Solution:-
Steps for marking the points on graph:
1. As per the given data plot the points 
and 
on the graph.
2. Then plot point M whose vertices are 
and 
.
3. Condition given the question, taking P as the point of symmetry.
4. So, point symmetric to in the line is 
5. Point symmetric to 
in the line is 
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.