Question

The points A (4, - 2, 1), B (7, - 4, 7), C (2, - 5, 10) and D (- 1, - 3, 4) are the vertices of a

• A. tetrahedron
• B. parallelogram
• C. rhombus
• D. square

Answer 1

Answer: (B)

Here, the mid-point of [AC] is
$\left(\frac{4+2}{2},\frac{-2-5}{2},\frac{1+10}{2}\right)=\left(3,-\frac{7}{2},\frac{11}{2}\right)$
and that of [BD] is
$\left(\frac{7-1}{2},\frac{-4-3}{2},\frac{7+4}{2}\right)=\left(3,-\frac{7}{2},\frac{11}{2}\right)$
So, the diagonals [AC] and [BD] bisect each other
$\Rightarrow$ ABCD is a parallelogram.
As $|AB|=\sqrt{3^2+2^2+6^2}=7$ and
$|AD|=\sqrt{5^2+1^2+3^2}\ne |AB|,$
therefore. ABCD is not a rhombus and naturally, it cannot be a square.