prove the midpoint of the hypotenuse of a right angle is equidistant from its tree vertex P(-2,5),Q(1,3)and R(-1,0)
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by using distance formula that is.... AB=√(x2-x1)^+(y2-y1)^ we can find....for P And Q where , x2=1 , x1= -2 , y2=3 , y1=5...
PQ=√[1-(-2)]^+(3-5)^ = √(-3)^+(-2)^ = √13..........for Q and R , x2= -1 , x1=1 , y1=3 , y2=0....PQ =√(-1-1)^+(0-3)^=√(-2)^(-3)^ = √13 it is coming for PQ and QR equal but it is not coming for RP it is coming double of it = √26
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