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