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Find the joint equation of the pair of lines which bisect angles between the lines given by x2 + 3xy + 2y2 = 0 - Mathematics and Statistics

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प्रश्न

Find the joint equation of the pair of lines which bisect angles between the lines given by x2 + 3xy + 2y2 = 0 

बेरीज

उत्तर

x2 + 3xy + 2y2 = 0 

∴ x2 + 2xy + xy + 2y2 = 0

∴ x(x + 2y) + y(x + 2y) = 0

∴ (x + 2y)(x + y) = 0

∴ separate equations of the lines represented by x2 + 3xy + 2y2 = 0 are x + 2y = 0 and x + y = 0

Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P(x, y) from the line x + 2y = 0

= the distance of P(x, y) from the line x + y = 0

∴ `|("x" + "2y")/sqrt(1 + 4)| = |("x + y")/sqrt(1 + 1)|`

∴ `("x" + "2y")^2/5 = ("x + y")^2/2`

∴ 2(x + 2y)2 = 5(x + y)2 

∴ 2(x2 + 4xy + 4y2) = 5(x2 + 2xy + y2)

∴ 2x2 + 8xy + 8y2 = 5x2 + 10xy + 5y2

∴ 3x2 + 2xy - 3y2 = 0

This is the required joint equation of the lines which bisect the angles between the lines represented by x2 + 3xy + 2y2 = 0

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Angle between lines represented by ax2 + 2hxy + by2 = 0
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Pair of Straight Lines - Miscellaneous Exercise 4 [पृष्ठ १३१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 4 Pair of Straight Lines
Miscellaneous Exercise 4 | Q 6 | पृष्ठ १३१

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