Question

If A(2, –1), B(3, 0), C(3, 4) and D(2, 3) are the vertices of a quadrilateral ABCD, then ABCD is always a

• A. Kite
• B. Rhombus
• C. Square
• D. Parallelogram

Answer 1

The correct option is D Parallelogram

Given that A(2, –1), B(3, 0), C(3, 4) and D(2, 3) are the vertices of a quadrilateral ABCD.



$AB=\sqrt{(2-3{)}^2+(-1-0{)}^2}=\sqrt{(-1{)}^2+(-1{)}^2}=\sqrt{1+1}=\sqrt{2}$

$BC=\sqrt{(3-3{)}^2+(0-4{)}^2}=\sqrt{0^2+4^2}=\sqrt{4^2}=4$

$CD=\sqrt{(3-2{)}^2+(4-3{)}^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$DA=\sqrt{(2-2{)}^2+(3-(-1){)}^2}=\sqrt{0+4^2}=\sqrt{4^2}=4$

$\therefore AB=CDandBC=DA$

Thus, the opposite sides of quadrilateral ABCD are equal.

Therefore, ABCD is a parallelogram.

Hence, the correct answer is option (d).