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Question
A (–3, 4), B (3, –1) and C (–2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP : PC = 2 : 3.
Solution
BP : PC = 2 : 3
Co-ordinates of P are
`((2 xx (-2) + 3 xx 3)/(2 + 3),(2 xx 4 + 3 xx (-1))/(2 + 3))`
= `((-4 + 9)/5, (8 - 3)/5)`
= (1, 1) ...(i)
Using distance formula, we have:
`AP = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
Let A (–3 , 4) x1 = –3, y1 = 4,
(1, 1) x2 = 1, y2 = 1 ...[From (i) we get]
`AP = sqrt((1 + 3)^2 + (1 - 4)^2)`
= `sqrt(16 + 9)`
= `sqrt(25)`
= 5 units.
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