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Question

If two opposite vertices of a square are (5, 4) and (1, -6) then the coordinates of its remaining two vertices are

A
(2,2) & (5,3)
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B
(8,3) & (2,1)
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C
(8,6) & (3,5)
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D
(1,3) & (2,5)
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Solution

The correct option is D (8,3) & (2,1)
SolutionABCDisasquare.SoAB=BC=CD=DA&thediagonalsAC=BD..........(i).Weshallapplydistanceformulatogettheabovelengths.d=(x1x2)2+(y1y2)2.NowΔABCisarightonewithACashypotenuse(ABC=90o).AC=(51)2+(4+6)2units=116units=BD(byi)........(ii).Weknowthatsideofasquare=diagonal2.AB=BC=CD=DA=1162units=58units.........(iii).Now,usingthedistanceformulad=(x1x2)2+(y1y2)2,AB2=BC2(x15)2+(y14)2=(x11)2+(y1+6)28x1+20y14=02x1+5y11=0y=12x15........(iv).AB2+BC2=AC2(x15)2+(y14)2+58=116(fromii&iii)(x15)2+(y14)2=58(x15)2+(12x154)2=5829x12174x1464=0x126x116=0(x18)(x1+2)=0x1=(8,2).So,from(iv),y1=(12×85,12(2)5)=(3,1).B(x1,y1)=(8,3)and(2,1).SimilarlyAD2=DC2(x25)2+(y24)2=(x21)2+(y2+6)28x2+20y24=02x2+5y21=0y2=12x25........(iv).AD2+DC2=AC2(x25)2+(y24)2+58=116(fromii&iii)(x25)2+(y24)2=58(x25)2+(12x254)2=5829x22174x2464=0x226x216=0(x28)(x2+2)=0x2=(8,2).So,from(iv),y2=(12×85,12(2)5)=(3,1).D(x2,y2)=(8,3)and(2,1).Sothecoordinatesofothertwoverticesare(8,3)and(2,1).ansOptionB.
393430_291495_ans.png

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