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

**Notes:**

If ax^{2} + 2hxy + by^{2}= 0 is a joint equation of lines then the lines represented by joint equation are

- Real if and only if h
^{2}– ab ≥ 0 - Real and distinct if and only if h
^{2}– ab > 0 - Real and coincident if and only if h
^{2}– ab = 0 - Imaginary and can’t be drawn if and only if h
^{2}– ab < 0

**ALGORITHM:**

- Write the joint equation of lines in standard form.
- Compare with standard ax
^{2}+ 2hxy + by^{2}= 0. Find values of a, h and b. - Find the value of the quantity h
^{2}– ab - Decide the nature of the line using notes given above.

**Example – 1: **

- Determine the nature of lines represented by the joint equation x
^{2}+ 2xy + y^{2}= 0 **Solution:**

Given joint equation is x^{2} + 2xy + y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

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

Now, h^{2} – ab = (1)^{2} – (1)(1) = 1 – 1 = 0

Here h^{2} – ab = 0, hence the lines are real and coincident.

**Example – 2: **

- Determine the nature of lines represented by the joint equation x
^{2}+ 2xy – y^{2}= 0 **Solution:**

Given joint equation is x^{2} + 2xy + y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

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

Now, h^{2} – ab = (1)^{2} – (1)(-1) = 1 + 1 = 2

Here h^{2} – ab > 0, hence the lines are real and distinct.

**Example – 3: **

- Determine the nature of lines represented by the joint equation 4x
^{2}+ 4xy + y^{2}= 0 **Solution:**

Given joint equation is 4x^{2} + 4xy + y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 4, 2h = 4, h = 2, b = 1

Now, h^{2} – ab = (2)^{2} – (4)(1) = 4 – 4 = 0

Here h^{2} – ab = 0, hence the lines are real and coincident.

**Example – 4:**

- Determine the nature of lines represented by the joint equation x
^{2}– y^{2}= 0 **Solution:**

Given joint equation is x^{2} + 0xy – y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = 1, 2h = 0, h = 0, b = – 1

Now, h^{2} – ab = (0)^{2} – (1)(-1) = 0 + 1 = 1

Here h^{2} – ab > 0, hence the lines are real and distinct.

**Example – 5: **

- Determine the nature of lines represented by the joint equation x
^{2}+ 7xy + 2y^{2}= 0

**Solution:**

Given joint equation is x^{2 }+ 7xy + 2y^{2} = 0

Comparing with ax^{2}+ 2hxy + by^{2} = 0

a = 1, 2h = 7, h = 7/2, b = 2

Now, h^{2} – ab = (7/2)^{2} – (1)(2) = 41/4 – 2 = 41/4

Here h^{2} – 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 ax^{2} + 2hxy + by^{2} = 0

a = 0, 2h = 1, h = 1/2, b = 0

Now, h^{2} – ab = (1/2)^{2} – (0)(0) =1/4

Here h^{2} – ab > 0, hence the lines are real and distinct.

**Example – 7: **

- Determine the nature of lines represented by the joint equation x
^{2}+ 2xy + 2y^{2}= 0 **Solution:**

Given joint equation is x^{2} + 2xy + 2y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

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

Now, h^{2} – ab = (1)^{2} – (1)(2) = 1 – 2 = – 1

Here h^{2} – ab < 0, hence the lines are imaginary and can’t be drawn.

**Example – 8: **

- Determine the nature of lines represented by the joint equation px
^{2}+ 2qxy – py^{2}= 0. **Solution:**

Given joint equation is px^{2} + 2qxy – py^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = p, 2h = 2q, h = q, b = – p

Now, h^{2} – ab = (q)^{2} – (p)(-p) = q^{2} + p^{2}

- as p and q are real numbers, there squares are always positive. Thus sum of their squares is always positive. Here h
^{2}– ab > 0, hence the lines are real and distinct.

**Example – 9: **

- Determine the nature of lines represented by the joint equation px
^{2}– qy^{2}= 0 **Solution:**

Given joint equation is px^{2} + 0xy – qy^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2}= 0

a = p, 2h = 0, h = 0, b = – q

Now, h^{2} – ab = (0)^{2} – (p)(-q) = 0 + pq = pq

- If p and q are both positive, then the product pq is positive, then h
^{2}– ab > 0, hence the lines are real and distinct. - If p and q are both negative, then the product pq is positive, then h
^{2}– ab > 0, hence the lines are real and distinct. - If p and q have opposite signs then the product pq is negative, then h
^{2}– ab < 0, hence the lines are imaginary and can’t be drawn. - If p = q = 0, then pq = 0. Here h
^{2}– ab = 0, hence the lines are real and coincident.

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

**ALGORITHM :**

- Write the joint equation of lines in standard form.
- Compare with standard ax
^{2}+ 2hxy + by^{2}= 0. Find values of a, h and b. - Find the value of the quantity h
^{2}-ab. - 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 x
^{2}+ 2hxy + 4y^{2}= 0 are real and coincident, find h. **Solution:**

Given joint equation is x^{2} + 2hxy + 4y^{2} = 0

Comparing with Ax^{2} + 2Hxy + By^{2} = 0

A = 1, 2H = 2h, H = h, B = 4

Now, lines are real and coincident

H^{2} – AB = 0

h^{2} – (1)(4) = 0

h^{2} – 4 = 0

h^{2} = 4

h = ± 2

**Example – 11:**

- If the lines represented by equation px
^{2}+ 6xy + 9y^{2}= 0 are real and distinct, find p. **Solution:**

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

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = p, 2h = 6, h = 3, b = 9

Now, lines are real and distinct

h^{2} – ab > 0

32 – (p)(9) > 0

9 – 9p > 0

-9p > -9

∴ p < 1

∴ p ∈ (- ∞, 1)

**Example – 12:**

- If the lines represented by equation px
^{2}+ 4xy + 4y^{2}= 0 are real and distinct, find p. **Solution:**

Given joint equation is px^{2} + 4xy + 4y^{2} = 0

Comparing with ax^{2} + 2hxy + by^{2} = 0

a = p, 2h = 4, h = 2, b = 4

Now, lines are real and distinct

h^{2} – ab > 0

2^{2} – (p)(4) > 0

4 – 4p > 0

-4p > -4

∴ p < 1

∴ p ∈ (- ∞, 1)

**Example – 13:**

- If the lines represented by equation 3x
^{2}+ 2hxy + 3y^{2}= 0 are real , find h. **Solution:**

Given joint equation is 3x^{2} + 2hxy + 3y^{2} = 0

Comparing with Ax^{2} + 2Hxy + By^{2} = 0

A = 3, 2H = 2h, H = h, B = 3

Now, lines are real and coincident

H^{2} – AB = 0

h^{2} – (3)(3) = 0

h^{2} – 9 = 0

∴ h^{2} = 9

∴ h = ± 3

Science > Mathematics > Pair of Straight Lines > You are Here |

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