Given: Points are A(-1, 2, 1), B(1,-2, 5), C(4, -7, 8) and D(2, -3, 4)
To prove: the quadrilateral formed by these 4 points is a parallelogram Opposite sides of a parallelogram are equal, but diagonals are not equal.
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(-1, 2, 1) and B(1,-2, 5) is AB,
= \(\sqrt{(-1-1))^2+(2-(-2))^2+(1-5))^2}\)
= \(\sqrt{(-2)^2+4^2+(-4)^2}\)
= \(\sqrt{4+16+16}\)
= \(\sqrt{36}\)
= 6
Distance between B(1,-2, 5) and C(4, -7, 8) is BC,
= \(\sqrt{(1-4)^2+(-2-(-7))^2+(5-8)^2}\)
= \(\sqrt{(-3)^2+5^2+(-3)^2}\)
= \(\sqrt{9+25+9}\)
= \(\sqrt{43}\)
Distance between C(4, -7, 8) and D(2, -3, 4) is CD,
= \(\sqrt{(4-2)^2+((-7)-(-3))^2+(8-4)^2}\)
= \(\sqrt{2^2+(-4)^2+4^2}\)
= \(\sqrt{4+16+16}\)
= \(\sqrt{36}\)
= 6
Distance between A(-1, 2, 1) and D(2, -3, 4) is AD,
= \(\sqrt{(-1-(-2))^2+(2-(3))^2+(1-4)^2}\)
= \(\sqrt{(-3)^2+5^2+(-3)^2}\)
= \(\sqrt{9+25+9}\)
= \(\sqrt{43}\)
Clearly,
AB = CD
BC = AD
Opposite sides are equal
Now, we will find the length of diagonals
Distance between A(-1, 2, 1) and C(4, -7, 8) is AC,
= \(\sqrt{(-1-4)^2+(-2-(-7))^2+(1-8)^2}\)
= \(\sqrt{(-5)^2+9^2+(-7)^2}\)
= \(\sqrt{25+81+49}\)
= \(\sqrt{155}\)
Distance between B(1, -2, 5) and D(2, -3, 4) is BD,
= \(\sqrt{(1-2)^2+(-2-(-3))^2+(5-4)^2}\)
= \(\sqrt{(-1)^2+1^2+1^2}\)
= \(\sqrt{1+1+1}\)
= \(\sqrt{3}\)
Clearly,
AC ≠ BD
The diagonals are not equal, but opposite sides are equal
Thus, Quadrilateral formed by ABCD is a parallelogram
Hence Proved
