Question

The points \[A\left( {5,1,1} \right),B\left( {7,4,7} \right),C\left( {1,6,10} \right)\] and \[D\left( { - 1,3,4} \right)\] are the vertices of a
A. Parallelogram
B. Rectangle
C. Rhombus
D. Square

Answer 1

Hint: In this question we are asked to find out what the figure is by using the coordinates given. To determine the figure we will use the distance formula and calculate the length of the sides AB, BC, CD and DA. If all the sides are equal then the figure will be either square or rhombus and if the pair of sides are equal then the diagram will be either parallelogram or rectangle. We will then find the diagonals and then find the figure.

Formula used:
Distance can be calculated with the help of the formula:
$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $.

Complete Step-by-step solution
Given we are given the coordinates \[A\left( {5,1,1} \right),B\left( {7,4,7} \right),C\left( {1,6,10} \right)\]\[D\left( { - 1,3,4} \right)\]

To solve this question firstly, we need to find the distance between AB, BC, CD, and AD.
 To find the distance we are using the following formula
Distance=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $
Finding AB,
$AB = \sqrt {{{\left( {7 - 5} \right)}^2} + {{\left( {4 - 1} \right)}^2} + {{\left( {7 - 1} \right)}^2}} $
$AB = \sqrt {4 + 9 + 36} = 7$..................................(1)
Similarly Finding value for BC, CD and AD
$\begin{align}
  & BC=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \\
 & =\sqrt{{{\left( 1-7 \right)}^{2}}+{{\left( 6-4 \right)}^{2}}+{{\left( 10-7 \right)}^{2}}} \\
 & =\sqrt{36+4+9} \\
 & =7 ..................................(2) \\
\end{align}$

Finding value for CD,
$\begin{align}
  & CD=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \\
 & =\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( 3-6 \right)}^{2}}+{{\left( 4-10 \right)}^{2}}} \\
 & =\sqrt{4+9+36} \\
 & =7..................................(3) \\
\end{align}$
Finding value for CD,
$\begin{align}
  & DA=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \\
 & =\sqrt{{{\left( 5-(-1) \right)}^{2}}+{{\left( 1-3 \right)}^{2}}+{{\left( 1-4 \right)}^{2}}} \\
 & =\sqrt{36+4+9} \\
 & =7 ..................................(4) \\
\end{align}$
 From above equation (1), (2), (3) and (4) we can say that
AB=BC=CD=AD=7.
 As all the sides are equal from this we can conclude that it can either be a Rhombus or a Square.
To find out further we need to find the diagonal AC and BD from the distance formula
$AC = \sqrt {16 + 25 + 81} $
$ \Rightarrow AC = \sqrt {122} $ ..................................(5)
$BD = \sqrt {64 + 1 + 9} $
$ \Rightarrow BD = \sqrt {74} $ ..................................(6)
According to the equation (5) and (6)
 Thus, $AC \ne BD$
Therefore according o the properties if the sides are equal but the diagonals are not equal then the figure is called as rhombus
Hence the correct option is C (Rhombus).

Note: As all the figures are different and have different properties, it is important to know all these properties. It is important as there are various types of polygons to choose from, like in this question where we found the side and diagonal; sometimes we may need to find the angle or use Pythagoras' theorem to reach a conclusion.