# Right-angled triangles on a graph

My question today is whether or not a concise formula has been discovered for the coordinates along the hypotenuse of a right-angled triangle when plotted on a graph.

I have been working on this and have discovered a formula which seems to work, but of course, it is not definite and I will need your help. If you test it on a few right-angled triangles, I would be very grateful.

However, it does depend on where you plot the triangle and which side the right angle is.

If the right angle sits at ($0, 0$), the formula is $a(b-x)/b$ where $a$ is the side that sits on the y-axis, $b$ is the side that sits on the x-axis and $x$ is the x-coordinate to which you want to find the y-coordinate of along the hypotenuse.

This formula is changed only slightly when the right angle is situated on the other side. I believe this formula to be $ax/b$.

Sorry for such a long-winded question.

Thank you.

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What is the question? The two coordinates are $(0,a)$, $(b,0)$ and the line joining these two points can be characterized in many ways such as the convex sum $t(b,0)+(1-t)(0,a) = (tb, (1-t)a)$ for $t\in [0,1]$. –  copper.hat Jun 13 '14 at 19:56
If you call the vertices of the triangle which are adjacent to the hypotenuse $p$ and $q$. Then what you are looking for is an equation for the line segment joining $p$ and $q$. @copper.hat I don't think the poster is advanced enough to know what the convex sum is. –  James Jun 13 '14 at 19:58

The corners you describe are $(0,a),(b,0)$. A formula for a line through these points is $x \mapsto (1- { x \over b}) a$ (which is the same as the formula you have written above).