Locus of the Points Equidistant From a Given Point
A straight li...
Question
A straight line through the vertex P of the △PQR intersects the side QR at S and the circumcircle of the△PQR at T. If S is not the centre of the circumcircle, then
A
1PS+1ST<2√QS⋅SR
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B
1PS+1ST>2√QS⋅SR
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C
1PS+1ST<4QR
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D
1PS+1ST>4QR
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Solution
The correct options are B1PS+1ST>2√QS⋅SR D1PS+1ST>4QR Since S is not the centre of the circumcenter PS≠ST,QS≠SR Since A.M.>G.M. 12(1PS+1ST)>1√PS×ST⇒1PS+1ST>2√QS×ST[∵PS×ST=QS×SR] We know that if x+y=k, a constant then xy is maximum if x=y=k2 Thus ⇒1QS×SR>4(QS)2⇒1PS+1ST>2√QS×SR>4QR