# 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 ?

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by shivam2000

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by shivam2000

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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

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

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Distance between two points =

AB =

BC =

CD =

DA =

Since all sides are equal, its a parallelogram

It could be a rhombus or a square

AB =

BC =

CD =

DA =

Since all sides are equal, its a parallelogram

It could be a rhombus or a square