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**Type – III: ****To Find Separate Equations of Lines When Joint Equation is Given**

**The Algorithm of Method – I:**

- Check if lines exist. use the same method used in the case to find nature of lines (do this orally).
- If factors of the equation can be found directly and easily then factorize the joint equation
- Write each factor = 0. Thus you will get two separate equations.
- Note that this method is useful only if the joint equation can be factorized.

**The Algorithm of Method – II:**

- Check if lines exist. use the same method used in the case to find nature of lines.
- Divide both sides of the equation by x
^{2}. - Simplify the equation and substitute y/x = m in it.
- Find two roots m
_{1}and m_{2}of quadratic equation in m - Find separate equations of lines by y = m
_{1}x and y = m_{2}x - Note that this method is applicable to any problem.

**Example – 1:**

- Obtain the separate equations of the lines represented by 3x
^{2}– 4xy + y^{2}= 0. **Solution by****Method – I (Factorisation Method):**

The given joint equation is 3x^{2} – 4xy + y^{2} = 0

∴ 3x^{2} – 3xy – xy + y^{2} = 0

∴ 3x(x – y) – y(x – y)= 0

∴ (x – y)(3x – y) = 0

∴ (x – y) = 0 and (3x – y) = 0

**Ans:** Separate equations of the lines are x – y = 0 and 3x – y = 0

**Example – 2:**

- Obtain the separate equations of the lines represented by 11x
^{2}+ 8xy + y^{2}= 0 **Solution by Method – II:**

The given joint equation is 11x^{2} + 8xy + y^{2} = 0

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in equation (1) we get

11 + 8m + m^{2} = 0

m^{2} + 8m + 11 = 0

This is a quadratic equation in m and has two roots say m_{1} and m_{2}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 8, c = 11

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (-4 + √5) x

∴ y = – (4 – √5) x

∴ (4 – √5) x + y = 0

The equation of the second line is

y = m_{2}x i.e. i.e.

∴ y = (-4 – √5) x

∴ y = – (4 + √5) x

∴ (4 + √5) x + y = 0

**Ans:** The separate equations of the lines are (4 – √5) x + y = 0 and (4 + √5) x + y = 0

**Example – 3:**

- Obtain separate equations of lines represented by joint equation y
^{2}– 5x^{2}= 0.

**Solution:**

Given joint equation is y^{2} – 5x^{2} = 0.

5x^{2} – y^{2} = 0.

∴ (√5x)^{2} – y^{2 }= 0

∴ (√5x + y) (√5x – y) = 0.

**Ans:** The separate equations of the lines are √5x + y = 0 and 5x – y = 0

**Example – 3:**

- Obtain separate equations of lines represented by joint equation x
^{2}– 5xy + 6y^{2}= 0. **Solution:**

Given joint equation is x^{2} – 5xy + 6y^{2} = 0.

∴ x^{2} – 3xy – 2xy + 6y^{2} = 0

∴ x(x – 3y) – 2y(x – 3y) = 0

∴ (x – 3y)(x – 2y) = 0

**Ans:** The separate equations of the lines are x – 3y = 0 and x – 2y = 0

**Example – 4:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}+ 10xy + 8y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} + 10xy + 8y^{2 }= 0.

∴ 3x^{2} + 6xy + 4xy + 8y^{2} = 0

∴ 3x(x + 2y) + 4y(x + 2y) = 0

∴ (x + 2y)(3x + 4y) = 0

**Ans:** The separate equations of the lines are x + 2y = 0 and 3x + 4y = 0

**Example – 4:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}+ 10xy + 8y^{2}= 0. **Solution:**

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

x^{2} – xy – xy + y^{2} = 0

∴ x(x – y) – y(x – y)= 0

∴ (x – y) (x – y)= 0

∴ (x – y)^{2}= 0

∴ (x – y) = 0

∴ x = y

**Ans:** These are coincident lines x = y

**Example – 5:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}+ 5xy – 2y^{2}= 0. **Solution:**

The given joint equation is 3x^{2} + 5xy – 2y^{2} = 0

∴ 3x^{2}+ 6xy – xy – 2y^{2} = 0

∴ 3x(x + 2y) – y(x + 2y)= 0

∴ (x + 2y) (3x – y)= 0

**Ans:** The separate equations of the lines are x + 2y = 0 and 3x – y = 0

**Example – 6:**

- Obtain separate equations of lines represented by joint equation x
^{2}– 4xy + y^{2}= 0. **Solution:**

The given joint equation is x^{2} – 4xy + y^{2} = 0

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 – 4m + m^{2} = 0

∴ m^{2} – 4m + 1= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = -4, c = 1

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (2 + √3) x

∴ (2 + √3) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (2 – √3) x

∴ (2 – √3) x – y = 0

**Ans:** The separate equations of the lines are (2 + √3) x – y = 0 and (2 – √3) x – y = 0

**Example – 7:**

- Obtain separate equations of lines represented by joint equation 10x
^{2}+ xy – 3y^{2}= 0. **Solution:**

Given joint equation is 10x^{2} + xy – 3y^{2} = 0.

∴10x^{2} + 6xy – 5xy – 3y^{2} = 0

∴ 2x(5x + 3y) – y(5x + 3y) = 0

∴ (5x + 3y)(2x – y) = 0

**Ans:** The separate equations of the lines are 5x + 3y = 0 and 2x – y = 0

**Example – 8:**

- Obtain separate equations of lines represented by joint equation 22x
^{2}– 10xy + y^{2}= 0.

**Solution:**

The given joint equation is 22x^{2} – 10xy + y^{2} = 0.

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

22 – 10m + m^{2} = 0

∴ m^{2} – 10m + 22= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 10, c = 22

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (5 + √3) x

∴ (5 + √3) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (5 – √3) x

∴ (5 – √3) x – y = 0

**Ans:** The separate equations of the lines are (5 + √3) x – y = 0 and (5 – √3) x – y = 0

**Example – 9:**

- Obtain separate equations of lines represented by joint equation x
^{2}– 4y^{2}= 0. **Solution:**

Given joint equation is x^{2} – 4y^{2} = 0.

∴ x^{2} – (2y)^{2} = 0

∴ (x + 2y)(x – 2y) = 0

Ans: The separate equations of the lines are x + 2y = 0 and x – 2y = 0

**Example – 10:**

- Obtain separate equations of lines represented by joint equation 5x
^{2}-3y^{2}= 0. **Solution:**

Given joint equation is 5x^{2} -3y^{2} = 0.

∴ (√5x)^{2} – (√3y)^{2} = 0

∴ (√5x + √3y)(x – √3y) = 0

Ans: The separate equations of the lines are √5x + √3y = 0 and √5x – √3y = 0

**Example – 11:**

- Obtain separate equations of lines represented by joint equation 6x
^{2}– 5xy + y^{2}= 0. **Solution:**

Given joint equation is 6x^{2} – 5xy + y^{2} = 0.

∴ 6x^{2} – 3xy – 2xy + y^{2} = 0

∴ 3x(2x – y) – y(2x – y) = 0

∴ (2x – y)(3x – y) = 0

Ans: The separate equations of the lines are 2x – y = 0 and 3x – y = 0

**Example – 12:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}– y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} – y^{2} = 0.

∴ (√3x)^{2} – y^{2} = 0

∴ (√3x + y)(√3x – y) = 0

Ans: The separate equations are √3x + y = 0 and √3x – y = 0

**Example – 13:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}+ 2xy + 7y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} + 2xy + 7y^{2} = 0.

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

a = 3, 2h = 2, h = 1, b = 7

Now, h^{2} – ab = (1)^{2} – (3)(7) = 1 – 21 = – 20

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

**Example – 14:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}– 7xy + 4y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} – 7xy + 4y^{2} = 0.

∴ 3x^{2} – 3xy – 4xy + 4y^{2} = 0

∴ 3x(x – y) – 4y(x – y) = 0

∴ (3x – 4y)(x – y) = 0

AQns: The separate equations of the lines are 3x – 4y = 0 and x – y = 0

**Example – 15:**

- Obtain separate equations of lines represented by joint equation x
^{2}+ 2xy – y^{2}= 0. **Solution:**

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

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2m – m^{2} = 0

∴ m^{2} – 2m – 1= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -1

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (1 + √2) x

∴ (1 + √2) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (1 – √2) x

∴ (1 – √2) x – y = 0

**Ans:** The separate equations of the lines are (1 + √2) x – y = 0 and (1 – √2) x – y = 0

**Example – 16:**

- Obtain separate equations of lines represented by joint equation 3y
^{2}+ 7xy = 0. **Solution:**

Given joint equation is 3y^{2} + 7xy = 0.

∴ y(3y + 7x) = 0

Ans: The separate equations of the lines are 7x + 3y = 0 and y = 0

**Example – 17:**

- Obtain separate equations of lines represented by joint equation 5x
^{2}– 9y^{2}= 0. **Solution:**

Given joint equation is 5x^{2} -9y^{2} = 0.

∴ (√5x)^{2} – (3y)^{2} = 0

∴ (√5x + 3y)(√5x – 3y) = 0

Ans: The separate equations of the lines are √5x + 3y = 0 and √5x – 3y = 0

**Example – 18:**

- Obtain separate equations of lines represented by joint equation x
^{2}– 4xy = 0. **Solution:**

Given joint equation is x^{2} – 4xy = 0.

∴ x(x – 4y) = 0

Ans: The separate equations of the lines are x – 4y = 0 and x = 0

**Example – 19:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}– 10xy – 8y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} – 10xy – 8y^{2} = 0.

∴ 3x^{2} – 12xy + 2xy – 8y^{2} = 0

∴ 3x(x – 4y) + 2y(x – 4y) = 0

∴ (3x + 2y)(x – 4y) = 0

Ans: The separate equations of the lines are 3x + 2y = 0 and x – 4y = 0

**Example – 20:**

- Obtain separate equations of lines represented by joint equation 3x
^{2}– 2xy – 3y^{2}= 0. **Solution:**

Given joint equation is 3x^{2} – 2xy – y^{2} = 0.

∴ 3x^{2} – 3xy + xy – y^{2} = 0

∴ 3x(x – y) + y(x – y) = 0

∴ (3x + y)(x – y) = 0

Ans: The separate equations of the lines are 3x + y = 0 and x – y = 0

**Example – 21:**

- Obtain separate equations of lines represented by joint equation 6x
^{2}– 5xy – 6y^{2}= 0. **Solution:**

Given joint equation is 6x^{2} – 5xy – 6y^{2} = 0.

∴ 6x^{2} – 9xy+ 4xy – 6y^{2} = 0

∴ 3x(2x – 3y) + 2y(2x – 3y) = 0

∴ (3x + 2y)(2x – 3y) = 0

Ans: The separate equations of the lines are 3x + 2y = 0 and 2x – 3y = 0

**Example – 22:**

- Obtain separate equations of lines represented by joint equation 2x
^{2}+ 2xy – y^{2}= 0.

**Solution:**

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

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

2 + 2m – m^{2} = 0

∴ m^{2} – 2m – 2= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2, c = -2

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (1 + √3) x

∴ (1 + √3) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (1 – √3) x

∴ (1 – √3) x – y = 0

**Ans:** The separate equations of the lines are (1 + √3) x – y = 0 and (1 – √3) x – y = 0

**Example – 23:**

- Obtain separate equations of lines represented by joint equation x
^{2}+ 2(tanα) xy – y^{2}= 0. **Solution:**

The given joint equation is x^{2} + 2(tanα) xy – y^{2} = 0

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2tanα m – m^{2} = 0

∴ m^{2} – 2tanα m – 1= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = – 2tanα, c = -1

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (tanα + secα) x

∴ (tanα + secα) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (tanα – secα) x

∴ (tanα – secα) x – y = 0

**Ans:** The separate equations of the lines are (tanα + secα) x – y = 0 and (tanα – secα) x – y = 0

**Example – 24:**

- Obtain separate equations of lines represented by joint equation x
^{2}+ 2(cosecα) xy + y^{2}= 0. **Solution:**

The given joint equation is x^{2} + 2(cosecα) xy + y^{2} = 0.

Dividing both sides of the equation by x^{2}

The equation of a line passing through the origin is of the form y = mx

Substituting y/x = m in the equation (1) we get

1 + 2cosecα m + m^{2} = 0

∴ m^{2} + 2cosecα m + 1= 0

This is a quadratic equation in m and has two roots say m_{1} and m_{1}

which gives slopes of the two lines represented by the joint equation.

a = 1, b = 2cosecα, c = 1

The roots are given by

The equation of the first line is

y = m_{1}x

∴ y = (- cosec α + cot α) x

∴ y = – (cosec α – cot α) x

∴ (cosec α – cot α) x – y = 0

The equation of the second line is

y = m_{2}x

∴ y = (- cosec α – cot α) x

∴ y = – (cosec α + cot α) x

∴ (cosec α + cot α) x – y = 0

**Ans:** The separate equations of the lines are (cosec α – cot α) x – y = 0 and (cosec α + cot α) x – y = 0

**Example – 25:**

- Obtain separate equations of lines represented by joint equation 4x
^{2}– y^{2}+ 2x + y = 0. Also find point of intersection.. **Solution:**

The given joint equation is 4x^{2} – y^{2}+ 2x + y = 0

∴ (2x)^{2} – y^{2}+ (2x + y) = 0

∴ (2x + y)(2x – y) + (2x + y) = 0

∴ (2x + y)(2x – y + 1] = 0

∴ Separate equations of the lines are 2x + y = 0 and 2x – y + 1 = 0

Let, 2x + y = 0 ……….. (1)

2x – y = -1 ……….. (2)

Adding equations (1) and (2)

4x = -1 i.e x = -1/4

Substituting in equation (1) we get

2(-1/4) + y = 0

∴ -1/2 + y = 0

∴ y = 1/2

Hence the point of intersection is (-1/4, 1/2)

**Ans:** The separate equations of the lines are 2x + y = 0 and 2x – y + 1 = 0

and their point of intersection is (-1/4, -1/2)

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

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