Why is the ratio of the slope of a line always equal? Imagine you have one triangle and the hypotenuse is a line $AB$. 
Now the slope of this triangle is defined as:
$$ m =\frac{y_i-y_1}{x_i-x_1} $$
$i$ = any number on the line $AB$
Now instead if you were to take a few points on this line $AB$ and performed the operation:
$$ \frac{y_5}{x_5} =k_5, \frac{y_4}{x_4}=k_4, \frac{y_3}{x_3}=k_3....... $$
It turns out that
$$ k_5 =k_4 = k_3.......$$
Now my question is why is the ratio of the slope at every point on the line $AB$, equal?
Why does $k_5=k_4=k_3$ and so on
 A: The equation of any line can be written in slope-point form.
$$y-y_1=m(x-x_1)$$
Leaving it in this form, we simply divide over the $(x-x_1)$
$$\frac{y-y_1}{x-x_1} = m$$ 
Holds true for any line. Because $(y_1, x_1)$ was any fixed point on the line, and $(x,y)$ is defined as a variable point, we can see this holds true for all points on the line.
Thus, there is always a constant ratio, $m$ for each pair of points on the line.
A: I think the comment from N.S.John is particularly useful, so let me expand it a bit for you.  Suppose you compute the slope of $AB$ using two points on it, say $P$ and $Q$, so the slope you get is the quotient $y/x$ of the vertical distance $y$ by the horizontal distance $x$ between those points. Notice that these distances are the two sides of a right triangle whose hypotenuse is the segment $PQ$ of the line $AB$.  You get this triangle by drawing a horizontal line through $P$ and a vertical line through $Q$, and finding their intersection point, which I'll call $X$. The triangle is $PXQ$ and the slope you computed is the ratio of lengths $(XQ)/(XP)$.  
Now suppose I come along and compute the slope of the same line $AB$ using two other points on it, say $P'$ and $Q'$.  As before, form a right triangle $P'X'Q'$, where the slope that I computed is the ratio of lengths $(X'Q')/(X'P')$. Your question is why my slope and your slope are the same.
The reason is that the triangles $PXQ$ and $P'X'Q'$ are similar.  The angles at $X$ and $X'$ are equal because they are both right angles.  The angles at $P$ and $P'$ are equal because they are corresponding angles where the parallel (horizontal) lines $PX$ and $P'X'$ are cut by the transversal $AB$.  Those equalities between angles imply similarity of the triangles.  And similarity implies, in turn, equality of ratios of corresponding sides --- which is exacly what you wanted.
A: A line is defined by all the points $(x,y)$ of the plane verifying $ax+by+c=0$ for given $a,b,c$. So the slope of a line is always constant ($\frac{b}{a}$). I am guessing that in your case, you have $c=0$ (your line goes through $(0,0)$), so calculating $\frac{y_i}{x_i}$ is equivalent to calculating the slope (hence gives a constant ratio).
