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प्रश्न
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.
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उत्तर
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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