Example

Ex 1:

Show that the lines joining the origin to the points of intersection of the curve x² – xy + y² + 3x + 3y – 2 = 0 and the straight line x – y – √2 = 0 are mutually perpendicular.

Sol:

Equation of the curve is x² – xy + y² + 3x + 3y – 2 = 0 ... (i)

Equation of the line is x – y – √2 = 0

x – y = √2

= 1 ... (ii)

Homogenising, (i) with the help of (ii)

x² – xy + y² + 3x.1 + 3y.1 – 2.1² = 0

x² – xy + y² + 3(x + y)

– 2

= 0

x² – xy + y² + (3/√2) (x² – y²) – (x² – 2xy + y²) = 0

x² – xy + y² + (3/√2) x² – (3/√2) y² – x² + 2xy – y² = 0

(3/√2) x² + xy – (3/√2) y² = 0

Comparing with ax² + 2hxy + by² = 0

a = (3/√2), b = –(3/√2) and h =

⇒ a + b = (3/√2) – (3/√2) = 0

∴ The given lines are mutually perpendicular.