Question

Find the joint equation of the pair of a line through the origin and perpendicular to the lines given by

x² + xy - y² = 0

Answer 1

Comparing the equation x² + xy - y² = 0 with ax₂ + 2hxy + by² = 0, we get,

a = 1, 2h = 1, b = - 1

Let m₁ and m₂ be the slopes of the lines represented by x² + xy - y² = 0 

∴ m₁ + m₂ = `(-"2h")/"b" = (-1)/-1 = 1`  and  m₁m₂ = `"a"/"b" = 1/-1 = -1`    ...(1)

Now, required lines are perpendicular to these lines

∴ their slopes are `(-1)/"m"_1` and `- 1/"m"_2`

Since these lines are passing through the origin, their separate equations are

y = `(-1)/"m"_1 "x"` and y = `(-1)/"m"_2 "x"`

i.e. m₁y = - x and m₂y = - x

i.e. x + m₁y  = 0 and x + m₂y = 0

∴ their combined equation is

(x + m₁y)(x + m₂y) = 0

∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0

∴ `"x"^2 + 1"xy" + (-1)"y"^2 = 0`   ....[By(1)]

∴ `"x"^2 + "xy" - "y"^2 = 0`