Answer:
D(-1,0)
Step-by-step explanation:
Plot points A(-3, 2), B(-1, 4), and C(1, 2) on the coordinate plane (see attached diagram).
Two sides AB and BC are of equal length and are perpendicular, because
[tex]AB=\sqrt{(-3-(-1))^2+(2-4)^2}=\sart{(-2)^2+(-2)^2}=\sqrt{4+4}=2\sqrt{2}\ units\\ \\BC=\sqrt{(-1-1)^2+(4-2)^2}=\sart{(-2)^2+2^2}=\sqrt{4+4}=2\sqrt{2}\ units\\ \\\text{Slope AB}=\dfrac{4-2}{-1-(-3)}=\dfrac{2}{2}=1\\ \\\text{Slope BC}=\dfrac{2-4}{1-(-1)}=\dfrac{-2}{2}=-1\\ \\\text{Slope AB}\cdot \text{Slope BC}=1\cdot (-1)=-1\Rightarrow \text{lines AB and BC are perpendicular}[/tex]
If the quadrilateral ABCD has all sides perpendicular, then ABCD is a square. The diagonals of the square bisects each other. Find the coordinates of the point of intersection of these diagonals. This is the midpoint of segment AC:
[tex]x=\dfrac{-3+1}{2}=-1\\ \\y=\dfrac{2+2}{2}=2[/tex]
If point O(-1,2) is the point of intersection of diagonals, then it is the midpoint of the diagonal BD. Find coordinates of point D:
[tex]x_O=\dfrac{x_B+x_D}{2}\Rightarrow -1=\dfrac{-1+x_D}{2},\ -1+x_D=-2,\ x_D=-1\\ \\y_O=\dfrac{y_B+y_D}{2}\Rightarrow 2=\dfrac{4+y_D}{2}, 4+y_D=4,\ y_D=0[/tex]
Thus, D(-1,0)
