
Verify the following:
$\left( -1,2,1 \right),\left( 1,-2,5 \right),\left( 4,-7,8 \right)$ and $\left( 2,-3,4 \right)$ are vertices of a parallelogram.
Answer
431.7k+ views
Hint: To verify that the given coordinates are the vertices of a parallelogram, we will use the property of the parallelogram in which it is said that the opposite sides are equal to each other. If we name the parallelogram as \[\text{ABCD}\], then $AB=CD$ and $BC=DA$. The formula for measuring the length of the sides is:
$=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$
Where, $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ are points.
Complete step by step answer:
Since, the given points of vertices are $\left( -1,2,1 \right),\left( 1,-2,5 \right),\left( 4,-7,8 \right)$ and $\left( 2,-3,4 \right)$.
Let’s suppose that they are the points of a parallelogram named \[\text{ABCD}\]. Then from the property of parallelogram, $AB=CD$ and $BC=DA$.
Here, we assumed that:
$\begin{align}
& \Rightarrow A=\left( -1,2,1 \right) \\
& \Rightarrow B=\left( 1,-2,5 \right) \\
& \Rightarrow C=\left( 4,-7,8 \right) \\
& \Rightarrow D=\left( 2,-3,4 \right) \\
\end{align}$
Now, we will find the length of the side $AB$ with help of the formula as:
$\Rightarrow AB=\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( 1-5 \right)}^{2}}}$
Solve the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow AB=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\
& \Rightarrow AB=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 4 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\
\end{align}$
Here, we will do the squaring off $-2,4$ and $-4$ and will get $4,16$ and $16$ respectively as:
$\Rightarrow AB=\sqrt{4+16+16}$
The sum of $4,16$ and $16$ is $36$. So, the above step will be:
$\Rightarrow AB=\sqrt{36}$
As we know that the under root of $36$ is $6$.
$\Rightarrow AB=6$
Here, we will find the length of $CD$ with the help of the formula as:
$\Rightarrow CD=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( -7-\left( -3 \right) \right)}^{2}}+{{\left( 8-4 \right)}^{2}}}$
After solving the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow CD=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -7+3 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\
& \Rightarrow CD=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\
\end{align}$
Now, we will square $2,-4$ and $4$ and will get $4,16$ and $16$ respectively as:
$\Rightarrow CD=\sqrt{4+16+16}$
And the sum of $4,16$ and $16$ is $36$. So, the above step will be:
$\Rightarrow CD=\sqrt{36}$
We know that the under root of $36$ is $6$.
$\Rightarrow CD=6$
Thus, $AB=CD=6$
Similarly, we will follow the procedure for $BC$ and $DA$ as:
Here, we will find the length of the side $BC$ with help of the formula as:
$\Rightarrow BC=\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( -2-\left( -7 \right) \right)}^{2}}+{{\left( 5-8 \right)}^{2}}}$
Solve the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -2+7 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
& \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 5 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
\end{align}$
Here, we will square $-3,5$ and $-3$ and will get $9,25$ and $9$ respectively as:
$\Rightarrow BC=\sqrt{9+25+9}$
Since, The sum of $9,25$ and $9$ is $43$. So, the length of $BC$ be:
$\Rightarrow BC=\sqrt{43}$
Here, we will find the length of $DA$ with the help of the formula as:
$\Rightarrow DA=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( -3-2 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}$
After solving the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow DA=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
& \Rightarrow DA=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
\end{align}$
Now, we will square square $-3,5$ and $-3$ and will get $9,25$ and $9$ respectively as:
$\Rightarrow DA=\sqrt{9+25+9}$
And the sum of $9,25$ and $9$ is $43$. So, the length of $DA$ will be:
\[\Rightarrow DA=\sqrt{43}\]
Thus, $BC=DA=\sqrt{43}$
Hence, $\left( -1,2,1 \right),\left( 1,-2,5 \right),\left( 4,-7,8 \right)$ and $\left( 2,-3,4 \right)$ are vertices of a parallelogram.
Note: Since, a parallelogram is a quadrilateral diagram with parallel sides in which opposite sides are equal and parallel to each other and similarly opposite angles are also equal and sum of all angles is equal to $360{}^\circ $. So, we use this property to verify the given question. We can also use the area of parallelogram to verify the given question by finding the area of parallelogram with the help of given points of vertices with different methods.
$=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$
Where, $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ are points.
Complete step by step answer:
Since, the given points of vertices are $\left( -1,2,1 \right),\left( 1,-2,5 \right),\left( 4,-7,8 \right)$ and $\left( 2,-3,4 \right)$.
Let’s suppose that they are the points of a parallelogram named \[\text{ABCD}\]. Then from the property of parallelogram, $AB=CD$ and $BC=DA$.
Here, we assumed that:
$\begin{align}
& \Rightarrow A=\left( -1,2,1 \right) \\
& \Rightarrow B=\left( 1,-2,5 \right) \\
& \Rightarrow C=\left( 4,-7,8 \right) \\
& \Rightarrow D=\left( 2,-3,4 \right) \\
\end{align}$
Now, we will find the length of the side $AB$ with help of the formula as:
$\Rightarrow AB=\sqrt{{{\left( -1-1 \right)}^{2}}+{{\left( 2-\left( -2 \right) \right)}^{2}}+{{\left( 1-5 \right)}^{2}}}$
Solve the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow AB=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\
& \Rightarrow AB=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 4 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\
\end{align}$
Here, we will do the squaring off $-2,4$ and $-4$ and will get $4,16$ and $16$ respectively as:
$\Rightarrow AB=\sqrt{4+16+16}$
The sum of $4,16$ and $16$ is $36$. So, the above step will be:
$\Rightarrow AB=\sqrt{36}$
As we know that the under root of $36$ is $6$.
$\Rightarrow AB=6$
Here, we will find the length of $CD$ with the help of the formula as:
$\Rightarrow CD=\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( -7-\left( -3 \right) \right)}^{2}}+{{\left( 8-4 \right)}^{2}}}$
After solving the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow CD=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -7+3 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\
& \Rightarrow CD=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( 4 \right)}^{2}}} \\
\end{align}$
Now, we will square $2,-4$ and $4$ and will get $4,16$ and $16$ respectively as:
$\Rightarrow CD=\sqrt{4+16+16}$
And the sum of $4,16$ and $16$ is $36$. So, the above step will be:
$\Rightarrow CD=\sqrt{36}$
We know that the under root of $36$ is $6$.
$\Rightarrow CD=6$
Thus, $AB=CD=6$
Similarly, we will follow the procedure for $BC$ and $DA$ as:
Here, we will find the length of the side $BC$ with help of the formula as:
$\Rightarrow BC=\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( -2-\left( -7 \right) \right)}^{2}}+{{\left( 5-8 \right)}^{2}}}$
Solve the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -2+7 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
& \Rightarrow BC=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 5 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
\end{align}$
Here, we will square $-3,5$ and $-3$ and will get $9,25$ and $9$ respectively as:
$\Rightarrow BC=\sqrt{9+25+9}$
Since, The sum of $9,25$ and $9$ is $43$. So, the length of $BC$ be:
$\Rightarrow BC=\sqrt{43}$
Here, we will find the length of $DA$ with the help of the formula as:
$\Rightarrow DA=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( -3-2 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}$
After solving the mathematical operation within the bracket as:
$\begin{align}
& \Rightarrow DA=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
& \Rightarrow DA=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
\end{align}$
Now, we will square square $-3,5$ and $-3$ and will get $9,25$ and $9$ respectively as:
$\Rightarrow DA=\sqrt{9+25+9}$
And the sum of $9,25$ and $9$ is $43$. So, the length of $DA$ will be:
\[\Rightarrow DA=\sqrt{43}\]
Thus, $BC=DA=\sqrt{43}$
Hence, $\left( -1,2,1 \right),\left( 1,-2,5 \right),\left( 4,-7,8 \right)$ and $\left( 2,-3,4 \right)$ are vertices of a parallelogram.
Note: Since, a parallelogram is a quadrilateral diagram with parallel sides in which opposite sides are equal and parallel to each other and similarly opposite angles are also equal and sum of all angles is equal to $360{}^\circ $. So, we use this property to verify the given question. We can also use the area of parallelogram to verify the given question by finding the area of parallelogram with the help of given points of vertices with different methods.
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