Question

Find the joint equation of the pair of lines through the origin and perpendicular to the lines given by 2x² + 7xy + 3y² = 0

Answer 1

2x² + 7xy + 3y² = 0   ...(1)

Let m₁ and m₂ be the slopes of lines given by equation (1)

∴ m₁ + m₂ = `(-2h)/b = (-7)/3`,

m₁m₂ = `a/b = 2/3`

Now the required lines are perpendicular to the given lines.

∴ Their slopes are `-1/m_1` and `-1/m_2`

∴ Separate equations are

y = `-1/m_1x` and y = `-1/m_2x`

i.e. x + m₁y = 0 and x + m₂y = 0

∴ Combine equation is

(x + m₁y)(x + m₂y) = 0

i.e. x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴ `x^2 + ((-7)/3)xy + 2/3y^2` = 0

∴ 3x² – 7xy + 2y² = 0