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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: 23rd Feb 2024
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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
$\Rightarrow AB = \sqrt {4 + 1}$
$\Rightarrow AB = \sqrt 5$ ………………….. $(1)$
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
$\Rightarrow BC = \sqrt {1 + 4}$
$\Rightarrow BC = \sqrt 5$ ……………….$\left( 2 \right)$
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
$\Rightarrow CD = \sqrt {4 + 1}$
$\Rightarrow CD = \sqrt 5$ ……………………. $\left( 3 \right)$
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
$\Rightarrow DA = \sqrt {1 + 4}$
$\Rightarrow DA = \sqrt 5$ …………………. $\left( 4 \right)$
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
$\Rightarrow AC = \sqrt {9 + 9}$
$\Rightarrow AC = \sqrt {18}$ ……………… $\left( 5 \right)$
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
$\Rightarrow BD = \sqrt {1 + 1}$
$\Rightarrow BD = \sqrt 2$ ………………….$(6)$
Therefore, $(4, - 1),(6,0),(7,2)$ and $(5,1)$ are the vertices of a rhombus and it is not a square.