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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.

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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.

Sum
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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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Chapter 7: Coordinate Geometry - Exercise 7.3 [Page 84]

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NCERT Exemplar Mathematics [English] Class 10
Chapter 7 Coordinate Geometry
Exercise 7.3 | Q 20 | Page 84
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