# Show that the point A(1,2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

74 views

closed

Show that the point A(1,2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

+1 vote
by (32.4k points)
selected

Given: Points are A(1,2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6)

To prove: the quadrilateral formed by these 4 points is a parallelogram but not a rectangle

Opposite sides of both parallelogram and rectangle are equal

But diagonals of a parallelogram are not equal whereas they are equal for rectangle

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, 3) and B(-1, -2, -1) is AB, Distance between B(-1, -2, -1) and C(2, 3, 2) is BC, Distance between C(2, 3, 2) and D(4, 7, 6) is CD, The distance between A(1, 2, 3) and D(4, 7, 6) is AD, Clearly, AB = CD

Opposite sides are equal

Now, we will find the length of diagonals

The distance between A(1, 2, 3) and C(2, 3, 2) is AC,

=$\sqrt{(1-2)^2+(2-3)^2+(3-2)^2}$

$\sqrt{(-1)^2+(-1)^2+1^2}$

$\sqrt{1+1+1}$

= $\sqrt{3}$

Distance between B(-1, -2, -1) and D(4, 7, 6) is BD,

=$\sqrt{(-1-4)^2+(-2-7)^2+(-1-6)^2}$

$\sqrt{(-5)^2+(-9)^2+(-5)^2}$

$\sqrt{25+81+25}$

= $\sqrt{131}$

Clearly,

AC ≠ BD

The diagonals are not equal, but opposite sides are equal

Thus, Quadrilateral formed by ABCD is a parallelogram but not a rectangle

Hence Proved