Question

find equation of pair of lines intersecting at (2,-1) and perpendicular to pair 6x^2-13xy-5y^2=0​

Answer 1

Given : a pair of equation of lines interesting at (2, -1) and perpendicular to pair 6x² - 13xy - 5y² = 0

To find : we have to find the equation of the pair of equation of lines.

solution : here 6x² - 13xy - 5y² = 0

we know, m₁ + m₂ = -2h/b and m₁m₂ = a/b if pair of equation of lines ax² + 2hxy + by² = 0

here h = -13/2 , a = 6 , b = - 5

so, m₁ + m₂ = -(-13/2)/5 = 13/10 .......(1)

m₁m₂ = -6/5 ........(1)

as pair of equation of lines is perpendicular to 6x² - 13xy - 5y² = 0

so, m₁' = -1/m₁ , m₂' = -1/m₂

and equation of lines ....

(y + 1) = -1/m₁ (x - 2)

⇒m₁y + m₁ + x - 2 = 0

similarly, (y + 1) = -1/m₂ (x - 2)

⇒m₂y + m₂ + x - 2 = 0

now, pair of equation of lines : (m₁y + m₁ + x - 2)(m₂y + m₂ + x - 2) = 0

⇒ m₁m₂y² + x² + (m₁ + m₂)xy + (m₁ + m₂)x - 2(m₁ + m₂)y + 2m₁m₂y - 4x + 4 = 0

now putting equations (1) and (2)

⇒-6/5y² + x² + 13/10 xy + 13/10 x - 2(13/10)y + 2 × -6/5 y - 4x + 4 = 0

⇒-12y² + 10x² + 13xy + 13x - 26y - 24y - 4x + 4 = 0

⇒10x² - 12y² + 13xy + 9x - 50y + 4 = 0

Therefore the pair of equation of lines is 10x² - 12y² + 13xy + 9x - 50y + 4 = 0