ANGLE BETWEEN THE PAIR OF LINES

Let y = m₁ x and y = m₂ x be the lines represented by ax² + 2hxy + by² = 0. Then,

m₁ + m₂ = -2h/ b and m₁m₂ = a/b … (i)

Let  be the angle between the lines y = m₁ x and y = m₂ x. Then,

⇒ tan θ = m₁ - m₂/ (1 + m₁m₂)

⇒ tan θ = √ [(m₁ + m₂)² - 4 m₁m₂]/ 1 + m₁m₂

⇒ tan θ = √ [4h²/b²] - [4a/b]/ [1 + a/b]   [Using (i)]

⇒ tan θ = 2√ (h² - ab)/ (a + b)

⇒ θ = tan⁻¹ {2√ (h² - ab)/ (a + b)}

Thus, the angle θ between the pair of lines represented by ax² + 2hxy + by² = 0 is given by tan θ = 2√ (h² - ab)/ (a + b)

PROBLEM:

Prove that the angle between the straight lines given by (x cosα – y sinα)² = (x² + y²) sin²α is 2α.

SOLUTION:

We have,

(x cos α – y sin α)² = (x² + y²) sin² α

⇒ x² (cos² α – sin² α) – xy sin 2α + 0.y² = 0.

Comparing this equation with ax² + 2hxy + by² = 0, we get

a= cos²α – sin² α, b = 0 and 2h = -sin 2α

Let  be the angle between the given lines. Then,

⇒ tan θ = 2√ (h² - ab)/ (a + b)

⇒ tan θ = 2√ [(sin²2α)/4 - 0]/ (cos²α - sin²α)

⇒ tan θ = (sin2α)/ (cos2α) = tan2α

⇒ θ = 2α