Question

9. Prove that A(2,-1), B(3,4), C(-2,3) and D(-3,-2) are the vertices of a rhombus ABCD. Is it a square?​

Answer 1

Dear the side would be equal but if both diagonals are not equal then it is a rhombus if they are equal then it is a square.

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Answer 2

Step-by-step explanation:

Given:

Four vertices of a quadrilateral A(2,1); B(3,4); C(-2,3); and D(-3,-2)

Required to Find:

A, B, C, D are the four vertices of a rhombus.

Also check whether it is square or not.

Solution:

To prove that A(2,1); B(3,4); C(-2,3); and D(-3,-2) are the four vertices of a rhombus,

We have to prove that the length of the all sides of the quadrilateral are equal i.e. AB = BC = CD = DA

As we know that the length of the two points = $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

∴$AB=\sqrt{(3-2)^{2}+(4+1)^{2}}\\=\sqrt{1+25}\\=\sqrt{26}units$

∴$BC=\sqrt{(-2-3)^{2}+(3-4)^{2}}\\=\sqrt{25+1}\\=\sqrt{26}units$

∴$CD=\sqrt{(-3+2)^{2}+(-2-3)^{2}}\\=\sqrt{1+25}\\=\sqrt{26}units$

∴$DA=\sqrt{(2+3)^{2}+(-1+2)^{2}}\\=\sqrt{25+1}\\=\sqrt{26}units$

∵ AB = BC = CD = DA = √26 units.

∴ A(2,1); B(3,4); C(-2,3); and D(-3,-2) are the four vertices of a rhombus

Now,  if the diagonals of the rhombus is equal then it is a square i.e. if AC = BD then it is a square.

∴ $AC=\sqrt{(-2-2)^{2}+(3+1)^{2}}\\=\sqrt{16+16}\\=\sqrt{32}units$

∴ $BD=\sqrt{(-3-3)^{2}+(-2-4)^{2}}\\=\sqrt{36+36}\\=\sqrt{72}units$

∵ AC ≠ BD

∴ It is not a square.

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