Question

Show that the points (2,3) (3,4) (5,6) (4,5) are the verticesof parallelogram

Answer 1

Given to show the vertices of parallelogram are (2, 3) , (3, 4) , (5 ,6) , (4,5)

Required Concept :-

If they  are vertices of the parallelogram then their opposite sides should be same . So, we find the distance between the points If the opposite sides were equal then given vertices are the vertices of parallelogram . We find distance between the points by using distance formula .

According to that we have to show ,

AB=CD and AD =BC

Formula Used :-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

So, Lets  do !

Let,

A=( 2, 3)

B =( 3, 4)

C =( 5, 6)

D = ( 4, 5)

So Finding Distance between AB, BC, CD, AD

$AB =\sqrt{(2-3)^2+(3-4)^2}$

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

$AB=\sqrt{1+1}$

$\red {AB =\sqrt{2}}$

$BC = \sqrt{(3-5)^2+(4-6)^2}$

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

$BC =\sqrt{4+4}$

$BC = \sqrt{8}$

$\red{BC = 2 \sqrt{2}}$

$CD =\sqrt{(4-5)^2+(5-6)^2}$

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

$CD =\sqrt{1+1}$

$\red{CD =\sqrt{2}}$

$AD =\sqrt{(2-4)^2+(3-5)^2}$

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

$AD =\sqrt{4+4}$

$AD = \sqrt{8}$

$\red{AD =2\sqrt{2}}$

So,

$AB= CD \\AD= BC$

Opposite sides are equal Hence , the given vertices are vertices of parallelogram.

Proved !

Note :-

Refer attachment for the diagram .