Question

The product of the length of the perpendiculars drawn from the point (1,1) to the pair of lines x^2 + xy - 6y^2 = 0 is?

Answer 1

i) x² + xy - 6y² = (x + 3y)(x - 2y), by factorization.

ii) So the two lines represented by x² + xy - 6y² = 0 are"
One: x + 3y = 0 and the other is x - 2y = 0

iii) Distance of one perpendicular from (1, 1) on to the line x + 3y = 0 is:
|1 + 3|/√10 and on to the line x - 2y = 0 is = |1 - 2|/√5
==> The perpendicular distances are: 4/√10 and 1/√5

[Applying: "Distance of a point (x₁, y₁), from the line ax + by + c= 0, is given by: |ax₁ + by₁ + c|/√(a² + b²)."]

iv) So product of perpendicular distances = (4/√10)*(1/√5) = 4/√50

Thus the product of perpendiculars = 4/√50 = 2√2/5