Prediction of Nature of Line From the Joint Equation

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Unit – II A: Nature of Lines From the Joint Equation

Notes:

If ax2 + 2hxy + by2= 0 is a joint equation of lines then the lines represented by joint equation are

  • Real if and only if h2 – ab ≥ 0
  • Real and distinct if and only if h2 – ab > 0
  • Real and coincident if and only if h2 – ab = 0
  • Imaginary and can’t be drawn if and only if h2 – ab < 0

ALGORITHM:

  1. Write the joint equation of lines in standard form.
  2. Compare with standard ax2 + 2hxy + by2 = 0. Find values of a, h and b.
  3. Find the value of the quantity h2 – ab
  4. Decide the nature of the line using notes given above.

Example – 1:  

  • Determine the nature of lines represented by the joint equation x2 + 2xy + y2 = 0
  • Solution:

Given joint equation is x2 + 2xy + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 1, 2h = 2, h = 1, b = 1

Now, h2 – ab = (1)2 – (1)(1) = 1 – 1 = 0

Here h2 – ab = 0, hence the lines are real and coincident.

Example – 2:  

  • Determine the nature of lines represented by the joint equation x2 + 2xy – y2 = 0
  • Solution:

Given joint equation is x2 + 2xy + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 1, 2h = 2, h = 1, b = -1

Now, h2 – ab = (1)2 – (1)(-1) = 1 + 1 = 2

Here h2 – ab > 0, hence the lines are real and distinct.

Example – 3:  

  • Determine the nature of lines represented by the joint equation 4x2 + 4xy + y2 = 0
  • Solution:

Given joint equation is 4x2 + 4xy + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 4, 2h = 4, h = 2, b =  1

Now, h2 – ab = (2)2 – (4)(1) = 4 – 4 = 0

Here h2 – ab = 0, hence the lines are real and coincident.

Example – 4:

  • Determine the nature of lines represented by the joint equation x2 – y2 = 0
  • Solution:

Given joint equation is x2 + 0xy – y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 1, 2h = 0, h = 0, b = – 1

Now, h2 – ab = (0)2 – (1)(-1) = 0 + 1 = 1

Here h2 – ab > 0, hence the lines are real and distinct.

Example – 5:  

  • Determine the nature of lines represented by the joint equation x2 + 7xy + 2y2 = 0

Solution:

Given joint equation is x+ 7xy + 2y2 = 0

Comparing with ax2+ 2hxy + by2 = 0

a = 1, 2h = 7, h = 7/2, b =  2

Now, h2 – ab = (7/2)2 – (1)(2) = 41/4 – 2 = 41/4

Here h2 – ab > 0, hence the lines are real and distinct.

Example – 6:  

  • Determine the nature of lines represented by the joint equation xy  = 0
  • Solution:

Given joint equation is xy  = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 0, 2h = 1, h = 1/2, b =  0

Now, h2 – ab = (1/2)2 – (0)(0) =1/4

Here h2 – ab > 0, hence the lines are real and distinct.

Example – 7:  

  • Determine the nature of lines represented by the joint equation x2 + 2xy + 2y2 = 0
  • Solution:

Given joint equation is x2 + 2xy + 2y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = 1, 2h = 2, h = 1, b =  2

Now, h2 – ab = (1)2 – (1)(2) = 1 – 2 = – 1

Here h2 – ab < 0, hence the lines are imaginary and can’t be drawn.

Example – 8:  

  • Determine the nature of lines represented by the joint equation px2 + 2qxy – py2 = 0.
  • Solution:

Given joint equation is px2 + 2qxy – py2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = p, 2h = 2q, h = q, b = – p

Now, h2 – ab = (q)2 – (p)(-p) = q2 + p2

  • as p and q are real numbers, there squares are always positive. Thus sum of their squares is always positive. Here h2 – ab > 0, hence the lines are real and distinct.

Example – 9:  

  • Determine the nature of lines represented by the joint equation px2 – qy2 = 0
  • Solution:

Given joint equation is px2 + 0xy – qy2 = 0

Comparing with ax2 + 2hxy + by2= 0

a = p, 2h = 0, h = 0, b = – q

Now, h2 – ab = (0)2 – (p)(-q) = 0 + pq = pq

  • If p and q are both positive, then the product pq is positive, then h2 – ab > 0, hence the lines are real and distinct.
  • If p and q are both negative, then the product pq is positive, then h2 – ab > 0, hence the lines are real and distinct.
  • If p and q have opposite signs then the product pq is negative,  then h2– ab < 0, hence the lines are imaginary and can’t be drawn.
  • If p = q = 0, then pq = 0. Here h2 – ab = 0, hence the lines are real and coincident.

Type – II B: Nature of Lines is Given. To Find the Value of Constant

ALGORITHM :

  1. Write the joint equation of lines in standard form.
  2. Compare with standard ax2 + 2hxy + by2 = 0. Find values of a, h and b.
  3. Find the value of the quantity h2 -ab.
  4. Use the conditions depending upon the nature of the line use notes given at the top and find the value of the constant.

Example – 10:

  • If the lines represented by equation x2 + 2hxy + 4y2 = 0  are real and coincident, find h.
  • Solution:

Given joint equation is x2 + 2hxy + 4y2 = 0

Comparing with Ax2 + 2Hxy + By2 = 0

A = 1, 2H = 2h, H = h, B =  4

Now, lines are real and coincident

H2 – AB = 0

h2 – (1)(4) = 0

h2 – 4 = 0

h2 = 4

h = ± 2

Example – 11:

  • If the lines represented by equation px2 + 6xy + 9y2 = 0  are real and distinct, find p.
  • Solution:

Given joint equation is px2 + 6xy + 9y2  = 0

Comparing with ax2 + 2hxy + by2 = 0

a = p, 2h = 6, h = 3, b =  9

Now, lines are real and distinct

h2 – ab > 0

32 – (p)(9) > 0

9 – 9p > 0

-9p > -9

∴ p < 1

∴ p ∈ (- ∞, 1)

Example – 12:

  • If the lines represented by equation px2 + 4xy + 4y2 = 0  are real and distinct, find p.
  • Solution:

Given joint equation is px2 + 4xy + 4y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

a = p, 2h = 4, h = 2, b =  4

Now, lines are real and distinct

h2 – ab > 0

22 – (p)(4) > 0

4 – 4p > 0

-4p > -4

∴ p < 1

∴ p ∈ (- ∞, 1)

Example – 13:

  • If the lines represented by equation 3x2 + 2hxy + 3y2 = 0  are real , find h.
  • Solution:

Given joint equation is 3x2 + 2hxy + 3y2 = 0

Comparing with Ax2 + 2Hxy + By2 = 0

A = 3, 2H = 2h, H = h, B =  3

Now, lines are real and coincident

H2 – AB  = 0

h2 – (3)(3) =  0

h2 – 9  = 0

∴ h2 =  9

∴  h  = ± 3

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