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I have a triangle on the cartesian plane where I know the following: $$ A = (x_1,y_1), B = (x_2,y_2), C = (x_3,y_3) $$

I want to find the possible locations for $C$.

I know the location of $A$ and $B$.

I know that I want the angle $ABC$ will be $90^{\mkern1mu\mathrm{o}}$.

and for narrowing the search down, I know that if: $$ AB = x $$ then:

$$ AC = 2x $$ example triangle

maryfrank
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    If $AB$ is the hypotenuse of the rectangle triangle $ACB$ it is impossible that $AB=AC/2$. – Emilio Novati Jan 27 '20 at 17:22
  • Try https://math.stackexchange.com/questions/158679/how-to-calculate-coordinates-of-third-point-in-a-triangle-2d-knowing-2-points or https://math.stackexchange.com/questions/2156851/calculate-the-coordinates-of-the-third-vertex-of-triangle-given-the-other-two-an – David K Jan 27 '20 at 23:38

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There can be no solution: if $\widehat{ACB}$ is a right angle, C is on the circle with diameter $AB$, and if $AC=2AB$, it is on the circle with centre $A$ and radius $2AB$. These two circles have no intersection.

Bernard
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