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Question
Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.
Solution
Let the line 2x + 3y – 5 = 0 divides the line segment joining the points A(8, –9) and B(2, 1) in the ratio λ : 1 at point P.
∴ Coordinates of P = `{(2λ + 8)/(λ + 1), (λ - 9)/(λ + 1)}` ...`[∵ "Internal division" = {(m_1x_2 + m_2x_1)/(m_1 + m_2), (m_1y_2 + m_2y_1)/(m_1 + m_2)}]`
But P lies on 2x + 3y – 5 = 0
∴ `2((2λ + 8)/(λ + 1)) + 3((λ - 9)/(λ + 1)) - 5` = 0
⇒ 2(2λ + 8) + 3(λ – 9) – 5(λ + 1) = 0
⇒ 4λ + 16 + 3λ – 27 – 5λ – 5 = 0
⇒ 2λ – 16 = 0
⇒ λ = 8
⇒ λ : 1 = 8 : 1
So, the point P divides the line in the ratio 8 : 1.
∴ Point of division P = `{(2(8) + 8)/(8 + 1), (8 - 9)/(8 + 1)}`
= `((16 + 8)/9, - 1/9)`
= `(24/9, (-1)/9)`
= `(8/3, (-1)/9)`
Hence, the required point of division is `(8/3, (-1)/9)`.
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