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**Type – V: ****To Find Value of Constant When Relation Between Slopes is Given**

**Type – V:**

**To Find Value of Constant When Relation Between Slopes is Given**

**Notes:**

- If m
_{1}and m_{2}are the slopes of the two lines represented by joint equation ax^{2}+ 2hxy + by^{2}= 0. Then m_{1}. m_{2}= a/b ; m_{1}+ m_{2}= -2h/b and remember (m_{1}– m_{2}) 2 = (m_{1}+ m_{2})^{2}– 4m_{1}.m_{2}.

**ALGORITHM:**

- Write given of joint equation.
- Compare with the standard equation.
- Find values of a, h and b.
- Use above relations between slopes
- Solve equations and get the value of constant

**Example – 1:**

- Show that the equation 3x
^{2}– 4xy + y^{2}= 0 represents the pair of lines whose slopes differ by 2. **Solution:**

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

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

Here a = 3, 2h = – 4, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -(-4)/1 = 4 and m_{1} . m_{2 }= a/b = 3/1 = 3

Using (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}

∴ (m_{1} – m_{2})^{2} = (4)^{2} – 4 (3)

∴ (m_{1} – m_{2})^{2} = 16 – 12

∴ (m_{1} – m_{2})^{2} = 4

Taking square roots of both sides we get,

| m_{1} – m_{2}| = 2

Hence the slopes differ by 2.

**Example – 2:**

- Show that the equation 77x
^{2}– 36xy + 4y^{2}= 0 represents the pair of lines whose slopes differ by 2. **Solution:**

The given joint equation is 77x^{2} – 36xy + 4y^{2} = 0

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

Here a = 77, 2h = – 36, and b = 4

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -(-36)/4 = 9 and m_{1} . m_{2 }= a/b = 77/4

Using (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}

∴ (m_{1} – m_{2})^{2} = (9)^{2} – 4 (77/4)

∴ (m_{1} – m_{2})^{2} = 81 – 77

∴ (m_{1} – m_{2})^{2} = 4

Taking square roots of both sides we get,

| m_{1} – m_{2}| = 2

Hence the slopes differ by 2.

**Example – 3:**

- Show that the difference of the slopes of the two lines represented by equation x
^{2}(tan^{2}θ + cos^{2}θ) – 2xy tanθ + y^{2}sin^{2}θ = 0 is 2. **Solution:**

The given joint equation is x^{2}(tan^{2}θ + cos^{2}θ) – 2xy tanθ + y^{2}sin^{2}θ = 0

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

Here a = (tan^{2}θ + cos^{2}θ), 2h = – 2tanθ, and b = sin^{2}θ

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -(- 2tanθ)/sin^{2}θ = 2tanθ)/sin^{2}θ

and m_{1} . m_{2 }= (tan^{2}θ + cos^{2}θ)/sin^{2}θ

Using (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}

Hence the slopes differ by 2.

**Example – 4:**

- Show that the difference of the slopes of the two lines represented by equation (tan
^{2}θ + 3)(tan^{2}θ – 1)x^{2}– 2xy sec^{2}θ + y^{2}= 0 is 4. **Solution:**

The given joint equation is (tan^{2}θ + 3)(tan^{2}θ – 1)x^{2} – 2xy sec^{2}θ + y^{2} = 0

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

Here a = (tan^{2}θ + 3)(tan^{2}θ – 1), 2h = – 2sec^{2}θ, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -(- 2sec^{2}θ)/1 = 2sec^{2}θ

and m_{1} . m_{2 }= (tan^{2}θ + 3)(tan^{2}θ – 1)/1 = (tan^{2}θ + 3)(tan^{2}θ – 1)

Using (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}

Hence the slopes differ by 4.

**Example – 5:**

- Find λ, if the difference of the slopes of the lines λx
^{2}+ 6xy – 4y^{2}= 0 is equal to their product. **Solution:**

The given joint equation is λx^{2} + 6xy – 4y^{2} = 0

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

Here a = λ, 2h = 6, and b = – 4

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -6/-4 = 3/2 and m_{1} . m_{2 }= a/b = λ/-4 = – λ/4

Given | m_{1} – m_{2}| = m_{1} . m_{2}

squaring both sides

(m_{1} – m_{2})^{2} = (m_{1} . m_{2})^{2}

∴ (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}= (m_{1} . m_{2})^{2}

∴ (3/2)^{2} – 4 (- λ/4)= (- λ/4)^{2}

∴ 9/4 + λ = λ^{2}/16

Multiplying both sides of equation by 16

∴ 36 + 16λ = λ^{2}

∴ λ^{2} – 16λ – 36 = 0

∴ (λ – 18)(λ + 2) = 0

∴ λ = 18 or λ = – 2

**Example – 6:**

- Find k, if the difference of the slopes of the lines 3x
^{2}+ kxy – y^{2}= 0 is 4. **Solution:**

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

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

Here a = 3, 2h = k, and b = – 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = -k/-1 = k and m_{1} . m_{2 }= a/b = 3/-1 = – 3

Given | m_{1} – m_{2}| = 4

squaring both sides

(m_{1} – m_{2})^{2} = 16

∴ (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}= 16

∴ (k)^{2} – 4 (- 3)= 16

∴ k^{2} + 12 = 16

∴ k^{2} = 4

∴ k = ± 2

**Example – 7:**

- Find k if the slopes of lines given by kx
^{2}+ 5xy + y^{2}= 0 differ by 1. **Solution:**

The given joint equation is kx^{2} + 5xy + y^{2} = 0

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

Here a = k, 2h = 5, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – 5/1 = – 5 and m_{1} . m_{2 }= a/b = k/1 = k

Given | m_{1} – m_{2}| = 1

squaring both sides

(m_{1} – m_{2})^{2} = 1

∴ (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}= 1

∴ (-5)^{2} – 4 k= 1

∴ 25 – 4 k= 1

∴ – 4 k= – 24

∴ k = 6

**Example – 8:**

- Find k, if the slope of one of the lines given by kx
^{2}+ 4xy – y^{2}= 0 exceeds the slope of other by 8. **Solution:**

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

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

Here a = k, 2h = 4, and b = – 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – 4/-1 = 4 and m_{1} . m_{2 }= a/b = k/-1 = – k

Given | m_{1} – m_{2}| = 8

squaring both sides

(m_{1} – m_{2})^{2} = 64

∴ (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}= 64

∴ (4)^{2} – 4(- k)= 64

∴ 16 + 4 k= 64

∴ – 4 k= 48

∴ k = 12

**Example – 9:**

- Find k, if the difference between the slopes of the lines 12x
^{2}+ k xy – y^{2}= 0 is 7. **Solution:**

The given joint equation is 12x^{2} + k xy – y^{2} = 0

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

Here a = 12, 2h = k, and b = – 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – k/-1 = k and m_{1} . m_{2 }= a/b = 12/-1 = – 12

Given | m_{1} – m_{2}| = 7

squaring both sides

(m_{1} – m_{2})^{2} = 49

∴ (m_{1} + m_{2})^{2} – 4 m_{1} . m_{2}= 49

∴ (k)^{2} – 4(- 12)= 49

∴ k^{2} + 48 = 49

∴ k^{2} = 1

∴ k = ± 1

**Example – 10:**

- Find k, if the slope of one of the lines given by 3x
^{2}+ 4xy + ky^{2}= 0 is three times the slope of the other. **Solution:**

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

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

Here a = 3, 2h = 4, and b = k

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – 4/k and m_{1} . m_{2 }= a/b = 3/k

Given m_{1} = 3m_{2}

∴ k^{2} = k

∴ k^{2} – k = 0

∴ k (k – 1) = 0

∴ k = 0 and k = 1

But k cannot be zero.

∴ k = 1

**Example – 11:**

- Find k, if the slope of one of the lines given by kx
^{2}+ 4xy – y^{2}= 0 is three times the slope of the other. **Solution:**

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

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

Here a = k, 2h = 4, and b = – 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – 4/-1 = 4, and m_{1} . m_{2 }= a/b = k/-1 = – k

Given m_{1} = 3m_{2}

Now, m_{1} + m_{2} = 4

∴ 3m_{2} + m_{2} = 4

∴ 4m_{2} = 4

∴ m_{2} = 1

∴ m_{1} = 3m_{2 }= 3 x 1 = 3

Now, m_{1} . m_{2 }= – k

∴ (3)(1) = – k

∴ k = – 3

**Example – 12:**

- Find k, if the slope of one of the lines given by 4x
^{2}+ kxy + y^{2}= 0 is four times the slope of the other. **Solution:**

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

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

Here a = 4, 2h = k, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – k/1 = – k, and m_{1} . m_{2 }= a/b = 4/1 = 4

Given m_{1} = 4m_{2}

Now, m_{1} + m_{2} = – k

∴ 4m_{2} + m_{2} = – k

∴ 5m_{2} = – k

∴ m_{2} = – k/5

Now, m_{1} . m_{2 }= 4

∴ 4m_{2} . m_{2 }= 4

∴ (m_{2})^{2} = 1

∴ (- k/5)^{2} = 1

∴ (- k/5) = ± 1

∴ k = ± 5

**Example – 13:**

- Find k, if the slope of one of the lines given by 5x
^{2}+ kxy + y^{2}= 0 is five times the slope of the other. **Solution:**

The given joint equation is 5x^{2} + kxy + y^{2} = 0

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

Here a = 5, 2h = k, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – k/1 = – k, and m_{1} . m_{2 }= a/b = 5/1 = 5

Given m_{1} = 5m_{2}

Now, m_{1} + m_{2} = – k

∴ 5m_{2} + m_{2} = – k

∴ 6m_{2} = – k

∴ m_{2} = – k/6

Now, m_{1} . m_{2 }= 5

∴ 5m_{2} . m_{2 }= 5

∴ (m_{2})^{2} = 1

∴ (- k/6)^{2} = 1

∴ (- k/6) = ± 1

∴ k = ± 6

**Example – 14:**

- Find k if the sum of the slopes of lines given by kx
^{2}+ 8xy + 5 y^{2}= 0 is twice the product of the slopes. **Solution:**

The given joint equation is kx^{2} + 8xy + 5 y^{2} = 0

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

Here a = k, 2h = 8, and b = 5

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – 8/5, and m_{1} . m_{2 }= a/b = k/5

Given m_{1} + m_{2} = 2 m_{1} . m_{2}

-8/5 = 2 x k/5

∴ k = – 4

**Example – 15:**

- Find k if the sum of the slopes of lines given by 2x
^{2}+ kxy -3y^{2}= 0 is equal to the product of the slopes. **Solution:**

The given joint equation is 2x^{2} + kxy -3y^{2} = 0

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

Here a = 2, 2h = k, and b = -3

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – k/-3 = k/3, and m_{1} . m_{2 }= a/b = 2/-3 = – 2/3

Given m_{1} + m_{2} = m_{1} . m_{2}

k/3 = (-2/3)

∴ k = – 2

**Example – 16:**

- Find k, if the sum of the slopes of lines given by 3x
^{2}+ kxy + y^{2}= 0 is zero. **Solution:**

The given joint equation is 2x^{2} + kxy -3y^{2} = 0

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

Here a = 3, 2h = k, and b = 1

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b = – k/1 = – k, and m_{1} . m_{2 }= a/b = 3/1 = 3

Given m_{1} + m_{2} = 0

∴ – k = 0

∴ k = 0

**Example – 17:**

- Find k, if the slope of one of the lines given by 3x
^{2}– 4xy + ky^{2}= 0 is 1. **Solution:**

Given joint equation of lines is 3x^{2} – 4xy + ky^{2} = 0

It is in the form ax^{2} + 2hxy + by^{2} = 0.

a = 3, 2h = – 4 and b= k

The auxiliary equation of given line is of the form

bm^{2} + 2hm + a = 0

∴ km^{2} – 4m + 3 = 0 ……… (1)

Slope of one of the line is 1

Substituting m = 1 in equation (1) we get

k(1)^{2} – 4(1) + 3 = 0

∴ k – 1 = 0

∴ k = 1

**Example – 18:**

- If the slope of one of the lines given by ax
^{2}+ 2hxy + by^{2}= 0 is k times the slope of the other, prove that 4kh^{2}= ab(1+ k)^{2}. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

Given m_{1} = km_{2}

Now, m_{1} + m_{2} = – 2h/b

∴ km_{2} + m_{2} =- 2h/b

∴ (k + 1)m_{2} =- 2h/b

∴ m_{2} = – 2h/b(k + 1)

Now, m_{1} . m_{2 }= a/b

∴ km_{2} . m_{2 }= a/b

∴ k(m_{2})^{2} = a/b

∴ (m_{2})^{2} = a/bk

4kh^{2} = ab(1+ k)^{2} (proved as required)

**Example – 19:**

- If the slope of one of the lines given by ax
^{2}+ 2hxy + by^{2}= 0 is 5 times the slope of the other, prove that 5h^{2}= 9ab. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

Given m_{1} = 5m_{2}

Now, m_{1} + m_{2} = – 2h/b

∴ 5m_{2} + m_{2} =- 2h/b

∴ 6m_{2} =- 2h/b

∴ m_{2} = – h/3b

Now, m_{1} . m_{2 }= a/b

∴ 5m_{2} . m_{2 }= a/b

∴ 5(m_{2})^{2} = a/b

5h^{2} = 9ab (proved as required)

**Example – 20:**

- If the slope of one of the lines given by ax
^{2}+ 2hxy + by^{2}= 0 is 4 times the slope of the other, prove that 16h^{2}= 25ab. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

Given m_{1} = 4m_{2}

Now, m_{1} + m_{2} = – 2h/b

∴ 4m_{2} + m_{2} =- 2h/b

∴ 5m_{2} =- 2h/b

∴ m_{2} = – 2h/5b

Now, m_{1} . m_{2 }= a/b

∴ 4m_{2} . m_{2 }= a/b

∴ 4(m_{2})^{2} = a/b

16h^{2} = 25ab (proved as required)

**Example – 21:**

- If the slope of one of the lines given by ax
^{2}+ 2hxy + by^{2}= 0 is 3 times the slope of the other, prove that 3h^{2}= 4ab. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

Given m_{1} = 3m_{2}

Now, m_{1} + m_{2} = – 2h/b

∴ 3m_{2} + m_{2} =- 2h/b

∴ 4m_{2} =- 2h/b

∴ m_{2} = – h/2b

Now, m_{1} . m_{2 }= a/b

∴ 3m_{2} . m_{2 }= a/b

∴ 3(m_{2})^{2} = a/b

3h^{2} = 4ab (proved as required)

**Example – 22:**

- Find the condition that slope of one of the lines represented by ax
^{2}+ 2hxy + by^{2}= 0 is square of the slope of another line. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

Given m_{1} = (m_{2})^{2}

Now, m_{1} . m_{2 }= a/b

∴ (m_{2})^{2} . m_{2 }= a/b

∴ (m_{2})^{3} = a/b

Now, m_{1} + m_{2} = – 2h/b

∴ (m_{2})^{2} + m_{2} = – 2h/b

Making cube of both sides

((m_{2})^{2} + m_{2})^{3} = (- 2h/b)^{3}

Multiplying both sides by b^{3}

a^{2}b – 6abh + ab^{2} = -8h^{3}

a^{2}b + ab^{2} + 8h^{3 }= 6abh

This is the required condition

**Example – 23:**

- Show that one of the straight line given by ax
^{2}+ 2hxy + by^{2}= 0 bisects an angle between the co-ordinate axes if and only if (a + b)^{2}= 4h^{2}. **Solution:**

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

- Now given that One of the straight line represented by the joint equation bisects the angle between the co-ordinate axes. Thus the slope of the line is ± 1. let m
_{2}= ± 1

**Example – 24:**

- If the straight lines represented by a joint equation ax
^{2}+ 2hxy + by^{2}= 0 make angles of equal measures with the co-ordinate axes. Then prove that a = ± b. **Solution:**

Let α be the angle made by the lines with coordinate axes.

** **

Let OA be the line making an angle α with the x-axis and

OB be the line making an angle α with y-axis.

The angle made by the line with the positive direction of x-axis is α. Hence its slope is

m_{1} = tan α

The angle made by OB with the positive direction of x-axis is (π/2 ± α). Hence its slope is

m_{2} = tan (π/2 ± α) = ± cot α

The given joint equation is ax^{2} + 2hxy + by^{2} = 0

Let m_{1} and m_{2} be the slopes of the two lines represented by given joint equation

(m_{1} + m_{2}) = – 2h/b, and m_{1} . m_{2 }= a/b

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