Question

The four points (1,-2), (3,6),(5,10) and (3,2) are the vertices of a

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

Answer 1

The correct option is D Parallelogram

Let A(1,2), B(3,6), C(5,10) and D(3,2) are given points
$AB=\sqrt{(1-3{)}^2+(-2-6{)}^2}=\sqrt{4+64}=\sqrt{68}$
$BC=\sqrt{(3-5{)}^2+(6-10{)}^2}=\sqrt{4+16}=\sqrt{20}$
$CD=\sqrt{(5-3{)}^2+(10-2{)}^2}=\sqrt{4+64}=\sqrt{68}$
$AD=\sqrt{(3-1{)}^2+(2-(-2){)}^2}=\sqrt{4+16}=\sqrt{20}$
$AC=\sqrt{(1-5{)}^2+(-2-10{)}^2}=\sqrt{16+144}=\sqrt{160}$
$BD=\sqrt{(3-3{)}^2+(6-2{)}^2}=\sqrt{0+16}=\sqrt{16}$
$AB=CD$ & $BC=AD$
but $AC\ne BD$
So, $ABCD$ is a parallelogram,
Thus option d is correct