Question

A classmate plotted the following points:

A(-3, 2), B(-1, 4), and C(1, 2).
Where should the classmate plot point
D so that the quadrilateral formed has
perpendicular sides?

Answer 1

D(-1,0)

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

`AB=√((-3-(-1))^2+(2-4)^2)=\sart{(-2)^2+(-2)^2}=√(4+4)=2√(2)\ units\\ \\BC=√((-1-1)^2+(4-2)^2)=\sart{(-2)^2+2^2}=√(4+4)=2√(2)\ units\\ \\\text{Slope AB}=(4-2)/(-1-(-3))=(2)/(2)=1\\ \\\text{Slope BC}=(2-4)/(1-(-1))=(-2)/(2)=-1\\ \\\text{Slope AB}\cdot \text{Slope BC}=1\cdot (-1)=-1\Rightarrow \text{lines AB and BC are perpendicular}`

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:

`x=(-3+1)/(2)=-1\\ \\y=(2+2)/(2)=2`

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:

`x_O=(x_B+x_D)/(2)\Rightarrow -1=(-1+x_D)/(2),\ -1+x_D=-2,\ x_D=-1\\ \\y_O=(y_B+y_D)/(2)\Rightarrow 2=(4+y_D)/(2), 4+y_D=4,\ y_D=0`

Thus, D(-1,0)