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  • Prove that the points (3,0) (6,4) and (-1,3) are the vertices of a right-angled isosceles triangle         

Prove that the points (3,0) (6,4) and (-1,3) are the vertices of a right-angled isosceles triangle 

 

 

 

 

 
 
 
 
 

Answers (1)

Let A = (3,0) , B = (6,4) ; c = (-1,3) 

A B = \sqrt { (3-6)^2+ ( 0-4)^2} = \sqrt { 3^2 + 4 ^2 } = \sqrt { 25}\\\\ BC = \sqrt {( 6 - (-1))^2 + ( 4-3)^2} = \sqrt { 7 ^2 + 1 ^2} = \sqrt { 50 } \\\\ AC = \sqrt { 3 - ( -1 )^2+ (0-3 )^2} = \sqrt { 4^2 + 3 ^2 } = \sqrt {25}

Now were can say,  AB = AC = \sqrt {25}

AC ^2 = \sqrt { 50} ^ 2 = 50 ---( 1) \\\\ AB ^2 = \sqrt { 25}^ 2 = 25 --- (2) \\\\ BC ^ 2 = \sqrt { 25} ^2 = 25 --- (3) \\\\ AC ^ 2 = AB ^2 + BC ^ 2 

using (1) (2) (3)

Hence using Pythagoras theorem it is proved that \Delta ABC right-angled isosceles triangle 

 

Posted by

Ravindra Pindel

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