Let $A(-2;-2)$ and $B(4;4)$ are vertices of the square $ABCD$

Question

Let $A(-2;-2)$ and $B(4;4)$ are vertices of the square $ABCD$. If the intersection of the diagonals of the square lies in the second quadrant, find its coordinates.

I am not sure what exactly we are supposed to use in order to find the center $O$ of the square.

I found that the equation of the line through $A,B$ is $$AB:y=x$$ $AD$ lies on a line perpendicular to $AB$ and passing through $A$, so it has an equation $$AD:y=-x+n$$ and $-2=2+n\Rightarrow n=-4\Rightarrow AB:y=-x-4.$. Therefore we can write the coordinates of $D$ as $D=(x;-x-4)$. $O$ is the midpoint of $BD$, so $O=\left(\dfrac{x+4}{2};\dfrac{-x}{2}\right)$. Now $\vec{OA}.\vec{OB}=0$ and from here we can actually find $x$ and from there the coordinates of $O$, but I seem to always mess up the calculations as I got $x=-16$ or $x=0$ which isn't the case (we can see it from the diagram). Is there something more direct?

Answer 1

If $O(x,y)$ is the center, we must have $\vec {OA}\cdot \vec{OB}=0$.

Which, given that $\vec {OA}=(x+2,y+2),\vec{OB}=(x-4,y-4),$ is equivalent to $(x+2)(x-4)+(y+2)(y-4)=0$.

Let's call this relationship $(1)$.

But also it must hold that $$\vert\vec {OA} \vert=\vert \vec{OB}\vert \Rightarrow\\\sqrt{(x+2)^2+(y+2)^2}=\sqrt{(x-4)^2+(y-4)^2}\Rightarrow\\(x+2)^2+(y+2)^2=(x-4)^2+(y-4)^2\Rightarrow\\(x+2)^2-(x-4)^2=(y-4)^2-(y+2)^2\Rightarrow\\ (x+2-x+4)(x+2+x-4)=(y-3-y-2)(y-4+y+2)\Rightarrow\\6(2x-2)=-6(2y-2)\Rightarrow\\x-1=-y+1\Rightarrow\\x=-y+2$$

Now if we substitute $x$ on $(1)$ we obtain $$(-y+4)(-y-2)+(y+2)(y-4)=0\Rightarrow\\(y-4)(y+2)+(y+2)(y-4)=0\Rightarrow\\(y-4)(y+2)=0\Rightarrow$$

$y=4$ or $y=-2$.

But we can't have $y=-2$ since $O(x,y)$ lies in the second quadrant.

Thus $y=4\Rightarrow x=-4+2=-2$ and the center of the square is $O(-2,4)$.

Answer 2

Length of the line segment AB from your diagram is :$\sqrt{(4-(-2)^2+4-(-2)^2}=6\sqrt{2}$

Mid point $(M)$ of this line segment is at: $(\frac{(4-2)}{2},\frac{(4-2)}{2})=(1,1)$

$AM=MB=3\sqrt2$

$MO=\frac{AD}{2}=\frac{AB}{2}=3\sqrt2$; since $AD=AB$

Equation of $AB$ as you found it is $y=x$

Now perpendicular distance of the point M from the center is $3\sqrt2$, i.e M is the foot of perpendicular of the center on line AB.

Thus:$$\frac{1-x_1}{1}=\frac{1-y_1}{1}=-\frac{(y_1-x_1)}{1^2+1^2}$$

Where $(x_1,y_1)$ are the coordinates of the center.

$1-\frac{3x_1}{2}=-\frac{y_1}{2} \ \ \ ....(1)$

$1-\frac{y_1}{2}=\frac{x_1}{2} \ \ \ ....(2)$

These will be sufficient to get you an answer.

Answer 3

Complex numbers work perfectly here. The numbers $a=-2-2i$ and $b=4+4i$ coresponds to vertices $A$ and $B$. Then $d= a+(b-a)i$ corespond to vertex $D$ (we get $D$ by rotating $B$ around $A$) and finally the center corespond to \begin{align}\sigma &= {1\over 2}(b+d)\\ &= {1\over 2}(2+2i +6i-6)\\ & = -2+4i\end{align}

So the center is at point $(-2,4)$

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Rotation of $z$ to $z'$ around point $a$ for angle $90^{\circ}$ is described by equation $$z'-a = i(z-a)$$