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प्रश्न
The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
उत्तर
Given that, the line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2.
∴ Coordinate of point P = `{(5(1) + 3(2))/(1 + 2), (1(1) + 2(2))/(1 + 2)}` ...`[∵ "By section formula for internal ratio" = ((m_1x_2 + m_2x_1)/(m_1 + m_2), (m_1y_2 + m_2y_1)/(m_1 + m_2))]`
= `((5 + 6)/3, (1 + 4)/3)`
= `(11/3, 5/3)`
But the point `P(11/3, 5/3)` lies on the line 3x – 18y + k = 0 ...[Given]
∴ `3(11/3) - 18(5/8) + k` = 0
⇒ 11 – 30 + k = 0
⇒ k – 19 = 0
⇒ k = 19
Hence, the required value of k is 19.
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