{1}{b}}dx\right)^2+\left(\int^{y_2}_{y_1}{a\cfrac{1}{a}}dy\right)^2}\)
\(L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{b^2+a^2}\)
which is what is expected. Why the fuss? We started with the origin as a point, the rest of the graph is with this point as the lowest denominator. A infinite line is a point with a velocity; a finite length on the graph is a point with a velocity after a finite integral over an arbitrary time interval.
Both length and the axes have a direction and are vectors. A curve is a point with changing velocity. Intersections are paths that crosses irrespective of time. But a collision is in both space and time.
Still what's the fuss? The time dimension enters into the discussion because the issue at hand is space and time in spacetime diagrams.
