Question

Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of the parallelogram.

Answer 1

Hint: Here, we have to prove that ABCD is a parallelogram by showing that the opposite sides are equal. That is to prove: AB = CD and BC = AD. Here, to prove these we have to apply the distance formula:
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$

Complete Step-by-Step solution:
Here, we are given four points (-7, -3), (5, 10), (15, 8) and (3, -5). Now we have to prove these are the vertices of a parallelogram.
Now, let us take the four vertices as A (-7, -3), B (5, 10), C (15, 8) and D (3, -5).
First let us consider the figure:

We know that in a parallelogram the opposite sides are equal. So, here we have to prove that:
AB = CD
BC = AD
Here, to prove all these things we have to apply the distance formula. For that consider two points \[P({{x}_{1}},{{y}_{1}})\] and $Q({{x}_{2}}.{{y}_{2}})$ then the distance between them is given by the distance formula, d:
$d=PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Now, consider the points A (-7, -3) and B (5, 10), where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -7,-3 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 5,10 \right)$. By applying the distance formula we will get:
$\begin{align}
  & AB=\sqrt{{{(5-(-7))}^{2}}+{{(10-(-3))}^{2}}} \\
 & AB=\sqrt{{{(5+7)}^{2}}+{{(10+3)}^{2}}} \\
 & AB=\sqrt{{{12}^{2}}+{{13}^{2}}} \\
 & AB=\sqrt{144+169} \\
 & AB=\sqrt{313} \\
 & AB=17.69 \\
\end{align}$
Now, consider the points C (15, 8) and D (3, -5), where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 15,8 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,-5 \right)$. By applying the distance formula we will get:
$\begin{align}
  & CD=\sqrt{{{(3-15)}^{2}}+{{(-5-8)}^{2}}} \\
 & CD=\sqrt{{{(-12)}^{2}}+{{(-13)}^{2}}} \\
 & CD=\sqrt{144+169} \\
 & CD=\sqrt{313} \\
 & CD=17.69 \\
\end{align}$
Hence, we can say that:
$AB=CD$
Now, let us consider the points A (-7, -3) and D (3, -5), where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -7,-3 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,-5 \right)$. By applying the distance formula we will get:
$\begin{align}
  & AD=\sqrt{{{(3-(-7))}^{2}}+{{(-5-(-3))}^{2}}} \\
 & AD=\sqrt{{{(3+7)}^{2}}+{{(-5+3)}^{2}}} \\
 & AD=\sqrt{{{10}^{2}}+{{(-2)}^{2}}} \\
 & AD=\sqrt{100+4} \\
 & AD=\sqrt{104} \\
 & AD=10.19 \\
\end{align}$
Next, let us consider the points B (5, 10) and C (15, 8) where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 5,10 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 15,8 \right).$By applying the distance formula we will get:
$\begin{align}
  & BC=\sqrt{{{(15-5)}^{2}}+{{(8-10)}^{2}}} \\
 & BC=\sqrt{{{10}^{2}}+{{(-2)}^{2}}} \\
 & BC=\sqrt{100+4} \\
 & BC=\sqrt{104} \\
 & BC=10.19 \\
\end{align}$
Therefore, we can say that:
\[BC\text{ }=\text{ }AD\]
Hence, we have proved that the opposite sides are equal.
Therefore, we can say that ABCD is a parallelogram.

Note: We know that the diagonals of a parallelogram bisect each other. So, alternatively we can prove that ABCD is a parallelogram by showing that the midpoints of the diagonals AC and BD are equal.