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Combined Equation of Lines When Their Inclinations are Given

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Inclinations are Given

In this article, we shall study to find a combined equation of pair of lines when the inclinations of lines are given.

Algorithm:
  1. If θ is the inclination of a line, then its slope is given by m = tan θ.
  2. Find slopes m1 and m2 of the two lines.
  3. Use y = mx form to find equations of the two lines.
  4. Write equations of lines in the form u = 0 and  v = 0.
  5. Find u.v = 0.
  6. Simplify the L.H.S. of the joint equation.

Example 01:

  • Find the combined equation of the lines passing through the origin and making an angle of 30° with the positive direction of the x-axis
  • Solution:
Combined Equation

 Slope of the first line = m1 = tan 30° =  1/√3

The equation the first line is y = m1 x

y = 1/√3 x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m2 = tan (-30°)= – tan 30°  =  -1/√3

The equation the second line is y = m2 x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 02 :

  • Find the combined equation of the lines passing through the origin and making an angle of 45° with the positive direction of the x-axis
  • Solution:
Combined Equation

Slope of the first line = m1 = tan 45° =  1

The equation the first line is y = m1 x

y =  1.x

x – y  = 0     …………. (1)

 Slope of the second line = m2 = tan (-45°)= – tan 45°  =  -1

The equation the second line is y = m2 x

y = – 1 x

x +  y  = 0     …………. (2)

Their joint equation is (x – y) (x + y) = 0

x² – y² = 0

This is the required combined equation.

Example 03:

  • Find the combined equation of the lines passing through the origin and making an angle of 60° with the positive direction of the x-axis.
  • Solution:
blank

 Slope of the first line = m1 = tan 60° =  √3

The equation the first line is y = m1 x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m2 = tan (-60°)= – tan 60°  =  -√3

The equation the second line is y = m2 x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 04:

  • Find the combined equation of the lines passing through the origin and making an angle of 60° with the y-axis
  • Solution:
blank

 Slope of the first line = m1 = tan 30° =  1/√3

The equation the first line is y = m1 x

y = 1/√3x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m2 = tan (150°)= tan(180° – 30°) = -tan 30°  =  -1/√3

The equation the second line is y = m2 x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 05:

  • Find the joint equation of a pair of lines through the origin having and having inclination 30° and 150°.
  • Solution:

 Slope of the first line = m1 = tan 30° =  1/√3

The equation the first line is y = m1 x

y = 1/√3x

√3 y  = x

x – √3 y  = 0     …………. (1)

 Slope of the second line = m2 = tan (150°)= tan(180° – 30°) = -tan 30°  =  -1/√3

The equation the second line is y = m2 x

y =-  1/√3x

√3 y  = – x

x + √3 y  = 0     …………. (2)

Their joint equation is (x – √3 y) (x + √3 y) = 0

x² – 3y² = 0

This is the required combined equation.

Example 06:

  • Find the joint equation of a pair of lines through the origin having and having inclination 60° and 120°
  • Solution:

 Slope of the first line = m1 = tan 60° =  √3

The equation the first line is y = m1 x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m2 = tan (120°)= tan (180° – 60°) = – tan 60°  =  -√3

The equation the second line is y = m2 x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 07:

Find the joint equation of a pair of lines through the origin having and having inclination π/3 and 5π/3.

Solution:

 Slope of the first line = m1 = tan π/3 =  √3

The equation the first line is y = m1 x

y = √3x

√3 x  – y = 0     …………. (1)

 Slope of the second line = m2 = tan (5π/3)= tan (2π – π/3) = – tan π/3  =  -√3

The equation the second line is y = m2 x

y =-  √3x

√3 x  + y = 0     …………. (2)

Their joint equation is (√3 x  – y) (√3 x + y) = 0

3x² – y² = 0

This is the required combined equation.

Example 08:

  • Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line x = 1.
  • Solution:
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 Let OA and OB be two lines such that triangle OAB is an equilateral triangle.

Equation of AB is x = 1 and AB, is parallel to the y-axis.

Now triangle OAB is an equilateral triangle,

by symmetry, m ∠ MOA  = m ∠MOB = 30°

Slope of OA  = m1 = tan(-30°) =  – tan 30° = – 1/3  and

Slope of OB = m2 = tan( 30°)  =  1/3

Equation of OA is passing through the origin, is

y =  m1x. i.e.   y = – 1/3  x

∴  √3 y = –  x

∴  x + √3 y = 0  …….. (1)

As OB is also passing through the origin, the form of the equation of OB is

y =  m2x. i.e.   y = 1/3  x

∴  √3 y =  x

∴  x – √3 y = 0  …….. (2)

Hence, the combined equation of the pair OA and OB is

(x + √3 y)(x – √3 y ) = 0

∴  x 2 – 3y 2 = 0.

This is the required combined equation.

Note: The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line x = k is always x 2 – 3y 2 = 0.

Example 09:

  • Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line y  =  2.
  • Solution:
blank

Let OA and OB be two lines such that the triangle OAB is an equilateral triangle.

The equation of AB is y  =  2 and AB is parallel to the x-axis.

Now since triangle OAB is an equilateral triangle,

m ∠ XOA = 60° and m ∠ XOB = 120°

∴  Slope of OA =  tan 60° =  3 and

∴  slope of OB = tan 120° – 3 

As OA is passing through the origin, the form of equation of OA is

 y = m1x.  i.e. y = 3 x

∴  √3 x – y = 0  ………. (1)

As OB is also passing through the origin, the form of equation of OB is y = mx.

 y = m2x.  i.e. y = – 3 x

∴  √3 x +  y = 0  ………. (2)

Hence their combined equation is

( 3 x – y)(  3 x +  y) = 0

∴  3x2 – y2 = 0.

This is the required combined equation.

Note: The joint equation of the straight lines through the origin which forms the equilateral triangle with any straight line y = k is always  3x2 – y2 = 0.

Science > Mathematics > Pair of Straight Lines > Combined Equation of Lines When Inclinations are Given

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