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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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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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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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