# Verify the following: (5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) are the vertices of a rhombus

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Verify the following:

(5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) are the vertices of a rhombus

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Given: Points are A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4)

To prove: the quadrilateral formed by these 4 points is a rhombus All sides of both square and rhombus are equal But diagonals of a rhombus are not equal whereas they are equal for square

Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by,

$\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}$

Therefore,

Distance between A(5, -1, 1) and B(7, -4, 7) is AB

$\sqrt{(5-7)^2+(-1-(-4))^2+(1-7)^2}$

$\sqrt{(-2)^2+3^2+(-6)^2}$

$\sqrt{4+9+36}$

$\sqrt{49}$

= 7

Distance between B(7, -4, 7) and C(1, -6, 10) is BC

$\sqrt{(7-1)^2+(-4-(-6))^2+(7-10)^2}$

$\sqrt{6^2+2^2+(-3)^2}$

$\sqrt{36+4+9}$

$\sqrt{49}$

= 7

Distance between C(1, -6, 10) and D(-1, -3, 4) is CD,

$\sqrt{(1-(-1))^2+(-6-(-3))^2+(10-4)^2}$

$\sqrt{2^2+(-3)^2+6^2}$

$\sqrt{4+9+36}$

$\sqrt{49}$

= 7

Distance between A(5, -1, 1) and D(-1, -3, 4) is AD,

$\sqrt{(5-(-2))^2+(1-(3))^2+(1-4)^2}$

$\sqrt{6^2+2^2+(-3)^2}$

$\sqrt{36+4+9}$

$\sqrt{49}$

= 7

Clearly,

AB = BC = CD = AD

All sides are equal

Now, we will find length of diagonals

Distance between A(5, -1, 1) and C(1, -6, 10) is AC,

$\sqrt{(5-1)^2+(-1-(-6))^2+(1-10)^2}$

$\sqrt{6^2+5^2+(-9)^2}$

$\sqrt{36+25+81}$

$\sqrt{142}$

Distance between B(7, -4, 7) and D(-1, -3, 4) is BD,

$\sqrt{(7-(-1))^2+(-4-(-3))^2+(7-4)^2}$

$\sqrt{8^2+(-1)^2+3^2}$

$\sqrt{64+1+9}$

$\sqrt{74}$

Clearly,

AC ≠ BD

The diagonals are not equal, but all sides are equal

Thus, Quadrilateral formed by ABCD is a rhombus

Hence Proved