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Monday, December 5, 2016

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

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