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Question

Let P be an interior point of ΔABC such that 4PA+3PB+5PC=0. If Area(ΔABC)=k×Area(ΔAPC), then k is

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Solution


Let a=PA, b=PB, c=PC
4a+3b+5c=0 (1)

Area(ΔAPC)=12c×a
Area(ΔABC)=Area(ΔAPB)+Area(ΔBPC)+Area(ΔAPC)
Area(ΔABC)=12a×b+b×c+c×a

Taking cross product of equation (1) with a and c one at a time, we get
3a×b=5c×a
and 3b×c=4c×a
a×b=53(c×a)
and b×c=43(c×a)

Area(ΔABC)=42c×a =4×Area(ΔAPC)
Hence, k=4

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