If points (0,0), (1,0), and (x,y) are the vertices of a right triangle, determine any equations that x and/or y must satisfy

Question

I know that the x can be 0, 0.5, 1 and while x is 0 or 1 y can be anything be when x is 0.5 y is 0.5 as well, but I can't find the equation that satisfies all of them.

Answer 1

There is not one equation. There are two lines that $(x,y)$ can be on a circle. You should find an equation for each line and the circle. There is a way to combine the equations, but it obscures what is going on a little.

The lines are $x=0$ and $x=1$. The circle is centered at $(\frac 12,0)$ with radius $\frac 12$, so is $(x-\frac 12)^2+y^2=\frac 14$ We can combine all that into $$x(x-1)\left[\left(x-\frac 12\right)^2+y^2-\frac 14\right]=0$$

Answer 2

For $A(0,0);B(1,0);C(x,y)$

If $m_{AB}$ is the gradient of $AB$

We exactly one of the following to be $-1$

$m_{AB}\cdot m_{BC}$

$m_{AB}\cdot m_{CA}$

$m_{BC}\cdot m_{CA}$

Answer 3

By Pytagora's theorem, the triangle is right exactly when one of the following equations holds $$x^2+y^2=1+(x-1)^2+y^2 \Leftrightarrow 1-x =0 \mbox{ or }\\ x^2+y^2+1=(x-1)^2+y^2 \Leftrightarrow x =0 \mbox{ or }\\ x^2+y^2+(x-1)^2+y2=1 \Leftrightarrow x^2-x+y^2=0 $$

Therefore, the triangle is right exactly when $$(1-x) \cdot x \cdot \left( x^2-x+y^2 \right) =0$$

Of course, as some of the comments point out, you need to know that $(x,y) \neq (0,0)$ and $(x,y) \neq (1,1)$, as these are the solutions which do lead to "degenerate" right triangles.

Answer 4

There are three possibilities for which side of the triangle $\vec{(0,0),(1,1)}$ is.

The first two are trivial, they have $x=0$ or $x=1$ with $y\neq0$.

The third is when the line is the hypotenuse. Observe the diagram. I've rotated the triangle, $(0,0)$ is at the angle marked $theta$, $(1,1)$ is at the other non-right angle. $S=\sin\theta, C=\cos\theta$ for reference.

We use the Inverse Pythagorean Theorem, $$\frac{1}{h^2}=\frac{1}{S^2}+\frac{1}{C^2}$$ Which yields $$\frac{1}{h^2}=\frac{S^2+C^2}{(SC)^2}\to h=SC\because S^2+C^2=1$$

Then we find $t$ using the regular Pythagorean Theorem, $$t^2=C^2-(SC)^2\to t^2=C^2(1-S^2)=C^4\to t=C^2\implies 1-t=S^2$$

Hence the third triangle looks like the below, we have: $$x=\cos^2\theta$$ $$y=\cos\theta\sin\theta$$