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Question
A (2, 5), B (–1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that : AP : PB = AQ : QC = 1 : 2.
- Calculate the co-ordinates of P and Q.
- Show that : `PQ = 1/3 BC`.
Solution
i. Co-ordinates of P are
`((1 xx (-1) + 2 xx 2)/(1 + 2),(1 xx 2 + 2 xx 5)/(1 xx 2))`
= `(3/3, 12/3)`
= (1, 4)
Co-ordinates of Q are
`((1 xx 5 + 2 xx 2)/(1 + 2),(1 xx 8 + 2 xx 5)/(1 + 2))`
= `(9/3, 18/3)`
= (3, 6)
ii. Using distance formula, we have:
`BC = sqrt((5 + 1)^2 + (8 - 2)^2)`
`BC = sqrt(36 + 36)`
`BC = 6sqrt(2)`
`PQ = sqrt((3 - 1)^2 + (6 - 4)^2)`
`PQ = sqrt(4 + 4)`
`PQ = 2sqrt(2)`
Hence, `PQ = 1/3 BC`.
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