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Mathematics Question on distance between two points

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.

Answer

(i) Let points (0, 7, -10), (1, 6, -6), and (4, 9, -6) be denoted by A, B, and C respectively
AB=(1−0)2+(6−7)2+(−6+10)2\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}
= (1)2+(−1)2+(4)2\sqrt{(1)^2+(-1)^2+(4)^2}
= 1+1+16\sqrt{1+1+16}
= 18\sqrt{18}
= 32\sqrt{2}
BA = (4−1)2+(9−6)2+(−6+6)2\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}
=(3)2+(3)2\sqrt{(3)^2+(3)^2}
= 9+9\sqrt{9+9}=18\sqrt{18}=32\sqrt{2}
CA = (0−4)2+(7−9)2+(−10+6)2\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}
= (−4)2+(−2)2+(−4)2\sqrt{(-4)^2+(-2)^2+(-4)^2}
= 16+4+16\sqrt{16+4+16}=36\sqrt{36}=6
Here, AB = BC ≠ CA
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.
AB=(−1−0)2+(6−7)2+(6−10)2\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}
= (−1)2+(−1)2+(−4)2\sqrt{(-1)^2+(-1)^2+(-4)^2}
= 1+1+16\sqrt{1+1+16}= 18\sqrt{18}
= 3√2
BC=(−4+1)2+(9−6)2+(6−6)2\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}
= (−3)2+(3)2+(0)2\sqrt{(-3)^2+(3)^2+(0)^2}
= 9+9\sqrt{9+9}= 18\sqrt{18}
= 32\sqrt2
CA = (0+4)2+(7−9)2+(10−6)2\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}
=(4)2+(−2)2+(4)2\sqrt{(4)^2+(-2)^2+(4)^2}
= 16+4+16\sqrt{16+4+16}
= 36\sqrt{36}
= 6
Now, AB2+BC2 = (32)2+(32)2(3\sqrt2)^2+(3\sqrt2)^2 = 18+18=36= AC2
Therefore, by Pythagoras theorem, ABC 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.
AB=(1+1)2+(−2−2)2+(5−1)2\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}
= 4+16+16\sqrt{4+16+16}
=36\sqrt{36} = 6
BC= (4−1)2+(−7+2)2+(8−5)2\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}
= 9+25+9=43\sqrt{9+25+9}=\sqrt{43}
CD=(2−4)2+(−3+7)2+(4−8)2\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}
=4+16+16\sqrt{4+16+16}
=36\sqrt{36}
=6
DA = (−1−2)2+(2+3)2+(1−4)2\sqrt{(-1-2)^2+(2+3)^2+(1-4)^2}
= 9+25+9=43\sqrt{9+25+9}=\sqrt{43}
Here, AB = CD = 6, BC = AD =43\sqrt{43}
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.