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I have a right triangle $ABC$. I am given the coordinates of the two points $A(x_1, y_1)$ and $C(x_2, y_2)$. Given points $A$ and $C$, I want to determine the coordinates of $B$. I know there are two solutions for this. I want to find them both.

This image

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    $\begingroup$ Are the legs supposed to be parallel to the axes, as the picture kind of suggests? $\endgroup$ Mar 18, 2016 at 11:47
  • $\begingroup$ No, not always . $\endgroup$
    – Marox Tn
    Mar 18, 2016 at 11:48
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    $\begingroup$ Then there are infinitely many such triangles. $\endgroup$ Mar 18, 2016 at 11:49
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    $\begingroup$ @MaroxTn you want to find $B$ $\endgroup$ Mar 18, 2016 at 11:54
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    $\begingroup$ For completeness, if we do require the legs (catheti) to be parallel to the coordinate axes, there are clearly two solutions, $(x_2,y_1)$ and $(x_1,y_2)$. The latter solution is shown in the illustration. $\endgroup$ Mar 19, 2016 at 9:54

2 Answers 2

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There are many solutions for $B$:

Draw a circle trough the points $A$ and $C$, with diameter $|AC|$, then all points on that circle except $A$ and $C$ are solutions

This is called Thales' theorem

illustration

We can find the points on this circle by first finding the equation of that circle with center

$$O=\dfrac{A+C}{2}$$

Taking your example: $A=(4,3)$ and $C=(2,1)$ we find that

$$O=\dfrac{(4,3)+(2,1)}{2}=\dfrac{(6,4)}{2}=(3,2)$$

And the radius of the circle is $|AO|=\sqrt{1^2+1^2}=\sqrt{2}$

So the equation of that circle is $$(x-3)^2+(y-2)^2=2$$

All points $B=(x,y)$ that satisfy this equation, except $A$ and $C$ make a right triangle with your given points.

Let's solve the equation for $y$:

$$y=\pm \sqrt{2-(x-3)^2}+2$$

So choose a value for $x$ but make sure the part under the square root will not be negative, and this will give you two valid values for $y$!

Example: choose $x=3$ then the formula gives $y=\pm \sqrt{2}+2$ so $$B=(3,\sqrt{2}+2)$$

Is one of many solutions.

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  • $\begingroup$ A and B are constant points, not variable points, is there any relationship between them and C ? $\endgroup$
    – Marox Tn
    Mar 18, 2016 at 11:49
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    $\begingroup$ You mean $A$ and $C$ are constant points? these are the points that you know, and $B$ is the one you are looking for right? The drawing illustrates that all points on the circle ar possible solutions for $B$ $\endgroup$ Mar 18, 2016 at 11:51
  • $\begingroup$ Sorry I meant I have A and C, and I want to know B $\endgroup$
    – Marox Tn
    Mar 18, 2016 at 12:07
  • $\begingroup$ Is it solvable ? $\endgroup$
    – Marox Tn
    Mar 18, 2016 at 12:09
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    $\begingroup$ certainly, I solved it for you $\endgroup$ Mar 18, 2016 at 12:17
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Practically... between two push pins placed $ d= 2 \sqrt 2 $ distance apart press two sides (not hypotenuse) of a set square or triangle touching and same time rotating it. Notice that the vertex making a right angle can be moved to many points, in fact around a circle of diameter $d$.

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