Solution:-
Steps for marking the points on graph:
1. As per the given data plot the points 
2. Then plot point M whose vertices are 
3. Condition given the question, taking P as the point of symmetry.
4. So, point symmetric to 
5. Point symmetric to 
6. Now join AP, PC, BP and PD
By using the distance formula,
![\[AD\text{ }=\text{ }\surd ({{\left( 8\text{ }\text{ }6 \right)}^{2}}~+\text{ }{{\left( 2\text{ }\text{ }0 \right)}^{2}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fb0f7615f8a9581ad55bad4cd2f5731e_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }\surd ({{2}^{2}}~+\text{ }{{2}^{2}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-01c4e257a0f13276142a58e44bd88bea_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }\surd \left( 4\text{ }+\text{ }4 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00abf3952ee02a7d72af8847a6c59219_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }\surd 8\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f85c4fcd7b213812685e62191648faf3_l3.png "Rendered by QuickLaTeX.com")
Then,
![\[AB\text{ }=\text{ }\surd ({{\left( 8\text{ }\text{ }6 \right)}^{2}}~+\text{ }{{\left( 2\text{ }\text{ }4 \right)}^{2}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e0af9ee4735172cc3b276348a9e94e0_l3.png "Rendered by QuickLaTeX.com")
![\[=\text{ }\surd ({{2}^{2}}~+\text{ }(-{{2}^{2}}))\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9369993f1f082338ec434eaa66ea704e_l3.png "Rendered by QuickLaTeX.com")
So, from Pythagoras theorem
![\[B{{D}^{2}}~=\text{ }A{{D}^{2}}~+\text{ }A{{B}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a74dc338b0806ae51c02863056acbd21_l3.png "Rendered by QuickLaTeX.com")
![\[{{4}^{2}}~=\text{ }{{\left( \surd 8 \right)}^{2}}~+\text{ }{{\left( \surd 8 \right)}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1310a2f4b46abdb286ff934810a86867_l3.png "Rendered by QuickLaTeX.com")
![\[16\text{ }=\text{ }8\text{ }+\text{ }8\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ff5629ad75ce5f07604346b5b07912ed_l3.png "Rendered by QuickLaTeX.com")
![\[16\text{ }=\text{ }16\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a0e0d7631c7b32619e9e1bf2a68eb3ca_l3.png "Rendered by QuickLaTeX.com")
Therefore, 
Hence, it is clear that AB = BC = CD = DA, AC and BD bisect each other at right angles, so ABCD is a square.