Question

The vertices of quadrilateral ABCD are A(3,0), B(4,3), C(7,3), D(6,0).

(a) Prove by means of coordinate geometry, that quadrilateral ABCD is a parallelogram
(b) Prove that ABCD is not a rhombus

Answer 1

(a) If AB = CD and AB || CD then ABCD is a parallelogram.


`A(3,0), B(4,3), C(7,3), D(6,0) \\d(A,B)= √((4-3)^2+(3-0)^2) = √(10) \\d(C,D)= √((6-7)^2+(0-3)^2) = √(10) \\AB=CD \\ \\m_(AB)= (3-0)/(4-3) = \\ \\m_(CD)= (0-3)/(6-7)=3 \\ \\m_(AB)=m_(CD)\Rightarrow AB||CD`

(b) If AB ≠ BC then ABCD is not a rhombus.


`B(4,3), C(7,3) \\d(B,C)= √((7-4)^2+(3-3)^2)=3 \\BC \\eq AB`