show that the lines represented by (lx+my)^2 - 3(mx-ly)^2 is equal to 0 and lx+my+n is equal to 0 form an equilateral triangle with area n^2 divided by √3(l square + m square)
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Answer:
Answer:
l + m + n = 0 option (1)
Explanation:
Suppose we have three straight lines whose equations are:
a₁x + b₁y + c₁ = 0,
a₂x + b₂y + c₂ = 0
a₃x + b₃y + c₃ = 0.
These lines are said to be concurrent if the following condition holds:
Determinant of
a₁ b₁ c₁
a₂ b₂ c₂ = 0
a₃ b₃ c₃
Now
l m n
m n l = 0
n l m
l(nm - l²) - m(m² - nl) + n(ml - n²) = 0
lmn - l³ - m³ + lmn + lmn - n³ = 0
l³ + m³ + n³ = 3lmn
this condition true if an only if
l + m + n = 0 (In case of l ≠ m ≠ n)
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