I was given this problem:
Without drawing a graph and given the following 3 vertices, find the coordinates of the last vertex of the square: $(3,2)$$(0,5)$$(-3,2)$
So first, I found the length of the sides and the diagonal of the square, which are $\sqrt{18}$ and $6$ respectively. By graphing, I know the solution is $(0, -1)$. Then, I assume that since the length between $(3,2)$ and $(-3,2)$ is the diagonal, then the distance between $(0,5)$ and the remaining vertex must be the diagonal too. And since the length of the side is 6, then the distance between the vertex and either $(3,2)$ or $(-3,2)$ must be 6. So:
$$\sqrt{(x-3)^2+(y-2)^2}=\sqrt{18}$$ $$\sqrt{(x-0)^2+(y-5)^2}=6$$
Which gives (after a bit of cleaning up):
$$x^2+y^2-10y=11$$ $$x^2-6x+y^2-4y=5$$
Then, replacing the second expression into the first one:
$$x^2-6x+y^2-4y=5 \Rightarrow x^2=5+6x-y^2+4y$$ $$\require{cancel}5+6x-\cancel{y^2}+4y+\cancel{y^2}-10y=11$$ $$5+6x+4y-10y=11$$ $$6x-6y+6$$ $$x-y=1$$ $$x=1+y$$
Up to this point, I know I'm not entirely wrong because the expression is true for the actual coordinates of the vertex, because $0=1+(-1)$ is true. But I wouldn't know how to proceed if I hadn't known the answer beforehand. I need to find both $x$ and $y$, is there a linear equation I'm missing to find the exact coordinates of the last vertex? Is my process okay or is there a simpler way to do it?
$\endgroup$ 23 Answers
$\begingroup$You're almost there. What you have found is a relationship between $x$ and $y$, namely $$x=1+y.$$ But there are infinitely many pairs satisfying this relation! On the other hand, you also have that $$x^2+y^2-10y=11.$$ You can plug in the first expression: $$(1+y)^2+y^2-10y=11.$$ Solving this quadratic equation should be easy, but it will give you two roots. On the other hand, only one of the roots can give you the $y$-coordinate of the fourth vertex.
$\endgroup$ 1 $\begingroup$Hint: The coordinates are $(3,2),(0,5),(-3,2),(x,y)$. The segment connecting $(3,2),(0,5)$ is parallel to the segment connecting $(x,y),(-3,2)$, and the segment connecting $(0,5),(x,y)$ is perpendicular to the line connecting $(3,2),(-3,2)$.
Another extremely simple way is that $(3,2)$ when reflected across the $y$-axis gives $(-3,2)$, and thus $(0,5)$ when reflected across the $x$-axis gives $(x,y)$.
$\endgroup$ $\begingroup$Solution without the calculation: You have three points, $A,B,C$. Calculate the distances between Every pair of points. You find side length $s$ and diagonal $d$ ($d$ will be larger and only show up as one distance). Let $A,C$ be the two points adjacent to the point you need to find (you know which to pick because they are the only pair $d$ apart). Write down the equation for a circle of radius $s$ centered at $A$ and $C$. Using the quadratic equation, solve one equation in terms of one of the variables, plug that into the other equation,and solve the resulting quartic. This lets you find both points of intersection (fun fact, these circles can't be tangent). The fourth point of the square is the point of intersection further away from $B$ (or, you can just check which one is a distance $d$ from $B$.
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