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.
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4Are the legs supposed to be parallel to the axes, as the picture kind of suggests? – André Nicolas Mar 18 '16 at 11:47
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No, not always . – Marox Tn Mar 18 '16 at 11:48
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Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 18 '16 at 11:49
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12Then there are infinitely many such triangles. – André Nicolas Mar 18 '16 at 11:49
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I think you guys, got me wrong, what I meant that A and C are two constant points they may be for example $A(4,3)$ a,d $C(2,1)$ and I want to find C – Marox Tn Mar 18 '16 at 11:51
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Could the author of this question please replace $B(x_2, y_2)$ with $C(x_2, y_2)$, otherwise it doesn't make sense? I mean this part <<I want to find only using them the coordinates of $B$>> – rtybase Mar 18 '16 at 11:53
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1@MaroxTn you want to find $B$ – Jens Renders Mar 18 '16 at 11:54
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2For 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. – Jeppe Stig Nielsen Mar 19 '16 at 09:54
2 Answers
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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
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.
Jens Renders
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A and B are constant points, not variable points, is there any relationship between them and C ? – Marox Tn Mar 18 '16 at 11:49
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2You 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 $B$ – Jens Renders Mar 18 '16 at 11:51
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Is every triangle with point B on this circle a right triangle though with a 90 degree angle between segments AB and BC? – Peter Smith Mar 18 '16 at 17:28
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@PeterSmith Yes. It's covered in the answer already: follow the link to "Thales' theorem". – hvd Mar 18 '16 at 17:30
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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 $d$.
Narasimham
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