Given: Points are A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1)
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,
\(\sqrt{(a-m)^2+(b-n)^2+(c-0)^2+R}\)
Therefore,
Distance between A(1, 3, 4) and B(-1, 6, 10) is AB,
Distance between B(-1, 6, 10) and C(-7, 4, 7) is BC,
Distance between C(-7, 4, 7) and D(-5, 1, 1) is CD,
Distance between A(1, 3, 4) and D(-5, 1, 1) is AD,
Clearly,
AB = BC = CD = AD
All sides are equal
Now, we will find length of diagonals
Distance between A(1, 3, 4) and C(-7, 4, 7) is AC,
Distance between B(-1, 6, 10) and D(-5, 1, 1) is BD,
Clearly,
AC ≠ BD
The diagonals are not equal but all sides are equal
Thus, Quadrilateral formed by ABCD is a rhombus but not square Hence Proved
