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प्रश्न
Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) and B(1, 2). Also find k.
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उत्तर
Let A(x1, y1), B(x2, y2) and P(x, y) be the given points.
Here, x1 = 8, y1 = 9, x2 = 1, y2 = 2, x = k, y = 7
∴ By section formula,
`y = (my_2 + ny_1)/(m + n)`
∴ `7 = (2m + 9n)/(m + n)`
∴ 7(m + n) = 2m + 9n
∴ 7m + 7n = 2m + 9n
∴ 5m = 2n
∴ `m/n = 2/5`
m : n = 2 : 5
`x = (mx_2 + nx_1)/(m + n)`
∴ `k = (2(1) + 5(8))/(2 + 5)`
∴ `k = (2 + 40)/(7)`
∴ `k = (42)/(7)`
∴ k = 6
∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.
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Solution:
Point P divides segment AB in the ratio m : n.
A(8, 9) = (x1, y1), B(1, 2) = (x2, y2) and P(6, 7) = (x, y)
Using Section formula of internal division,
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∴ 7m + 7n = `square` + 9n
∴ 7m – `square` = 9n – `square`
∴ `square` = 2n
∴ `m/n = square`
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