Nature of Lines Represented by Joint Equation

Science > Mathematics > Pair of Straight Lines > Nature of Lines Represented by Joint Equation

In this article, we shall study to predict the nature of lines using the joint equations of lines.

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 the notes given above.

Example 01:

• 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 02:

• 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 03:

• 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 04:

• 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 05:

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

Solution:

Given joint equation is x² + 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 06:

• 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 07:

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 08:

• 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 09:

• 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.

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

Science > Mathematics > Pair of Straight Lines > Nature of Lines Represented by Joint Equation