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प्रश्न
Find the combined equation of bisectors of angles between the lines represented by 5x2 + 6xy - y2 = 0.
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उत्तर
Comparing the equation 5x2 + 6xy - y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 5, 2h = 6, b = -1
Let m1 and m2 be the slopes of the lines represented by 5x2 + 6xy - y2 = 0.
∴ m1 + m2 = `(-2"h")/"b" = (-6)/-1 = 6` and m1m2 = `"a"/"b" = 5/-1 = - 5` ...(1)
The separate equations of the lines are
y = m1x and y = m2x, where m1 ≠ m2
i.e. m1x - y = 0 and m2x - y = 0.
Let P(x, y) be any point on one of the bisector of the angles between the lines.
∴ the distance of P from the line m1x - y = 0 is equal to the distance of P from the line m2x - y = 0.
∴ `|("m"_1"x" - "y")/(sqrt("m"_1^2 + 1))| = |("m"_2"x" - "y")/(sqrt("m"_2^2 + 1))|`
Squaring both sides, we get,
`("m"_1"x" - "y")^2/("m"_1^2 + 1) = ("m"_2"x" - "y")/("m"_2^2 + 1)`
∴ `("m"_2^2 + 1)("m"_1"x" - "y")^2 = ("m"_1^2 + 1)("m"_2"x" - "y")`
∴ `("m"_2^2 + 1)("m"_1^2"x"^2 - 2"m"_1"xy" + "y"^2) = ("m"_1^2 + 1)("m"_2^2"x"^2 - 2"m"_2"xy" + "y"^2)`
∴ `"m"_1^2"m"_2^2"x"^2 - 2"m"_1"m"_2^2"y"^2"xy" + "m"_2^2"y"^2 + "m"_1^2"x"^2 - 2"m"_1"xy" + "y"^2 = "m"_1^2"m"_2^2"x"^2 - 2"m"_1^2"m"_2"xy" + "m"_1^2"y"^2 + "m"_2^2"x"^2 - 2"m"_2"xy" + "y"^2`
∴ `("m"_1^2 - "m"_2^2)"x"^2 + 2"m"_1"m"_2("m"_1 -"m"_2)"xy" - 2("m"_1 - "m"_2)"xy" - ("m"_1^2 - "m"_2^2)"y"^2 = 0`
Dividing throughout by `"m"_1- "m"_2` (≠ 0), we get,
`("m"_1 + "m"_2)"x"^2 + 2"m"_1"m"_2"xy" - 2"xy" - ("m"_1 + "m"_2)"y"^2 = 0`
∴ 6x2 - 10xy - 2xy - 6y2 = 0 ...[By (1)]
∴ 6x2 - 12xy - 6y2 = 0
∴ x2 - 2xy - y2 = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x2 + 6xy - y2 = 0.
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