Combined Equation of Pair of Lines – 02

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Type – IC:  When Lines are Forming Equilateral Triangle With x = k or y = k are Given

ALGORITHM:

  1. Find angle made by the lines with the positive direction of x-axis and calculate slopes m1 and m2 of the two lines.
  2. Use y = mx form to find equations of the two lines.
  3. Write equations of lines in the form u = 0 and  v = 0.
  4. Find u.v = 0.
  5. Simplify the L.H.S. of the joint equation.

Example – 17:

  • Find the joint equation of the straight lines through the origin which forms the equilateral triangle with the straight line x = 1.
  • Solution:

 

 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 18 :

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

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.

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.

Type – ID: When Slopes of Lines are Given

ALGORITHM :

  • Locate slopes m1 and m2 of the two lines.
  • Use y = mx form to find equations of the two lines.
  • Write equations of lines in the form u = 0 and  v = 0. Find u.v = 0.
  • Simplify the L.H.S. of the joint equation.

Example 19:

  • Find the joint equation of a pair of lines through the origin having slopes  1 and 3.
  • Solution:

Let l1 and l2  be the two lines. Slope of l1 is 1 and that of l2  is 3



Therefore the equation of line (l1) passing through origin and having slope 1 is

y = 1 (x)

∴ x  –  y = 0 ……. (1)

Similarly the equation of line (l2) passing through origin and having slope 3 is

y = 3 (x)

∴ 3 x – y = 0  …. (2)



From (1) and (2) the required combined equation is

(x – y) (3x – y) = 0

∴ x(3x – y) – y(3x – y) = 0

∴ 3x2 – xy – 3xy + y2 = 0

∴ 3x2 – 4xy + y2 = 0

This is the required cobined equation.

Example 20:

  • Find the joint equation of a pair of lines through the origin having slopes  4 and -1.

Solution:

Let l1 and l2  be the two lines. Slope of l1 is 4 and that of l2  is – 1

 

Therefore the equation of line (l1) passing through origin and having slope 4 is

 y = 4 (x)

∴   4x – y = 0  ……. (1)

Similarly the equation of line (l2) passing through origin and having slope -1 is

y = -1 (x)

∴ x + y = 0  …. (2)

From (1) and (2) the required combined equation is

(4x – y) (x + y) = 0

∴ 4x(x + y) – y(x + y) = 0

∴ 4x2 + 4xy – xy + y2 = 0

∴ 4x2 + 3xy + y2 = 0

This is the required cobined equation.

Example 21:

  • Find the joint equation of a pair of lines through the origin having slopes  2 and – 1/2.
  • Solution:

Let l1 and l2  be the two lines. Slope of l1 is 2 and that of l2  is – 1/2

Therefore the equation of line ( l1) passing through origin and having slope 2 is

y = 2 (x)

∴ 2x – y = 0 ……. (1)

Similarly the equation of line (l2) passing through origin and having slope  is

y =  -1/2 (x)

∴ 2y = – x

∴  x + 2y = 0  ………. (2)

From (1) and (2) the required combined equation is

(2x – y) (x + 2y) = 0

∴ 2x(x + 2y) – y(x + 2y) = 0

∴ 2x2 + 4xy – xy – 2y2 = 0

∴ 2x2 + 3xy – 2y2 = 0

This is the required cobined equation.

Example 22:

  • Find the joint equation of a pair of lines through the origin having slopes 1/3  and -1/2.

Solution:

Let l1 and l2  be the two lines. Slope of l1 is 1/3 and that of l2  is – 1/2

Therefore the equation of line (l1) passing through origin and having slope  is

y = 1/3 (x)

∴  3y = x

∴  x – 3y = 0  ……. (1)

Similarly the equation of line (l2) passing through origin and having slope  is

y =  (x)

∴  2y = -x

∴  x + 2y = 0  ……. (2)

From (1) and (2) the required combined equation is

(x – 3y) (x + 2y) = 0

∴ x(x + 2y) – 3y(x + 2y) = 0

∴ x2 + 2xy – 3xy – 6y2 = 0

∴ x2 – xy = 6y2 = 0

This is the required cobined equation.

Example 23:

  • Find the joint equation of a pair of lines through the origin having slopes 1 + 3  and 1 – 3.
  • Solution:

Let l1 and l2  be the two lines. Slope of l1 is 1 + 3 and that of l2  is 1 – 3

Therefore the equation of line (l1) passing through origin and having slope  is

y = (1 + 3)x

∴   (1 + 3)x – y = 0 …. (1)

Similarly, the equation of the line (l2) passing through the origin and having slope   is

y = ( 1 – 3)x

∴  ( 1 – 3)x – y = 0 …. (2)

From (1) and (2) the required combined equation is

This is the required combined equation.

Type – IE: When Point and Lines Parallel to  Co-ordinate Axes are Given

ALGORITHM :

  • When line passes through (h, k) and is parallel to co-ordinate axes are given. Equations of lines are x = h and y = k
  • Write equations of lines in the form u = 0 and  v = 0.
  • Find u.v = 0.
  • Simplify the L.H.S. of the joint equation.

Example 24:

  • Find the joint equation of a pair of lines which passes through (1, 2) and parallel to co-ordinate axes.
  • Solution:

 

Let l1 and l2  be the two lines.

Equation of l1 passing through (1, 2) and parallel to y-axis is x = 1

∴  x – 1 = 0    …………. (1)

Equation of l2 passing through (1, 2) and parallel to  x-axis is y = 2

∴  y – 2 = 0   …………. (2)

The required joint equation is

(x – 1)(y – 2) = 0

∴ xy – 2x – y + 2 = 0

This is the required cobined equation.

Example 25:

  • Find the joint equation of a pair of lines which passes through (-3, 4) and parallel to co-ordinate axes.
  • Solution:

Let l1 and l2  be the two lines.

Equation of l1 passing through (-3, 4) and parallel to y – axis is x = -3

∴  x + 3 = 0    …………. (1)

Equation of l2 passing through (-3, 4) and parallel to  x – axis is y = 4

∴   y – 4 = 0   …………. (2)

The required joint equation is

(x + 3)(y – 4) = 0

∴ xy – 4x + 3y – 12 = 0

This is the required cobined equation.

Example 27:

  • Find the joint equation of a pair of lines which passes through (3, 2) and parallel to the lines x = 2 and y = 3.
  • Solution:

Combined Equation

Let l1 and l2  be the two lines.

Equation of l1 passing through (3, 2) and parallel to x = 2 is x = 3

∴  x – 3 = 0    …………. (1)

Equation of l2 passing through (3, 2) and parallel to y = 3 is y = 2

∴   y – 2 = 0   …………. (2)

The required joint equation is

(x – 3)(y – 2) = 0

∴  xy – 2x – 3y + 6 = 0

This is the required cobined equation.

Example 27:

  • Find the joint equation of a pair of lines which passes through (2, 3) and parallel to co-ordinate axes.
  • Solution:

Let l1 and l2  be the two lines.

Equation of l1 passing through (1, 2) and parallel to y – axis is x = 2

∴  x – 2 = 0    …………. (1)

Equation of l2 passing through (1, 2) and parallel to x – axis is y = 3

∴  y – 3 = 0   …………. (2)

The required joint equation is

(x – 2)(y – 3) = 0

∴  xy – 3x – 2y + 6 = 0

This is the required cobined equation.

Example 28:

  • Find the joint equation of a pair of lines which are at a distance of 9 units from the y-axis and parallel to it.
  • Solution:

Let l1 and l2  be the two lines.

Equation of l1 which is at a distance of 9 from y – axis and parallel to it is x = – 9 i.e. x + 9 = 0  …. (1)

Equation of l2 which is at a distance of 9 from y – axis and parallel to it is x =  9 i.e. x – 9 = 0  …… (1)

The required joint equation is

(x + 9)(x – 9) = 0

∴  x2 – 81 = 0

This is the required cobined equation.

Example – 29:

  • Find the joint equation of a pair of lines which are at a distance of 5 units from the x-axis and parallel to it.
  • Solution:

Let l1 and l2  be the two lines.

Equation of l1 which is at a distance of 5 from x – axis and parallel to it is y = 5i.e. y – 5  = 0   ……… (1)

Equation of l2 which is at a distance of 5 from x – axis and parallel to it is y =  -5 i.e. y + 5 = 0   …… (2)

The required joint equation is

(y – 5)(y + 5) = 0

∴ y2 – 25 = 0

This is the required cobined equation.

Science > Mathematics > Pair of Straight LinesYou are Here
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