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I have the Cartesian coordinates of the hypotenuse 'corners' in a right angle triangle. I also have the length of the sides of the triangle. What is the method of determining the coordinates of the third vertex where the opposite & adjacent sides meet.

Thanks, Kevin.

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And the downvote is for ? –  Kevin Sep 21 '10 at 6:47
because it doesn't appear that you worked on the problem before asking it. –  anon Sep 21 '10 at 7:15
@maud - I spent 6 hours yesterday on this, going off in the wrong direction as it turns out. Sleeping on it brought no inspiration so I asked the question here. I can take a picture of increasingly muddled scribblings for your review, but a) It wouldn't improve the question, b) the FAQ doesn't indicate its necessary and c) I saw no precedent for this when I searched other related questions. –  Kevin Sep 21 '10 at 9:58
Kevin: mentioning something along the lines of "I tried method X and it went to crap" wouldn't have hurt. –  Guess who it is. Sep 21 '10 at 11:14

3 Answers 3

up vote 5 down vote accepted

You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$ at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$ and $|PB|^2=m^2$ that is $$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$ and $$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$ Subtracting (2) from (1) gives a linear equation. Use this to eliminate one variable from (1). This yields a quadratic equation in the other variable. Solving this will give the two possible positions for $P$.

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Geometrically, subtracting equation 2 from equation 1 is equivalent to finding the radical line (mathworld.wolfram.com/RadicalLine.html ) of the two circles represented by the two equations. –  Guess who it is. Sep 21 '10 at 9:11

While I'd use the same algebra as Robin Chapman's solution, my first thought on this problem yields a third circle equation. The circumcenter of a right triangle is at the midpoint of its hypotenuse. Given the endpoints of the hypotenuse, $A=(a,b)$ and $B=(c,d)$, and letting $h$ be the length of the hypotenuse, the circumcircle has equation $$\left(x-\frac{a+c}{2}\right)^2+\left(y-\frac{b+d}{2}\right)^2=\left(\frac{h}{2}\right)^2.$$ This may look a bit intimidating in symbols, but isn't really any different to work with for solving than the other two circles. I don't know that this offers any advantages, though.

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This is related to the geometric fact that an angle inscribed in a semicircle is a right angle. I suppose there are cases where the algebra with this approach is easier than Robin's, and conversely. –  Guess who it is. Sep 21 '10 at 14:44
@J.M.: This fact can also be thought of as the locus of vertices of right angles where the sides of the angles pass through two given points is a circle. It is interesting to consider this locus for angles of other measures. I doubt the algebra could be much different with this equation than either of Robin's--in either case, the simplest method of solution is probably to expand both circle equations, subtract to get the radical line, and substitute back into one circle to solve. –  Isaac Sep 21 '10 at 15:04

Let $AB$ the hypotenuse, let vector $\vec c=\overrightarrow{OB}-\overrightarrow{OA}$, its length $c$, the right angle at $C$, $DC=h$ the height of the triangle, $a$ and $b$ the given length of the legs, $q$ length of $AD$ as usual. Define $J\colon R^2\rightarrow R^2$, $(v_1,v_2)\mapsto (-v_2,v_1)$ the rotation by 90 degree.

We know by Euclid that $q=b^2/c$ and elementarily that $ab=ch$, so $h=ab/c$. Then $\vec c/c$ is the unit vector of $c$, thus one solution is $$\overrightarrow{OC}=\overrightarrow{OA}+\frac{b^2}{c}\frac{\vec c}{c}+\frac{ab}{c}J\Bigl(\frac{\vec c}{c}\Bigr)=\overrightarrow{OA}+\frac{b}{c^2}\bigl(b\vec c+ aJ(\vec c)\big).$$ Can you find the second solution?

Moral: Avoid coordinates!


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