Question

If A (6, 1), B (8, 2), C (9, 4) and D (7, 3) are the vertices of ABCD ,show that ABCD is a parallelogram

Answer 1

Question

If A(6,1), B(8,2), C(9,4), & D(7,3) are the vertices of ABCD, show that ABCD is a parallelogram .

-: ANSWER :-

Given : -

If A(6,1), B(8,2), C(9,4), & D(7,3) are the vertices of ABCD, show that ABCD is a parallelogram .

Required to find : -

• ABCD is a parallelogram

Formula used : -

Distance between 2 points = √[(x1-x)²+(y1-y)²]

Solution : -

Here,

We need to show that ABCD is a parallelogram

We already know that;

• In a parallelogram, opposite sides are equal

So,

AB = CD

BC = AD

Using this concept let's solve this question !

The formula which we are going to use is;

Distance between 2 points = √[(x1-x)²+(y1-y)²]

Here,

• x1 & x can be any co-ordinate

The co-ordinates of the parallelogram are;

• A(6,1)
• B(8,2)
• C(9,4)
• D(7,3)

Distance between the points A&B = √[(8-6)²+(2-1)²]

AB = √[(2)²+(1)²]

AB = √[4+1]

AB = √5 units

Now,

Let's find the distance between B&C = √[(9-8)²+(4-2)²]

BC = √[(1)²+(2)²]

BC = √[1+4]

BC = √5 units

Now,

Let's find the distance between C&D = √[(7-9)²+(3-4)²]

CD = √[(-2)²+(-1)²]

CD = √[4+1]

CD = √5 units

Let's find the distance between A & D = √[(7-6)²+(3-1)²]

AD = √[(1)²+(2)²]

AD = √[1+4]

AD = √5 units

Hence,

• AB = √5 units
• BC = √5 units
• CD = √5 units
• AD = √5 units

From the above we can conclude that;

ABCD is not a parallelogram, but actually it is a square

In a square, All sides are equal .

However,

The square is defined as;

It is a parallelogram which has all it's sides equal .

Therefore,

ABCD is not a parallelogram ✓

Answer 2

Question

If A (6, 1), B (8, 2), C (9, 4) and D (7, 3) are the vertices of ABCD ,show that ABCD is a parallelogram

Given

A(6,1) , B(8,2) , C(9,4) , D(7,3)

To prove

Is the figure parallelogram ?

Solution

$AB = \sqrt{ { (8 - 6)}^{2} } + {(2 - 1)}^{2}$

$distance = \sqrt{ {( x_{2} - x_{1} }^{2}) } +( y_{2} - { y_{1} })^{2}$

$BC = \sqrt{ {(9 - 8)}^{2} + {(4 - 2)}^{2} }$

$\implies \sqrt{ {1}^{2} + {2}^{2} } = \sqrt{5}$

$CD \sqrt{ {(7 - 9)}^{2} } + ({3 - 4})^{2}$

$\implies \sqrt{ {(2)}^{2} } + {( - 1)}^{2} = \sqrt{5}$

$DA \sqrt{ {(7 - 6)}^{2} } + {(3 - 1)}^{2}$

$\implies \sqrt{ {1}^{2} } + {2}^{2} = \sqrt{5}$

AB = BC = CA = DA

ABCD is not a parallelogram but a square

More

parallelogram has opposite sides parallel and equal in length also opposite angles are equal .

Note :

Squares, rectangles and Rhombuses are all Parallelogram.!!