Question

Show that A(1,0) , B(0,1), C(1,2) and D(2,1) are vertices of a parallelogram ABCD. Is ABCD a rectangle ?

Answer 1

a(1,0)
b(0,1)
c(1,2)
d(2,1)


AB²=(X2-X)²+(Y2-Y)²
     = (0-1)²+(1-0)²
         =1+1
AB=√2
BC²=(1-0)²+(2-1)²=2
BC=√2
CD²=(2-1)²+(1-2)²=2
CD=√2
AD²=(2-1)²+(1-0)²=2
AD=√2

USE PYTHAGORIAN THEROM
AD²+DC²=AC²
AC²=2+2=4
AC=2

THEN WE CAN SAY THAT ΔADC IS A RIGHT-ANGLED TRIANGLE

AB²+BC²=AC²
SAME
ΔABC IS A RIGHT-ANGLED TRIANGLE

SO ABCD IS A SQUARE




 

Answer 2

Distance between two points = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }$

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

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

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

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

Since all sides are equal, its a parallelogram

It could be a rhombus or a square