Question

For quadrilateral ABCD, the coordinates of vertices A and B are A(1,2) and B(2,-2). Match each set of coordinates for vertices C and D, that is the most specific way to classify the quadrilateral.?

C(-6,-4), D(-7,0)

C(6,-1), D(5,3)

C(-1,-4), D(-2,0)

C(1,-6), D(0,-2)

Answer 1

A - Rectangle B - Square
C - Parallelogram D - Rhombus

Explanation:

We are given `A(1,2), B(2,-2)` and hence `AB=sqrt((2-1)^2+(-2-2)^2)=sqrt17`. Further slope of `AB` is `(-2-2)/(2-1)=-4/1=-4`.

Case A - `C(-6,-4), D(-7,0)`

As `CD=sqrt((-7-(-6))^2+(0-(-4))^2)=sqrt17` and slope of `CD` is `(0-(-4))/(-7-(-6))=4/(-1)=-4`

As `AB=CD` and `AB`||`CD` slopes being equal, ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+6)^2+(y+4)^2-0.08)((x+7)^2+y^2-0.08)=0 [-10, 10, -5, 5]}

Case B - `C(6,-1), D(5,3)`

As `CD=sqrt((5-6)^2+(3-(-1))^2)=sqrt17` and slope of `CD` is `(0-(-4))/(-7-(-6))=4/(-1)=-4`

Further, `BC=sqrt((6-2)^2+(-1-(-2))^2)=sqrt17` and slope of `BC` is `(-1-(-2))/(6-2)=1/4`

As `BC=AB` and they are perpendicular (as product of slopes is `-1`), ABCD is a square.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-6)^2+(y+1)^2-0.08)((x-5)^2+(y-3)^2-0.08)=0 [-10, 10, -5, 5]}

Case C - `C(-1,-4), D(-2,0)`

As mid point of `AC` is `((1-1)/2,(2-4)/2)` i.e. `(0,-1)` and midpoint of `BD` is `((2-2)/2,(-2+0)/2` i.e. `(0,-1)` i.e. midpoints of `AC` and `BD` are same,

but, `BC=sqrt((2-(-1))^2+(-2-(-4))^2)=sqrt13` i.e. `AB!=BC` and hence ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x+2)^2+y^2-0.08)=0 [-10, 10, -5, 5]}

Case D - `C(1,-6), D(0,-2)`

As mid point of `AC` is `((1+1)/2,(2-6)/2)` i.e. `(1,-2)` and midpoint of `BD` is `((2+0)/2,(-2+(-2))/2` i.e. `(1,-2)` i.e. midpoints of `AC` and `BD` are same,

and, `BC=sqrt((2-1)^2+(-2-(-6))^2)=sqrt17` i.e. `AB=BC` and hence ABCD is a rhombus.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-1)^2+(y+6)^2-0.08)(x^2+(y+2)^2-0.08)=0 [-14, 14, -7, 7]}