verify that the points (-1,-6,10) ; (1,-3,4) ;(-5,-1,1) and (-7,-4,7) are the vertices of the rhombus.

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Solution

$If (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) are the two points then distance between them can be calculated as, D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} Now the given point is, A (- 1, - 6, 10), B (1, - 3, 4), C (- 5, 1, 1) and D (- 7, - 4, 7) Now calculating distance AB = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} = \sqrt{{(1 - (- 1))}^{2} + {(- 3 + 6)}^{2} + {(4 - 10)}^{2}} = \sqrt{4 + 9 + 36} = 7 BC = \sqrt{{(- 5 - 1)}^{2} + {(- 1 + 3)}^{2} + {(1 - 4)}^{2}} = \sqrt{36 + 4 + 9} = 7 CD = \sqrt{{(- 7 + 5)}^{2} + {(- 4 + 1)}^{2} + {(7 - 1)}^{2}} = \sqrt{4 + 9 + 36} = 7 Similarily, AD = \sqrt{{(- 7 + 1)}^{2} + {(- 4 + 6)}^{2} + {(7 - 10)}^{2}} = \sqrt{4 + 9 + 36} = 7 Now all the sides are equal so it willbe a square or a rhombus, Now calculating diagonal BD and AC, AC = \sqrt{{(- 5 + 1)}^{2} + {(- 1 + 6)}^{2} + {(1 - 10)}^{2}} = \sqrt{16 + 25 + 81} = \sqrt{122} BD = \sqrt{{(- 7 - 1)}^{2} + {(- 4 + 3)}^{2} + {(7 - 4)}^{2}} = \sqrt{64 + 1 + 9} = \sqrt{74} Since diagonal are not equal so it will be a Rhombus .$

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Similar questions

(5, - 1, 1), (7, - 4, 7), (1, - 6, 10)

(- 1, - 3, 4)

Verify the following :

(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are vertices of an isosceles triangle.

(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are vertices of a right-angled triangle.

(iii) (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are vertices of a parallelogram.

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

The points A(5, -1, 1); B(7, -4, 7); C(1, -6, 10) and D(-1, -3, 4) are vertices of a [RPET 2000]

(5, - 1, 1) (7, - 4, 7), (1, - 6, 10) (- 1, - 3, 4)

A = (5, - 1, 1), B = (7, - 4, 7), C = (1, - 6, 10), D = (- 1, - 3, 4)

A B C D

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