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Question
Given a triangle ABC in which A = (4, −4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Solution
Given, BP : PC = 3 : 2
Using section formula, the co-ordinates of point P are
`((3 xx 5 + 2 xx 0)/(3 + 2), (3 xx 10 + 2 xx 5)/(3 + 2))`
= `(15/5, 40/5)`
= (3, 8)
Using distance formula, we have:
`AP = sqrt((3 - 4)^2 + (8 + 4)^2)`
= `sqrt(1 + 144)`
= `sqrt(145)`
= 12.04
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