Answer:
JKLM represents all three(Rhombus, Square and Rectangle)
Step-by-step explanation:
Concepts Used:
Pythagorean Theorem: In a right triangle the square of longest side is equal to the sum of the squares of the other sides.
Distance between two points:
[tex]Distance\ between\ (x_1,y_1)(x_2,y_2)\ is\ given\ by\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\J(-1,1)\ K(4,1)\ L(4,6)\ M(-1,6)\\\\JK=\sqrt{(4+1)^2+(1-1)^2}\\\\JK=\sqrt{5^2+0^2}\\\\JK=\sqrt{25}\\\\JK=5\\\\KL=\sqrt{(4-4)^2+(6-1)^2}\\\\KL=\sqrt{0^2+5^2}\\\\KL=\sqrt{25}\\\\KL=5\\\\LM=\sqrt{(-1-4)^2+(6-6)^2}\\\\LM=\sqrt{(-5)^2+0^2}\\\\LM=\sqrt{25}\\\\LM=5\\\\JM=\sqrt{(-1+1)^2+(6-1)^2}\\\\JM=\sqrt{0^2+5^2}\\\\JM=\sqrt{25}\\\\JM=5\\\\[/tex]
[tex]all\ sides\ are\ equal\\\\Now\ find\ out\ diagonal\\\\JL=\sqrt{(4+1)^2+(6-1)^2}\\\\JL=\sqrt{5^2+5^2}\\\\JL=\sqrt{25+25}\\\\JL=\sqrt{50}\\\\JL=\sqrt{25\times 2}\\\\JL=5\sqrt{2}\\\\KM=\sqrt{(-1-4)^2+(6-1)^2}\\\\KM=\sqrt{(-5)^2+(5)^2}\\\\KM=\sqrt{25+25}\\\\KM=\sqrt{50}\\\\KM=\sqrt{25\times 2}\\\\KM=5\sqrt{2}[/tex]
[tex]Now\ in\ \triangle JKL\\\\JL^2=JK^2+KL^2\\\\Hence\ \triangle JKL\ is\ a\ right\ triangle\\\\and\ \angle K=90\textdegree\\\\Similarly\ in\ \triangle KLM\\\\KM^2=KL^2+LM^2\\\\Hence\ \triangle KLM\ is\ a\ right\ triangle\\\\\angle L=90\textdegree\\\\Similerly\ \triangle JML\\\\JL^2=JM^2+ML^2\\\\Pythagorean\ theorem\ satisfied\\\\It\ is\ a\ right\ triangle\\\\\angle M=90\textdegree\\\\\angle J=360-(\angle M+\angle L+\angle K)\\\\\angle J=360-(90+90+90)\\\\\angle J=90\textdegree\\\\[/tex]
[tex]All\ sides\ are\ equal\ and\ angles\ are\ 90\textdegree\\\\Hence\ it\ is\ a\ square\\\\ Every\ square\ is\ a\ rectangle\ and\ Every\ square\ is\ a\ rhombus.\\\\Hence\ it\ represents\ all\ three[/tex]
