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?
Best Answer
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.