Math, asked by marynissiesimon, 1 year ago

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)​

Answers

Answered by sanjeevaarav910
7

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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