Question

(-1,-2),(1,0),(-1,2),(-3,0)​

Answer 1

Ascending order:- (-3,0)<(-1,-2)<(-1,2)<(1,0)

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Answer 2

Answer :

• The given points are the vertices of  a Square

Explanation :

Let the points (-1, -2), (1, 0), (-1, 2), and (-3, 0) be representing the vertices

• A, B, C, and D of the given quadrilateral respectively.

$\begin{array}{l}\sf{A B=\sqrt{(-1-1)^{2}+(-2-0)^{2}}=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}} \\\\\sf{B C=\sqrt{(1-(-1))^{2}+(0-2)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} }\\\\\sf{C D=\sqrt{(-1-(-3))^{2}+(2-0)^{2}}=\sqrt{(2)^{2}+(2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}}\end{array}$

$\begin{array}{l}\sf{A D=\sqrt{(-1-(-3))^{2}+(-2-0)^{2}}=\sqrt{(2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}} \\\\\sf{\text { Diagonal } A C=\sqrt{(-1-(-1))^{2}+(-2-2)^{2}}=\sqrt{0^{2}+(-4)^{2}}=\sqrt{16}=4} \\\\\sf{\text { Diagonal } B D=\sqrt{(1-(-3))^{2}+(0-0)^{2}}=\sqrt{(4)^{2}+0^{2}}=\sqrt{16}=4}\end{array}$

It can be observed that all sides of this quadrilateral are of the same length and also,  the diagonals are of the same length. Therefore, the given points are the vertices of  a Square