P(x,y) forms with A(0,3) and B(0,-3) the vertices of a right triangle with P as the vertex of the right triangle. Find its possible coordinates.

Question

Okay, so this is what I did; I plotted the two given coordinates:

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   A
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---.---
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   |
   B
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and judging from the graph, it seems there are four possible locations...

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 1 A 2
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---.---
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 3 B 4
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Since it's a right triangle, that means that the possible coordinates have to either be (X, -3) or (X, 3)

Slope of AB is 0 based on $$ m_{ab}=\frac{(-3-3)}{(0 - 0)}=0 $$

then I could just find the slope's negative reciprocal (also 0) and substitute it into one of the formula I could use (in this instance, i chose point 1 or 2):

Since, $$ m_{a1}=0 $$

then $$ m_{a1}=\frac{(-3-3)}{(0-X)} $$ $$ 0=\frac{(-3-3)}{(0-X)} $$ $$ 0=\frac{(-6)}{(-X)} $$ $$ 0=\frac{(6)}{(X)} $$

but then what? Or was this question meant to have infinite coordinates?

Thanks in advance!

Answer 1

Hint:

Have you ever heard of the Thales theorem for semi circles?

Edit after comment:

This is then a classic locus exercise. Let $P(x,y)$ denote a possible point. Then according to Pythagoras, you have

• $|\overline{AP}|^2 + |\overline{BP}|^2 = |\overline{AB}|^2$
• $|\overline{AP}|^2 = x^2 + (y-3)^2$, $|\overline{BP}|^2 = x^2 + (y+3)^2$, $|\overline{AB}|^2 = 6^2$
• $\Rightarrow$ the locus of possible points $P(x,y)$ is given by $$x^2 + (y-3)^2 + x^2 + (y+3)^2 = 36 \Leftrightarrow 2(x^2+y^2+9)=36 \Leftrightarrow x^2+y^2 = 9$$ This is a circle around the origin with radius $r=3$. But you need to take out $A$ and $B$ ($x \neq 0$), as $A$,$B$,$P$ form a right triangle.

Answer 2

Hint: it must be $$m_{PA}=\frac{y+3}{x}=-\frac{1}{\frac{y+3}{x}}$$ where b$$m_{PB}=\frac{y+3}{x}$$ where $m$ denotes the slope