Question

Prove that the points (3,0),(4,5),(-1,4) and (-2,-1) taken in order form a rhombus.​

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

Let,

• A (3,0)
• B (4,5)
• C (-1,4)
• D (-2,-1)

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$\mathsf {AB = \sqrt{(4 - 3) {}^{2} + (5 - 0) {}^{2} } }$

$\mathsf { = \sqrt{26} }$

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$\mathsf {BC = \sqrt{ (- 1 - 4) {}^{2} + (4 - 5) {}^{2} } }$

$\mathsf {= \sqrt{26} }$

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$\mathsf {CD = \sqrt{ - 2 + 1) {}^{2} + ( - 1 - 4) {}^{2} } }$

$\mathsf { = \sqrt{26} }$

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$\mathsf {DA = \sqrt{(3 + 2) {}^{2} + (0 + 1) {}^{2} } }$

$\mathsf {= \sqrt{26} }$

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Since, AB = BC = CD = DA

Hence:

• ABCD is a Rhombus.

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