Step-by-step explanation:

Given:-

The points A(3, 1), B(0, -2), C(1, 1) and D(4, 4)

To show:-

Show that the points A(3, 1), B(0, -2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.

Solution:-

Method-1:-

Given points are A(3, 1), B(0, -2), C(1, 1) and D(4, 4)

We know that

In a Parallelogram," The diagonals are bisecting to each other"

AC and BD are the diagonals

=>Mid POINT of AC = Mid Point of BD

So To show that the given points are the points of a Parallelogram then we have to show that the mid point of AC is equal to the mid point of BD.

Mid point of AC:-

Let (x1, Y1)= A(3, 1) =>x1 = 3 and y1 = 1

(x2, y2)=C(1, 1) =>x2 = 1 and y2 = 1

The mid point of the line joining the points (x1 ,y1) and (x2 ,y2) is M(X,y)=[(x1+x2)/2,(y1+y2)/2]

=>[(3+1)/2,(1+1)/2]

=>(4/2,2/2)

=>(2,1)

Mid point of AC = (2,1)-----------------(1)

Mid Point of BD:-

Let (x1, y1)= B(0,-2) =>x1 = 0 and y1 = -2

(x2, y2)=D(4,4) =>x2 = 4 and y2 = 4

The mid point of the line joining the points (x1 ,y1) and (x2 ,y2) is M(x,y)=[(x1+x2)/2,(y1+y2)/2]

=>[(0+4)/2,(-2+4)/2]

=>(4/2,2/2)

=>(2,1)

Mid Point of BD = (2,1)---------------(2)

From (1)&(2)

Mid Point of AC = Mid point of BD

Given points A,B,C,D are the vertices of the paralellogram ABCD.

Method-2:-

Given points are A(3, 1), B(0, -2), C(1, 1) and D(4, 4)

To show that the given points are the points of a Parallelogram then we have to show that two pairs of the lengths of the opposite sides of the Parallelogram ABCD. i.e. AB= CD and BC = DA

Length of AB :-

Let (x1, y1)= A(3, 1) =>x1 = 3 and y1 = 1

Let (x2, y2)= B(0,-2) =>x2= 0 and y2 = -2

We know that

The distance between two points ( x1,y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2]

AB =√[(0-3)^2-(-2-1)^2]

=>AB =√[(-3)^2+(-3)^2]

=>AB= √(9+9)

=>AB=√18

=>AB=√(2×9)

AB=3√2 units ---------(1)

Length of BC :-

Let (x1, y1)= B(0,-2) =>x1 = 0 and y1 = -2

Let (x2, y2)= C(1,1) =>x2= 1 and y2 = 1

We know that

The distance between two points ( x1,y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2]

=>BC=√[(1-0)^2+(1-(-2))^2]

=>BC=√[1^2+3^2]

=>BC=√(1+9)

BC =√10 units --------------(2)

Length of CD:-

Let (x1, y1)= C(1,1)=>x1=1 and y1=1

Let (x2, y2)= D(4,4) =>x2= 4and y2 = 4

We know that

The distance between two points ( x1,y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2]

=>CD = √[(4-1)^2+(4-1)^2]

=>CD=√[3^2+3^2]

=>CD=√(9+9)

=>CD =√18

=>CD=√(2×9)

CD=3√2 units ---------------(3)

Length of DA:-

Let (x1, y1)= D(4,4) =>x1 = 4 and y1 = 4

Let (x2, y2)= A(3,1) =>x2= 3 and y2 = 1

We know that

The distance between two points ( x1,y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2]

=>DA =√[(3-4)^2+(1-4)^2]

=>DA=√[(-1)^2+(-3)^2]

=>DA=√[(-1)^2+(-3)^2]

=>DA=√(1+9)

DA=√10 units---------------(4)

From (1)&(3)

AB=BC

From (2)&(4)

BC=DA

Two pairs of opposite sides are equal.

Answer:-

Given points are the vertices of the paralellogram ABCD.

Used Concept:-

• .In a Parallelogram," The diagonals are bisecting to each other"

• Two pairs of opposite sides are equal.

Used formulae:-

1)

The mid point of the line joining the points (x1 ,y1) and (x2 ,y2 ) is M(x,y) =[(x1+x2)/2,(y1+y2)/2]

2)The distance between two points ( x1,y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2]