Math, asked by Anonymous, 5 months ago

If we plot the coordinate points A(-1, 0) , B(0, 1) , C(0,-1) and D(-1, 0) on Cartesian plane .
Which figure comes out after joining the Points? Also find the side of figure.

Answers

Answered by darksoul3
29

Given:

A(-1,0)

B(0,1)

C(0,-1)

D(1,0)

By using distance Formula,

 =  \sqrt{ {(x2 - x1)}^{2} +  {(y2 - y1)}^{2}  }

So,

AB =  \sqrt{ {(0 - ( -1)}^{2}  +  {(1 - 0)}^{2} }

 =  \sqrt{ {(1)}^{2} +  {(1)}^{2}  }

 =  \sqrt{1 + 1}

 =  \sqrt{2}

BC =  \sqrt{ {(0 - 0)}^{2} +  {( - 1 - 1)}^{2}  }

 =   \sqrt{{(0)}^{2}  +  {( - 2)}^{2} }

 =  \sqrt{0 + 4}

 =  \sqrt{4}

= 2 units

CD =  \sqrt{ {( - 1 - 0)}^{2} +  {(0 - ( -1 )}^{2}  }

 =  \sqrt{ {( - 1)}^{2}  + {(1)}^{2} }

 =  \sqrt{1 + 1}

 =  \sqrt{2}

DA =  \sqrt{ {( - 1  - 1)}^{2}  +  {(0  -  0)}^{2} }

 =  \sqrt{ {(2)}^{2} +  {(0)}^{2}  }

 =  \sqrt{4}

= 2 units

So,

AB = CD & BC = DA

Hence it is a rectangle. As we know that opposite side of a rectangle are equal.


Anonymous: ❤Thank uh! :Meow Sip:
Answered by Anonymous
57

\huge\orange\star\red{\textsf { \textbf{\underline{Answer :- }}}}

By using distance Formula,

 \sf{ \implies\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }}

So,

 \sf  {\pink{\underline{AB = \sqrt{ {(0 - ( -1)}^{2} + {(1 - 0)}^{2} } }}}

 \:  \:  \:  \:  \:  \sf \implies\sqrt{ {(1)}^{2} + {(1)}^{2} }

  \:  \:  \:  \:  \: \sf \implies \sqrt{1 + 1}

 \:  \:  \:  \:  \:  \sf\implies \sqrt{2}

 \sf \pink {\underline{BC = \sqrt{ {(0 - 0)}^{2} + {( - 1 - 1)}^{2} }}}

 \:  \:  \:  \:  \:   \sf\implies\sqrt{{(0)}^{2} + {( - 2)}^{2} }

 \:  \:  \:  \:  \:  \sf \implies\sqrt{0 + 4}

 \:  \:  \:  \:  \:  \sf \implies \sqrt{4}

  \:  \:  \:  \:  \: \sf \implies 2  \: units

 \sf \pink{ \underline{CD = \sqrt{ {( - 1 - 0)}^{2} + {(0 - ( -1 )}^{2} }}}

  \:  \:  \:  \:  \: \sf \implies \sqrt{ {( - 1)}^{2} + {(1)}^{2} }

  \:  \:  \:  \:  \: \sf \implies\sqrt{1 + 1} \:

 \:  \:  \:  \:  \:  \sf \implies \sqrt{2}  \:

 \sf \pink{ \underline{DA = \sqrt{ {( - 1 - 1)}^{2} + {(0 - 0)}^{2} }}}

 \:  \:  \:  \:  \:  \sf {\implies\sqrt{ {(2)}^{2} + {(0)}^{2} }}  \:

 \:  \:  \:  \:  \:  \sf\implies \sqrt{4}  \:

 \:  \:  \:  \:  \:  \sf \implies{2  \: units} \:

So,

AB = CD & BC = DA

Hence it is a rectangle. As we know that opposite side of a rectangle are equal.

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