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
