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If A and B are real numbers between 0 and 1 such that the points (a,1) ,(1,b)and (0,0) form an equilateral triangle then 2(a+b) -ab is equal to

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@shekhar

If A and B are real numbers between 0 and 1 such that the points (a,1) ,(1,b)and (0,0) form an equilateral triangle then 2(a+b) -ab is equal to

\\Since,(0,0),(a,1)\:\:and\:\:(1,b)\:form\:an\:equilateral\:triangle\\\therefore all\:three\:side\:of \:triangle\:are\:equal\\

{\color{DarkBlue} \\Distance\:between\:two\:point\:(x_1,y_1)\:and\:(x_2,y_2)\:is\:\:\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}

\\Distance\:b/w\:(0,0)\:and\:(a,1)=Distance\:b/w\:(0,0)\:and\:(1,b) \\\sqrt{a^2+1}=\sqrt{1+b^2}\Rightarrow a=b\\\because a,b\:lies\:between\:0\:and\:1\\Distance\:b/w\:(0,0)\:and\:(a,1)=Distance\:b/w\:(a,1)\:and\:(1,b)\\\sqrt{a^2+1}=\sqrt{(a-1)^2+(1-b)^2}\\a^2+1=\pm(a^2-2a+1+1-2b+b^2) \\ \because a=b, b^2-4b+1=0\\we\:need\:to\:find\:2(a+b)-ab=4b-b^2=1

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himanshu.meshram

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