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Question
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of triangle ABC. Point P and Q lie on AB and AC respectively, such that AP: PB = AQ: QC = 1: 2. Calculate the coordinates of P and Q. Also, show that 3PQ = BC.
Solution
AP : PB = 1 : 2
Coordinates of P are,
P (x , y) = P `((-1 + 4)/(2 + 1) , (10 + 2)/(2 + 1)) = "P" (1 , 4)`
AQ : QC = 1 : 2
Coordinates of Q are,
Q (a , b) = Q `((4 + 5)/(2 + 1) , (10+ 8)/(2 + 1))` = Q (3 , 6)
Coordinates of P and Q are ( 1, 4) and (3, 6)
PQ = `sqrt ((3 - 1)^2 + (6 - 4)^2) = sqrt (4 + 4) = 2 sqrt 2` units
BC = `sqrt ((5 +1)^2 + (8 - 2)^2) = sqrt (36 + 36) = 6 sqrt 2` units
Hence proved, 3PQ = BC
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