A triangle has two of its sides along the lines y=m1x and y=m2x where m1,m2 are the roots of the eq

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Scorching goIITian

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m1 + m2 = -10/3 ; m1m2 = 1/3

A(0,0) , B(t,m₂t) , C(p,m₁p)

equation of altitude through C : y - m₁p = -1/m₂ (x - p)

equation of altitude through B : y - m₂t = -1/m₁ (x - t)

They both, being altitudes, pass through (6,2)

2 - m₁p = -1/m₂ (6 - p) ; p = (2m₂ + 6) / (1 + m₁m₂) ; p = 3(m₂ + 3)/2

2 - m₂t = -1/m₁ (6 - t) ; t = (2m₁ + 6) / (1 + m₁m₂) ; t = 3(m₁ + 3)/2

Equation of BC : x(m₁p - m₂t) - y(p - t) + (m₂ - m₁)pt = 0

m₁p - m₂t = 9(m₁- m₂)/2

p - t = 3(m₂ -m₁)/2

Putting these 2 results in equation of BC

9x/2 - 3y/2 + pt = 0

pt = 9(m₁+3)(m₂+3)//4

= 9(m1m2 + m1 + m2 + 9)/4

= 9(1/3 -10/3 + 9)/4

= 27/2

so eqn of BC is 9x - 3y +27 = 0 i.e. 3x - y + 9 = 0 ; y = 3x + 9 Answer

Milind Gupta IIT KHARAGPUR

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