Answer to Question #299234 in Analytic Geometry for Philly

Question

In the Cartesian plane, two vertices of a square have coordinates (3 ; 4) and ( -2 ; - 1). One of the other two vertices, has coordibates

Answer 1

Two opposite vertices of a square are $(3,4)$ and $(-2,-1).$ Find the coordinates of other two vertices.

Assume $A(-2, -1)$ and $C(3, 4)$ are two opposite vertices of a square 

$\overrightarrow{AC}=(3-(-2), 4-(-1))$

$|\overrightarrow{AC}|=\sqrt{5^2+5^2}=5\sqrt{2}$

The side of the square is 5.

$slope_{AC}=\dfrac{4-(-1)}{3-(-2)}=1$

Then

$slope_{BD}=-1$

The equation of the $BD$ is $y=-x+b.$

The center of the square is $O(\dfrac{3-2}{2},\dfrac{4-1}{2} ).$

Substitute

$\dfrac{3}{2}=-\dfrac{1}{2}+b=>b=2$

$BD:y=-x+2$

The angle between $AB$ and $BD$ is $45\degree.$ Then the equation of $AB$ is $x=-2.$

$y=-(-2)+2=4$

$B(-2, 4)$

The angle between $CD$ and $BD$ is $45\degree.$ Then the equation of $CD$ is $x=3.$

$y=-3+2=-1$

$D(3, -1)$