Two consecutive vertices of a square are A(2, −1) and B(−1, 3). Determine the coordinates of the other two vertices.

Step-by-step solution

1. 1

   Turn the given side into a vector. Because $A$ and $B$ are consecutive vertices, the segment $AB$ is a full side of the square, not a diagonal. Its vector is

   $\overrightarrow{AB}=B-A=(-1-2,\;3-(-1))=(-3,\,4),$

   with length $|\overrightarrow{AB}|=\sqrt{(-3)^2+4^2}=5$. So the square has side $5$.

2. 2

   Rotate that vector by 90° to get the adjacent side. In a square, consecutive sides are perpendicular and equally long — exactly what a $90^\circ$ rotation produces. Rotating $(x,y)$ counter-clockwise gives $(-y,x)$ and clockwise gives $(y,-x)$, so from $(-3,4)$ the two candidates are

   $(-4,-3)\qquad\text{and}\qquad(4,3),$

   both of length $5$. Two rotations means two valid squares, one on each side of $AB$ — the problem as stated has two answers, and giving only one is incomplete.

3. 3

   Case 1: translate by $(4,3)$. Adding this vector to each given vertex slides the side $AB$ across to the opposite side of the square:

   $D=A+(4,3)=(6,2),\qquad C=B+(4,3)=(3,6).$

   The square is $A(2,-1)$, $B(-1,3)$, $C(3,6)$, $D(6,2)$.

4. 4

   Case 2: translate by $(-4,-3)$. Using the opposite rotation gives the mirror square:

   $D=A+(-4,-3)=(-2,-4),\qquad C=B+(-4,-3)=(-5,0).$

   The square is $A(2,-1)$, $B(-1,3)$, $C(-5,0)$, $D(-2,-4)$.

5. 5

   Verify both squares. In each case all four sides measure $5$, and both diagonals measure $5\sqrt{2}\approx 7.071$ — equal diagonals of length side$\times\sqrt2$ is what distinguishes a square from a rhombus. The dot product $\overrightarrow{AB}\cdot\overrightarrow{BC}=0$ in both cases confirms the right angles.