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Prove that $(4, - 1),(6,0),(7,2)$ and $(5,1)$ are the vertices of a rhombus. Is it a square?

Last updated date: 16th Jun 2024
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Answer
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Hint: Here, we have to prove that the given points are the vertices of a rhombus. We will prove this by using the distance between two points formula. If all the length of the sides are equal then the given points will form the vertices of a rhombus. If the length of the diagonals are equal, then the vertices of a rhombus will form a square.

Formula used:
We will use the formula of distance between two points which is given $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, where $({x_1},{y_1})$ and $({x_2},{y_2})$ be the two points.

Complete Complete Step by Step Solution:
We will first draw the diagrams showing all the points.

Let ABCD be the vertices of rhombus.
Now, we will be using the distance between two points formula for all the sides to prove that the given points are the vertices of a rhombus.
Now, we have to find the distance between A$(4, - 1)$ and B $(6,0)$ using the distance formula.
Substituting ${x_1} = 4$, ${x_2} = 6$, ${y_1} = - 1$ and ${y_2} = 0$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$AB = \sqrt {{{(6 - 4)}^2} + {{(0 - ( - 1))}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow AB = \sqrt {{{(2)}^2} + {{(1)}^2}}$
Applying the exponent on the terms, we get
$\Rightarrow AB = \sqrt {4 + 1}$
Adding the terms, we get
$\Rightarrow AB = \sqrt 5$ ………………….. $(1)$
Now, we have to find the distance between B$(6,0)$ and C $(7,2)$ using the distance formula.
Substituting ${x_1} = 6$, ${x_2} = 7$, ${y_1} = 0$ and ${y_2} = 2$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$BC = \sqrt {{{(7 - 6)}^2} + {{(2 - 0)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow BC = \sqrt {{1^2} + {2^2}}$
Applying the exponent on the terms, we get
$\Rightarrow BC = \sqrt {1 + 4}$
Adding the terms, we get
$\Rightarrow BC = \sqrt 5$ ……………….$\left( 2 \right)$
Now, we have to find the distance between C $(7,2)$ and D $\left( {5,1} \right)$ using the distance formula.
Substituting ${x_1} = 7$, ${x_2} = 5$, ${y_1} = 2$ and ${y_2} = 1$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$CD = \sqrt {{{\left( {5 - 7} \right)}^2} + {{(1 - 2)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow CD = \sqrt {{{( - 2)}^2} + {{( - 1)}^2}}$
Applying the exponent on the terms, we get
$\Rightarrow CD = \sqrt {4 + 1}$
Adding the terms, we get
$\Rightarrow CD = \sqrt 5$ ……………………. $\left( 3 \right)$
Now, we have to find the distance between D $\left( {5,1} \right)$and A$(4, - 1)$using the distance formula.
Substituting ${x_1} = 7$, ${x_2} = 5$, ${y_1} = 2$ and ${y_2} = 1$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$DA = \sqrt {{{(4 - 5)}^2} + {{( - 1 - 1)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow DA = \sqrt {{{( - 1)}^2} + {{( - 2)}^2}}$
Applying the exponent on the terms, we get
$\Rightarrow DA = \sqrt {1 + 4}$
Adding the terms, we get
$\Rightarrow DA = \sqrt 5$ …………………. $\left( 4 \right)$
Since all the length of the sides of a square are equal, the given points form the vertices of a rhombus.
Now, we have to check whether it is a square.
Now, we have to find the length of the diagonals A $(4, - 1)$ and C $(7,2)$ using the distance formula.
Substituting ${x_1} = 4$, ${x_2} = 7$, ${y_1} = - 1$ and ${y_2} = 2$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$\Rightarrow AC = \sqrt {{{(7 - 4)}^2} + {{(2 - ( - 1))}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow AC = \sqrt {{{(3)}^2} + {{(3)}^2}}$
Applying the exponent on the terms, we get
$\Rightarrow AC = \sqrt {9 + 9}$
Adding the terms, we get
$\Rightarrow AC = \sqrt {18}$ ……………… $\left( 5 \right)$
Now, we have to find the length of the diagonals B $(6,0)$ and D $\left( {5,1} \right)$ using the distance formula.
Substituting ${x_1} = 6$, ${x_2} = 5$, ${y_1} = 0$ and ${y_2} = 1$ in the formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$, we get
$\Rightarrow BD = \sqrt {{{\left( {5 - 6} \right)}^2} + {{\left( {1 - 0} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow BD = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}}$
Applying the exponent on the terms, we get
$\Rightarrow BD = \sqrt {1 + 1}$
Adding the terms, we get
$\Rightarrow BD = \sqrt 2$ ………………….$(6)$

Since the length of the diagonals are not equal. Hence the given points do not form a square.
Therefore, $(4, - 1),(6,0),(7,2)$ and $(5,1)$ are the vertices of a rhombus and it is not a square.

Note:
We can prove that the given vertices form a parallelogram by using the midpoint formula. The midpoint formula can be used for finding the midpoint of the diagonal. If the midpoint of both the diagonals are equal, then it forms a parallelogram. If the opposite sides of a parallelogram are equal, then it forms a rhombus. If the length of the diagonal are equal, then it forms a square.