Angle between Pair of Lines

Straight lines is an extremely important topic of IIT JEE Mathematics. It often fetches some direct questions in various competitions like the IIT JEE. Since the topic is quite vast, students are advised to spend sufficient time on grasping the various concepts. Angle between pair of straight lines is an important head under straight lines. We begin with the concept of angle between pair of lines and then discuss some of the illustrations on the same:

Angle between two Straight Lines

Suppose we have two straight lines y = m₁x + c₁ and y = m₂x + c₂, then the angle between these two lines is given by tan θ = |(m₁ – m₂)/ (1 + m₁m₂)|.

If m₁, m₂ and m₃ are the slopes of three lines L₁ = 0, L₂ = 0 and L₃ = 0, where m₁ > m₂ > m₃ then the interior angles of the triangle ABC formed by these lines are given by,

tan A = (m₁ - m₂)/(1 + m₁m₂), tan B = (m₂ - m₃)/(1 + m₂m₃) and tan C = (m₃ - m₁)/(1 + m₃m₁).

Remark:

• If one of the line is parallel to y-axis then the angle between two straight lines is given by tan θ = ±1/m where ‘m’ is the slope of the other straight line.

• If the two lines are a₁x + b₁y + c₁ = 0 and a₂x +  b₂y + c₂ = 0, then the formula becomes tan θ = |(a₁b₂ - b₁a₂)/(a₁a₂ + b₁b₂)|

• Generally speaking, the angle between these two lines is assumed to be acute and hence, the value of tan θ is taken to be positive.
   

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Result:

Angle between pair of lines represented by ax² + 2hxy + by² = 0

Comparing the coefficients of x², y² and xy, we get

b(y – m₁x) (y – m₂x) = ax² + 2hxy + by²

m₁ + m₂ = –2h/b and

m₁ m₂ = a/b

tan θ_acute = |(m₂ – m₁)/(1 + m₁m₂)|

              = |√(m₁–m₂)² – 4m₁m₂/(1 + m₁m₂)|

              = |2√(h² – ab)/(a + b)|

Note:

• If two lines through the origin are represented by y = m₁x and y = m₂x, we cannot write

  ⇒ (y – m₁x) (y – m₂x) ≡ ax² + 2hxy + by²

  Because coefficient of y² on left hand side is one on right hand side, it is b.

• The given equation represents real lines only when h² – ab > 0

• If two lines are coincident then tan θ = 0 ⇒ h² = ab

• If two lines are perpendicular then m₁m₂ = 1 ⇒ a + b = 0

  i.e. x² + 2hxy – y² always represents pair of mutually perpendicular lines through origin.

• Two lines are equally inclined to axes but are not parallel:

  For such a case let us take a line l₁ which is inclined at an angle θ, then l₂ is inclined at (π – θ).

  tan (π – θ) = – tan θ which is the condition for two lines inclined equally to axes.

  m₁ = –h + √(h²–ab)/2 and m₂ = –h – √(h²–ab)/2

Illustration:

What is the equation of the pair of lines through origin and perpendicular to ax² + 2hxy + by² = 0?

Solution:

Let ax² + 2hxy + by² = 0 represent the lines y = m₁x (i) and y = m₂x (ii)

Lines perpendicular to the lines (i) and (ii) are y = –1/m₁ x and y = –1/m₂ x respectively and passing through origin

i.e. m₁y + x = 0 and m₂y + x = 0

Their combined equation is given by

(m₁y + x) (m₂y + x) = 0

⇒ m₁m₂ y² + (m₁ + m₂) xy + x² = 0

⇒ a/2 y² – 2h/b xy + x² = 0

⇒ bx² – 2hxy + ay² = 0 is the equation of the pair of lines perpendicular to pair of lines ax² + 2hxy + by² = 0.

Illustration:

Find the angle between the lines represented by the equation 2x² – 7xy + 3y² = 0.

Solution:

The given equation is 2x² – 7xy + 3y² = 0.

This can be written as 2x² – 7xy + 3y² = (2x – y)(x-3y)

Therefore, the given lines are (2x – y) = 0 and (x-3y) = 0.

Now, here, a = 2, h = -7/2 and b = 3.

Hence, the angle between the lines is given by tan θ = 2√{(-7/2)² – 6}/5 = 1.

Hence, θ = 45°.

Related Resources

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• For getting an idea of the type of questions asked, refer the previous year papers.

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