Question

Show that the points A(4, 7), B(0, 6), C(4, 5) and D(8, 6) are the vertices of a rhombus.

Answer 1

SOLUTION :-

Given,

A = ( 4 , 7 )

B = ( 0 , 6 )

C = ( 4 , 5 )

D = ( 8 , 6 )

We have to prove that these points are the vertices of a rhombus.

If these points are the vertices of a rhombus, then

AB = BC = CD = AD

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

• AB = $\sf\sqrt{(0-4)^2+(6-7)^2}$

              =  $\sf \sqrt{4^2+1}$

              = $\sf \sqrt{16+1}$

              = $\sf \sqrt{17}$ units

• BC  = $\sf\sqrt{(4-0)^2+(5-6)^2}$

              =  $\sf\sqrt{4^2+1}$

              = $\sf\sqrt{16+1}$

              = $\sf \sqrt{17}$ units

• CD = $\sf\sqrt{(8-4)^2+(6-5)^2}$

              = $\sf\sqrt{4^2+1}$

              = $\sf \sqrt{16+1}$

              = $\sf\sqrt{17}$ units

• AD = $\sf\sqrt{(8-4)^2+(6-7)^2}$

              = $\sf\sqrt {4^2+1}$

              = $\sf\sqrt {16+1}$

              = $\sf\sqrt{17}$ units

   ∴ AB = BC = CD = AD

So the given points are the vertices of a rhombus.

Hence proved.

Answer 2

Given :-

• Four points A(4,7), B(0,6), C(4,5) and D(8,6).

To Prove :-

• These points are the vertices of a rhombus.

SoluTion :-

We have to prove AB = BC = CD = DA

• Distance between A and B

→√ (x2 - x1)² + (y2 - y1)²

→ √ (0 - 4)² + (6 - 7)²

→ √ (-4)² + (-1)²

→ √ 16 + 1

→ √17 units .....i)

• Distance between B and C

→ √ (x2 - x1)² + (y2 - y1)²

→ √ (4 - 0)² + (5 - 6)²

→ √ (4)² + (-1)²

→ √ 16 + 1

→ √17 units .....ii)

• Distance between C and D

→ √ (8 - 4)² + (6 - 5)²

→ √ (4)² + (1)²

→ √ 16 + 1

→ √17 units ......iii)

• Distance between D and A

→ √ (8 - 4)² + (6 - 7)²

→ √ (4)² + (-1)²

→ √ 16 + 1

→ √17 units ......iv)

From i) , ii) , iii) and iv)

AB = BC = CD = DA

It's clear that all four sides are equal.

So these points are the vertices of the rhombus.

Hence, ProVed.

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