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Show that the points (−4, −4), (−2, −4), (4,0) and (2,3) are not the vertices points of a rectangle.

Answer
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HINT- Proceed the solution of this question with the assumption that this polygon will be a rectangle, so it has to fulfill all the conditions of the rectangle, if any single condition doesn’t get fulfilled, then we can say it will not be a rectangle.

Complete step-by-step answer:
Let, rectangle be ABCD, where A= (−4, −4), B= (−2, −4), C= (4,0) and D = (2,3)
In Rectangle, opposite sides are equal and diagonals are equal.
Let the coordinates of point M $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$ and N $\left( {{{\text{x}}_2},{{\text{y}}_2}} \right)$
So Distance between two points M and N will be = \[\sqrt {{{\left( {{{\text{x}}_2} - {{\text{x}}_1}} \right)}^2} + {{\left( {{{\text{y}}_2} - {{\text{y}}_1}} \right)}^2}} \]​………….(1)
A= (−4, −4), B= (−2, −4)
Here, \[{{\text{x}}_1} = - 4,{{\text{y}}_1} = - 4,{{\text{x}}_2} = - 2,{{\text{y}}_2} = - 4\]
So length of side AB= $\sqrt {{{\left( { - 2 - ( - 4)} \right)}^2} + {{\left( { - 4 - ( - 4)} \right)}^2}} {\text{ = }}\sqrt {4 + 0} {\text{ = 2 }}$

using formula (1)
B= (−2, −4), C= (4,0)
Here, \[{{\text{x}}_1} = - 2,{{\text{y}}_1} = - 4,{{\text{x}}_2} = 4,{{\text{y}}_2} = 0\]
So length of side BC= $\sqrt {{{\left( {4 - ( - 2)} \right)}^2} + {{\left( {0 - ( - 4)} \right)}^2}} {\text{ = }}\sqrt {36 + 16} {\text{ = }}\sqrt {{\text{52 }}} = 2\sqrt {13} $ ​

using formula (1)
C= (4,0), D= (2,3)
Here, \[{{\text{x}}_1} = 4,{{\text{y}}_1} = 0,{{\text{x}}_2} = 2,{{\text{y}}_2} = 3\]
So length of side CD= $\sqrt {{{\left( {2 - (4)} \right)}^2} + {{\left( {3 - (0)} \right)}^2}} {\text{ = }}\sqrt {4 + 9} {\text{ = }}\sqrt {13} $

using formula (1)
 D= (2,3), A= (−4, −4),
Here, \[{{\text{x}}_1} = 2,{{\text{y}}_1} = 3,{{\text{x}}_2} = - 4,{{\text{y}}_2} = - 4\]
So length of side DA= $\sqrt {{{\left( { - 4 - (2)} \right)}^2} + {{\left( { - 4 - (3)} \right)}^2}} {\text{ = }}\sqrt {36 + 49} {\text{ = }}\sqrt {85} $

Here, after calculating the length of all the sides,
Opposite sides are AB =2 cm and CD= $\sqrt {13} $ cm which are not equal and other
Opposite sides are BC =2$\sqrt {13} $ cm and DA= $\sqrt {85} $cm which are also not equal.
Hence the side's condition failed to prove its rectangle.

using formula (1)
 A= (−4, −4), C= (4,0)
Here, \[{{\text{x}}_1} = - 4,{{\text{y}}_1} = - 4,{{\text{x}}_2} = 4,{{\text{y}}_2} = 0\]
So length of diagonal AC= $\sqrt {{{\left( {4 - ( - 4)} \right)}^2} + {{\left( {0 - ( - 4)} \right)}^2}} {\text{ = }}\sqrt {64 + 16} {\text{ = }}\sqrt {80} {\text{ = 4}}\sqrt 5 $

 using formula (1)

B= (−2, −4), D= (2,3)
Here, \[{{\text{x}}_1} = - 2,{{\text{y}}_1} = - 4,{{\text{x}}_2} = 2,{{\text{y}}_2} = 3\]

So length of diagonal BD= $\sqrt {{{\left( {2 - ( - 2)} \right)}^2} + {{\left( {3 - ( - 4)} \right)}^2}} {\text{ = }}\sqrt {16 + 49} {\text{ = }}\sqrt {65} {\text{ }}$

Here, after calculating the length of both the diagonals AC=$4\sqrt 5 $ and BD = $\sqrt {65} $ which are not equal.
Hence even diagonal conditions failed to prove it rectangle.
Hence this polygon will not be a rectangle.

Note- There are also other properties which prove its rectangle, like adjacent sides of rectangles, makes an interior angle of 90 degrees and its diagonals bisect each other. Hence we can also verify these conditions to prove it rectangle.
A diagonal of a rectangle is a diameter of its circumcircle so geometrically we can also prove it a rectangle.

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