Question

prove that the equation x²-6xy+9y²+4x+12y-5=0 represent a pair of parallel straight lines and finds the distance between them​

Answer 1

Answer:given, x² - 6xy + 9y² + 3x - 9y - 4 = 0

⇒ x² - 2.x.(3y) + (3y)² + 3(x - 3y) - 4 = 0

⇒ (x - 3y)² + 3(x - 3y) - 4 = 0

Let (x - 3y) = P

P² + 3P - 4 = 0

⇒P² + 4P - P - 4 = 0

⇒ P(P + 4) - (P + 4) = 0

⇒(P - 1)(P + 4) = 0

⇒ P = 1 , -4

Now, put P = (x - 3y)

e.g., we have two parallel equations x - 3y = -4 and x - 3y = 1

We know, distance between two parallel line is given by

|D₁ - D₂|/√(a² + b²) where equations are ax + by + D₁ = 0 and ax + by + D₂ = 0

∴ distance between two lines x - 3y = -4 and x - 3y = 1

= |-4 - 1|/√(1² + 3²)

= 5/√10 unit

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