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Question
Show that the lines x2 − 4xy + y2 = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is −1.
Let OA(or OB) has slope m.
∴ its equation is y − mx ...(1)
Also, tan 60° = `|("m" − (−1))/(1 + "m"(−1))|`
`therefore sqrt3 = |("m + 1")/(1 - "m")|`
Squaring both sides, we get,
`3 = ("m" + 1)^2/(1 - "m")^2`
∴ 3(1 − 2m + m2) = m2 + 2m + 1
∴ 3 − 6m + 3m2 = m2 + 2m + 1
∴ 2m2 − 8m + 2 = 0
∴ m2 − 4m + 1 = 0
∴ `("y"/"x") − 4("y"/"x") + 1 = 0` ...[By(1)]
∴ y2 − 4xy + x2 = 0
∴ x2 − 4xy + y2 = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10
∴ x2 − 4xy + y2 = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM perpendicular line AB whose question is x + y = 10
∴ OM = `|(-10)/sqrt(1 + 1)| = 5sqrt2`
∴ area of equilateral Δ OAB `= ("OM")^2/sqrt3 = (5sqrt2)^2/sqrt3`
`= 50/sqrt3` sq units.
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