Answer :

eshant

Answer:

Firstly,

Let A(-2 3), B(6 7), C(8 3), D(0 -1) be given points of rectangle.

Step-by-step explanation:

Then, AB =

[tex] \sqrt{(6 + 2)^{2} + \sqrt{(7 - 3)^{2} } } \\ = 4 \sqrt{5} [/tex]

Now

CD=

[tex] \sqrt{(0 - 8)^{2} + ( - 1 - 3)^{2} } \\ = 4 \sqrt{5} [/tex]

Here, AB=CD

Finally,

Diagonal of AC must be equal to Diagonal of BD to prove that the given points are the vertices of a rectangle.

so,

Diagonal of AC=

[tex] \sqrt{(8 + 2)^{2} + (3 - 3) ^{2} } \\ = 10[/tex]

Diagonal of BD=

[tex] \sqrt{(0 - 6) ^{2} + ( - 1 - 7) ^{2} } \\ = 10[/tex]

Finally, Diagonal of AC = Diagonal of BD.

The given points are the vertices of Rectangle.

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