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प्रश्न
Find the ratio in which point T(–1, 6)divides the line segment joining the points P(–3, 10) and Q(6, –8).
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उत्तर
Let the ratio be k : 1.
Using section formula we have
\[- 1 = \frac{6k - 3 \times 1}{k + 1}\]
\[ \Rightarrow - k - 1 = 6k - 3\]
\[ \Rightarrow - 1 + 3 = 6k + k\]
\[ \Rightarrow 2 = 7k\]
\[ \Rightarrow k = \frac{2}{7}\]
Thus, the required ratio is 2 : 7.
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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)
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∴ 7m + 7n = `square` + 9n
∴ 7m – `square` = 9n – `square`
∴ `square` = 2n
∴ `m/n = square`
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