Question

Name the type of quadrilateral formed if any, by the following points and give reasons for your answer.
i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
iii) (4, 5), (7, 6), (4, 3), (1, 2)

Answer 1

Hint: In this question, we will use the basic concept of distance formula between two consecutive points. If all the lengths are equal, then we will find the lengths of the alternate points because it can be a square or a rhombus. And if two alternate sides are equal in length, then there are possibilities that the quadrilateral may be either rectangle or parallelogram and so the diagonal lengths are calculated then.

Complete step-by-step answer:

In this question, we will use the basic concept of distance formula, that is distance between $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. Let us consider each of the options given in the question one by one.
i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
Let $A=\left( -1,-2 \right),B=\left( 1,0 \right),C=\left( -1,2 \right)$ and $D=\left( -3,0 \right)$. So, we will now calculate $AB,BC,CD$ and $DA$. After that we will calculate the length of$AC$ and $BD$, then decide the shape of the quadrilateral. So, we can write the distance between $A$ and $B=AB$. So, we get,
$\begin{align}
  & AB=\sqrt{{{\left[ 1-\left( -1 \right) \right]}^{2}}+{{\left[ 0-\left( -2 \right) \right]}^{2}}} \\
 & \Rightarrow AB=\sqrt{{{2}^{2}}+{{2}^{2}}} \\
 & \Rightarrow AB=\sqrt{4+4} \\
 & \Rightarrow AB=2\sqrt{2}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Similarly, for $BC$, we get,
$\begin{align}
  & BC=\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( 2-0 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{{{\left( -2 \right)}^{2}}+{{2}^{2}}} \\
 & \Rightarrow BC=\sqrt{4+4} \\
 & \Rightarrow BC=2\sqrt{2}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
And, for $CD$, we get,
$\begin{align}
  & CD=\sqrt{{{\left[ -3-\left( -2 \right) \right]}^{2}}+{{\left[ 0-\left( 2 \right) \right]}^{2}}} \\
 & \Rightarrow CD=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\
 & \Rightarrow CD=\sqrt{4+4} \\
 & \Rightarrow CD=2\sqrt{2}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
And similarly, for $DA$, we get,
$\begin{align}
  & DA=\sqrt{{{\left[ -1-\left( -3 \right) \right]}^{2}}+{{\left[ \left( -2 \right)-0 \right]}^{2}}} \\
 & \Rightarrow DA=\sqrt{{{2}^{2}}+{{\left( -2 \right)}^{2}}} \\
 & \Rightarrow DA=\sqrt{4+4} \\
 & \Rightarrow DA=2\sqrt{2}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
So, here we can see from equations (i), (ii), (iii) and (iv) that $AB=BC=CD=DA$. So, we can say that the quadrilateral is either a square or a rhombus. Now, we will calculate the length of its diagonals, that is, $AC$ and $BD$. So, by calculating the length for $AC$, we get,
$\begin{align}
  & AC=\sqrt{{{\left[ -1-\left( -1 \right) \right]}^{2}}+{{\left[ 2-\left( -2 \right) \right]}^{2}}} \\
 & \Rightarrow AC=\sqrt{{{0}^{2}}+{{4}^{2}}} \\
 & \Rightarrow AC=\sqrt{{{4}^{2}}} \\
 & \Rightarrow AC=4 \\
\end{align}$
Similarly, we will calculate the length for $BD$ and we get,
$\begin{align}
  & BD=\sqrt{{{\left[ -3-\left( +1 \right) \right]}^{2}}+{{\left[ 0-0 \right]}^{2}}} \\
 & \Rightarrow BD=\sqrt{{{4}^{2}}+0} \\
 & \Rightarrow BD=\sqrt{{{4}^{2}}} \\
 & \Rightarrow BD=4 \\
\end{align}$
We can see that the diagonals$AC=BD$ and earlier we had observed that $AB=BC=CD=DA$.
Therefore, we can say that the quadrilateral formed by these set of points is nothing but a square.
ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
Let $A=\left( -3,5 \right),B=\left( 3,1 \right),C=\left( 0,3 \right)$ and $D=\left( -1,-4 \right)$. So, we will now calculate $AB,BC,CD$ and $DA$. After that we will calculate $AC$ and $BD$, and with the lengths of the sides and the diagonals, we will decide the shape of the quadrilateral that is formed. So, we can write the distance between $A$ and $B=AB$. So, we get,
$\begin{align}
  & AB=\sqrt{{{\left[ 3-\left( -3 \right) \right]}^{2}}+{{\left[ 1-5 \right]}^{2}}} \\
 & \Rightarrow AB=\sqrt{{{6}^{2}}+{{\left( -4 \right)}^{2}}} \\
 & \Rightarrow AB=\sqrt{36+16} \\
 & \Rightarrow AB=\sqrt{52} \\
 & \Rightarrow AB=2\sqrt{13}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Similarly, for $BC$, we get,
$\begin{align}
  & BC=\sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 3-1 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{2}^{2}}} \\
 & \Rightarrow BC=\sqrt{9+4} \\
 & \Rightarrow BC=\sqrt{13}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
And, for $CD$, we get,
$\begin{align}
  & CD=\sqrt{{{\left( -1-0 \right)}^{2}}+{{\left( -4-3 \right)}^{2}}} \\
 & \Rightarrow CD=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -7 \right)}^{2}}} \\
 & \Rightarrow CD=\sqrt{1+49} \\
 & \Rightarrow CD=\sqrt{50}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
And similarly, for $DA$, we get,
$\begin{align}
  & DA=\sqrt{{{\left[ -3-\left( -1 \right) \right]}^{2}}+{{\left[ 5-\left( -4 \right) \right]}^{2}}} \\
 & \Rightarrow DA=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 9 \right)}^{2}}} \\
 & \Rightarrow DA=\sqrt{4+81} \\
 & \Rightarrow DA=\sqrt{85}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
In this set of points, we can see from equations (i), (ii), (iii) and (iv) that $AB\ne BC\ne CD\ne DA$.
Therefore, the quadrilateral that is formed by these set of points is not a uniform quadrilateral.
iii) (4, 5), (7, 6), (4, 3), (1, 2)
Let $A=\left( 4,5 \right),B=\left( 7,6 \right),C=\left( 4,3 \right)$ and $D=\left( 1,2 \right)$. So, we will now calculate the lengths of $AB,BC,CD$ and $DA$. After that, if needed, we will calculate the lengths of$AC$ and $BD$, and decide the shape of the quadrilateral that is formed. So, we can write the distance between $A$ and $B=AB$. So, we get,
$\begin{align}
  & AB=\sqrt{{{\left( 7-4 \right)}^{2}}+{{\left( 6-5 \right)}^{2}}} \\
 & \Rightarrow AB=\sqrt{{{3}^{2}}+{{1}^{2}}} \\
 & \Rightarrow AB=\sqrt{9+1} \\
 & \Rightarrow AB=\sqrt{10}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Similarly, for $BC$, we get,
$\begin{align}
  & BC=\sqrt{{{\left( 4-7 \right)}^{2}}+{{\left( 3-6 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{9+9} \\
 & \Rightarrow BC=3\sqrt{2}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
And, for $CD$, we get,
$\begin{align}
  & CD=\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}} \\
 & \Rightarrow CD=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
 & \Rightarrow CD=\sqrt{9+1} \\
 & \Rightarrow CD=\sqrt{10}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
And similarly, for $DA$, we get,
$\begin{align}
  & DA=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 5-2 \right)}^{2}}} \\
 & \Rightarrow DA=\sqrt{{{3}^{2}}+{{3}^{2}}} \\
 & \Rightarrow DA=\sqrt{9+9} \\
 & \Rightarrow DA=3\sqrt{2}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
Here, we can observe from equations (i), (ii), (iii) and (iv) that $AB=CD$ and $BC=DA$. So, we can say that the quadrilateral is either a rectangle or a parallelogram. So, we will calculate the length of the diagonals, that is $AC$ and $BD$. So, by calculating the length for $AC$, we get,
$\begin{align}
  & AC=\sqrt{{{\left( 4-4 \right)}^{2}}+{{\left( 3-5 \right)}^{2}}} \\
 & \Rightarrow AC=\sqrt{{{0}^{2}}+{{\left( -2 \right)}^{2}}} \\
 & \Rightarrow AC=\sqrt{4} \\
 & \Rightarrow AC=2 \\
\end{align}$
Similarly, we will calculate the length for $BD$ and we get,
$\begin{align}
  & BD=\sqrt{{{\left( 1-7 \right)}^{2}}+{{\left( 2-6 \right)}^{2}}} \\
 & \Rightarrow BD=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\
 & \Rightarrow BD=\sqrt{36+16} \\
 & \Rightarrow BD=\sqrt{52} \\
\end{align}$
WE can see here that $AC\ne BD$ and earlier we had found out that $AB=CD$ and $BC=DA$. Therefore, the quadrilateral formed by these set of points is a parallelogram.

Note: In this question, we have the most basic coordinate geometry formula, that is distance formula, which is for two points, $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. The possible mistakes one can make in this question is by not calculating the length of the diagonals and directly predicting the shape of the quadrilateral. One should also be careful with the numerical signs while doing the calculations.