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प्रश्न
Show that the line 3x + 4y + 5 = 0 and the lines (3x + 4y)2 - 3(4x - 3y)2 = 0 form the sides of an equilateral triangle.
उत्तर
The slope of the line 3x + 4y + 5 = 0 is m1 = `(-3)/4`
Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m1 is 60°.
∴ tan 60° = `|("m" - "m"_1)/(1 + "m"."m"_1)|`, where tan 60° = `sqrt3`
∴ `sqrt3 = |("m" - (-3/4))/(1 + "m"(-3/4))|`
∴ `sqrt3 = |("4m" + 3)/(4 - "3m")|`
On squaring both sides, we get,
`3 = ("4m" + 3)^2/(4 - "3m")^2`
∴ 3(4 - 3m)2 = (4m + 3)2
∴ 3(16 - 24m + 9m2) = 16m2 + 24m + 9
∴ 48 - 72m + 27m2 = 16m2 + 24m + 9
∴ 11m2 - 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = `"y"/"x"`.
∴ the combined equation of the two lines is
`11("y"/"x")^2 - 96("y'/"x") + 39 = 0`
∴ `"11y"^2/"x"^2 - "96y"/"x" + 39 = 0`
∴ 11y2 - 96xy + 39x2 = 0
∴ 39x2 - 96xy + 11y2 = 0
∴ 39x2 - 96xy + 11y2 = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0
The equation 39x2 - 96xy + 11y2 = 0 can be written as: - 39x2 + 96xy - 11y2 = 0
i.e. `(9"x"^2 - 48"x"^2) + (24"xy" + 72"xy") + (16"y"^2 - 27"y"^2) = 0`
i.e. `(9"x"^2 + 24"xy" + 16"y"^2) - 3(16"x"^2 - 24"xy" + 9"y"^2) = 0`
i.e. (3x + 4y)2 - 3(4x - 3y)2 = 0
Hence, the line 3x + 4y + 5 = 0 and the lines (3x + 4y)2 - 3(4x - 3y)2 form the sides of an equilateral triangle.
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