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Show that (3,2),(2,-3),(0,0) are the vertices of a right angled triangle using distance formula

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Assinging the points:-
(x_1,y_1)=(3,2) 
(x_2,y_2)=(2,-3)
(x_3,y_3)=(0,0)
	herefore Distance between two points=sqrt(x_2-x_1)^2+(y_2-y_1)^2
AB=sqrt(2-3)^2+(-3-2)^2=sqrt1+25=sqrt26, units
BC=sqrt(0-2)^2+(0+3)^2=sqrt4+9=sqrt13, units
CA=sqrt(0-3)^2+(0-2)^2=sqrt9+4=sqrt13, units
(sqrt13)^2+(sqrt13)^2=(sqrt26)^2
	herefore BC^2+CA^2=AB^2
Hence the given points form a right angled traingle.

Posted by

Deependra Verma

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