Verify the following
(i) $(0, 7, - 10), (1, 6, - 6)$ and $(4, 9, - 6)$ are the vertices of an isosceles triangle
(ii) $(0, 7, 10), (- 1, 6, 6)$ and $(- 4, 9, 6)$ are the vertices of a right angled triangle
(iii) $(- 1, 2, 1), (1, - 2, 5), (4, - 7, 8)$ and $(2, - 3, 4)$ are the vertices of a parallelogram

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Solution

(i) Let point $(0, 7, - 10), (1, 6, - 6)$ and $(4, 9, - 6)$ be denoted by $A, B$ and $C$ respectively
$A B = \sqrt{{(1 - 0)}^{2} + {(6 - 7)}^{2} + {(- 6 + 10)}^{2}}$
$= \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(4)}^{2}}$
$= \sqrt{1 + 1 + 16}$
$= \sqrt{18}$
$\Rightarrow A B = 3 \sqrt{2}$
$B C = \sqrt{{(4 - 1)}^{2} + {(9 - 6)}^{2} + {(- 6 + 6)}^{2}}$
= $\sqrt{{(3)}^{2} + {(3)}^{2}}$
= $\sqrt{9 + 9} = \sqrt{18}$
$\Rightarrow B C = 3 \sqrt{2}$
$C A = \sqrt{{(0 - 4)}^{2} + {(7 - 9)}^{2} + {(- 10 + 6)}^{2}}$
= $\sqrt{{(- 4)}^{2} + {(- 2)}^{2} + {(- 4)}^{2}}$
= $\sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Here $A B = B C$ $\neq$ $C A$
Thus the given points are the vertices of an isosceles triangle
(ii) Let $(0, 7, 10), (- 1, 6, 6)$ and $(- 4, 9, 6)$ be denoted by $A, B$ and $C$ respectively
$A B = \sqrt{{(- 1 - 0)}^{2} + {(6 - 7)}^{2} + {(6 - 10)}^{2}}$
$= \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + {(- 4)}^{2}}$
$= \sqrt{1 + 1 + 16} = \sqrt{18}$
$= 3 \sqrt{2}$
$B C = \sqrt{{(- 4 + 1)}^{2} + {(9 - 6)}^{2} + {(6 - 6)}^{2}}$
$= \sqrt{{(- 3)}^{2} + {(3)}^{2} + {(0)}^{2}}$
$= \sqrt{9 + 9} = \sqrt{18}$
$= 3 \sqrt{2}$
$C A = \sqrt{{(0 + 4)}^{2} + {(7 - 9)}^{2} + {(10 - 6)}^{2}}$
$= \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}}$
$= \sqrt{16 + 4 + 16}$
$= \sqrt{36}$
$= 6$
Now $A B^{2} + B C^{2} = {(3 \sqrt{2})}^{2} + {(3 \sqrt{2})}^{2} = 18 + 18 = 36 = A C^{2}$
Therefore by pythagoras theorem $A B C$ is a right triangle
Hence the given points are the vertices of a right-angled triangle
(iii) Let $(- 1, 2, 1), (1, - 2, 5), (4, - 7, 8)$ and $(2, - 3, 4)$ be denoted by $A, B, C$ and $D$ respectively
$A B = \sqrt{{(1 + 1)}^{2} + {(- 2 - 2)}^{2} + {(5 - 1)}^{2}}$
$= \sqrt{4 + 16 + 16}$
$A B = \sqrt{36}$
$A B = 6$
$B C = \sqrt{{(4 - 1)}^{2} + {(- 7 + 2)}^{2} + {(8 - 5)}^{2}}$
$= \sqrt{9 + 25 + 9} = \sqrt{43}$
$C D = \sqrt{{(2 - 4)}^{2} + {(- 3 + 7)}^{2} + {(4 - 8)}^{2}}$
$= \sqrt{4 + 16 + 16}$
$= \sqrt{36}$
$C D = 6$
$D A = \sqrt{{(- 1 - 2)}^{2} + {(2 + 3)}^{2} + {(1 - 4)}^{2}}$
$D A = \sqrt{9 + 25 + 9} = \sqrt{43}$
Here $A B = C D = 6$ , $B C = A D = \sqrt{43}$
Hence the opposite sides of quadrilateral $A B C D$ whose vertices are taken in order are equal
Therefore $A B C D$ is a parallelogram
Hence the given points are the vertices of a parallelogram

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Similar questions

Verify the following:

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

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

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

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.

(0, 7, - 10), (1, 6, - 6)

(4, 9, - 6)

(0, 7, - 10), (1, 6, - 6)

(4, 9, - 6)

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