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

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

#### Notes:

• If m1 and m2 are the slopes of the two lines represented by joint equation  ax2 + 2hxy  + by2 = 0. Then m1. m2 = a/b ; m1 + m2 = -2h/b  and remember (m1 – m2) 2 =  (m1+ m2)2 – 4m1.m2.

#### ALGORITHM:

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

#### Example – 1:

• Show that the equation 3x2 – 4xy +  y2 = 0 represents the pair of lines whose slopes differ by 2.
• Solution:

The given joint equation is 3x2 – 4xy + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = 3, 2h = – 4, and b = 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -(-4)/1 = 4 and  m1 . m= a/b = 3/1 = 3

Using  (m1 – m2)2 =  (m1 + m2)2 –  4 m1 . m2

∴   (m1 – m2)2 =  (4)2 –  4 (3)

∴   (m1 – m2)2 =  16 – 12

∴   (m1 – m2)2 =  4

Taking square roots of both sides we get,

| m1 – m2| = 2

Hence the slopes differ by 2.

#### Example – 2:

• Show that the equation 77x2 – 36xy +  4y2 = 0 represents the pair of lines whose slopes differ by 2.
• Solution:

The given joint equation is 77x2 – 36xy +  4y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = 77, 2h = – 36, and b = 4

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -(-36)/4 = 9 and  m1 . m= a/b = 77/4

Using  (m1 – m2)2 =  (m1 + m2)2 –  4 m1 . m2

∴   (m1 – m2)2 =  (9)2 –  4 (77/4)

∴   (m1 – m2)2 =  81 – 77

∴   (m1 – m2)2 =  4

Taking square roots of both sides we get,

| m1 – m2| = 2

Hence the slopes differ by 2.

#### Example – 3:

• Show that the difference of the slopes of the two lines represented by equation x2(tan2θ + cos2θ)  – 2xy tanθ + y2sin2θ = 0 is 2.
• Solution:

The given joint equation is x2(tan2θ + cos2θ)  – 2xy tanθ + y2sin2θ = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = (tan2θ + cos2θ), 2h = – 2tanθ, and b = sin2θ

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -(- 2tanθ)/sin2θ = 2tanθ)/sin2θ

and  m1 . m= (tan2θ + cos2θ)/sin2θ

Using  (m1 – m2)2 =  (m1 + m2)2 –  4 m1 . m2

Hence the slopes differ by 2.

#### Example – 4:

• Show that the difference of the slopes of the two lines represented by equation (tan2θ + 3)(tan2θ – 1)x2 – 2xy sec2θ   +  y2 = 0 is 4.
• Solution:

The given joint equation is (tan2θ + 3)(tan2θ – 1)x2 – 2xy sec2θ   +  y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = (tan2θ + 3)(tan2θ – 1), 2h = – 2sec2θ, and b = 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -(- 2sec2θ)/1 =  2sec2θ

and  m1 . m= (tan2θ + 3)(tan2θ – 1)/1 = (tan2θ + 3)(tan2θ – 1)

Using  (m1 – m2)2 =  (m1 + m2)2 –  4 m1 . m2

Hence the slopes differ by 4.

#### Example – 5:

• Find λ, if the difference of the slopes of the lines λx2 + 6xy –  4y2 = 0 is equal to their product.
• Solution:

The given joint equation is λx2 + 6xy –  4y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = λ, 2h = 6, and b = – 4

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -6/-4 = 3/2 and  m1 . m= a/b = λ/-4  = – λ/4

Given | m1 – m2| = m1 . m2

squaring both sides

(m1 – m2)2 =  (m1 . m2)2

∴   (m1 + m2)2 –  4 m1 . m2=  (m1 . m2)2

∴   (3/2)2 –  4 (- λ/4)=  (- λ/4)2

∴   9/4 +  λ =  λ2/16

Multiplying both sides of equation by 16

∴   36 +  16λ =  λ2

∴  λ2 –  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  3x2 + kxy  – y2 = 0 is 4.
• Solution:

The given joint equation is 3x2 + kxy  – y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = 3, 2h = k, and b = – 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = -k/-1 = k and  m1 . m= a/b  = 3/-1 = – 3

Given | m1 – m2| = 4

squaring both sides

(m1 – m2)2 = 16

∴   (m1 + m2)2 –  4 m1 . m2=  16

∴   (k)2 –  4 (- 3)=  16

∴   k2 + 12 =  16

∴   k2  = 4

∴   k = ± 2

#### Example – 7:

• Find k if the slopes of lines given by kx2 + 5xy  + y2 = 0 differ by 1.
• Solution:

The given joint equation is kx2 + 5xy  + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – 5/1 = – 5 and  m1 . m= a/b = k/1 = k

Given | m1 – m2| = 1

squaring both sides

(m1 – m2)2 = 1

∴   (m1 + m2)2 –  4 m1 . m2= 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 kx2 + 4xy  – y2 = 0 exceeds the slope of other by 8.
• Solution:

The given joint equation is kx2 + 4xy  – y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = k, 2h = 4, and b = – 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – 4/-1 =  4 and  m1 . m= a/b  = k/-1 = – k

Given | m1 – m2| = 8

squaring both sides

(m1 – m2)2 = 64

∴   (m1 + m2)2 –  4 m1 . m2= 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  12x2 + k xy  – y2 = 0 is 7.
• Solution:

The given joint equation is 12x2 + k xy  – y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = 12, 2h = k, and b = – 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – k/-1 = k and  m1 . m= a/b = 12/-1 = – 12

Given | m1 – m2| = 7

squaring both sides

(m1 – m2)2 = 49

∴   (m1 + m2)2 –  4 m1 . m2= 49

∴   (k)2 –  4(- 12)=  49

∴   k2 + 48 =  49

∴  k2 = 1

∴   k = ± 1

#### Example – 10:

• Find k, if the slope of one of the lines given by 3x2 + 4xy  + ky2 = 0 is three times the slope of the other.
• Solution:

The given joint equation is 3x2 + 4xy  + ky2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – 4/k and  m1 . m= a/b = 3/k

Given m1 = 3m2

∴  k2 = k

∴  k2 – 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 kx2 + 4xy  – y2 = 0 is three times the slope of the other.
• Solution:

The given joint equation is kx2 + 4xy  – y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

Here a = k, 2h = 4, and b = – 1

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – 4/-1 = 4, and  m1 . m= a/b = k/-1 = – k

Given m1 = 3m2

Now, m1 + m2 = 4

∴  3m2 + m2 = 4

∴  4m2 = 4

∴  m2 = 1

∴  m1 = 3m= 3 x 1 = 3

Now, m1 . m=  – k

∴  (3)(1) = – k

∴  k = – 3

#### Example – 12:

• Find k, if the slope of one of the lines given by 4x2 + kxy  + y2 = 0 is four times the slope of the other.
• Solution:

The given joint equation is 4x2 + kxy  + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – k/1 = – k, and  m1 . m= a/b = 4/1 = 4

Given m1 = 4m2

Now, m1 + m2 = – k

∴  4m2 + m2 = – k

∴  5m2 = – k

∴  m2 = – k/5

Now, m1 . m= 4

∴  4m2 . m= 4

∴  (m2)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 5x2 + kxy  + y2 = 0 is five times the slope of the other.
• Solution:

The given joint equation is 5x2 + kxy  + y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – k/1 = – k, and  m1 . m= a/b = 5/1 = 5

Given m1 = 5m2

Now, m1 + m2 = – k

∴  5m2 + m2 = – k

∴  6m2 = – k

∴  m2 = – k/6

Now, m1 . m= 5

∴  5m2 . m= 5

∴  (m2)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 kx2 + 8xy + 5 y2 = 0 is twice the product of the slopes.
• Solution:

The given joint equation is kx2 + 8xy + 5 y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – 8/5, and  m1 . m= a/b = k/5

Given m1 + m2 = 2 m1 . m2

-8/5 = 2 x k/5

∴  k = – 4

#### Example – 15:

• Find k if the sum of the slopes of lines given by 2x2 + kxy -3y2 = 0 is equal to the product of the slopes.
• Solution:

The given joint equation is 2x2 + kxy -3y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – k/-3 = k/3, and  m1 . m= a/b = 2/-3 = – 2/3

Given m1 + m2 = m1 . m2

k/3 = (-2/3)

∴  k = – 2

#### Example – 16:

• Find k, if the sum of the slopes of lines given by 3x2 + kxy  + y2 = 0 is zero.
• Solution:

The given joint equation is 2x2 + kxy -3y2 = 0

Comparing with ax2 + 2hxy + by2 = 0

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

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b = – k/1 = – k, and  m1 . m= a/b = 3/1 = 3

Given m1 + m2 = 0

∴  – k = 0

∴  k = 0

#### Example – 17:

• Find k, if the slope of one of the lines given by 3x2 – 4xy  + ky2 = 0 is 1.
• Solution:

Given joint equation of lines is 3x2 – 4xy  + ky2 = 0

It is in the form ax2 + 2hxy  + by2 = 0.

a = 3, 2h = – 4 and b= k

The auxiliary equation of given line is of the form

bm2 + 2hm + a = 0

∴   km2 – 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 ax2 + 2hxy + by2 = 0 is k times the slope of the other, prove that 4kh2 = ab(1+ k)2.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

Given m1 = km2

Now, m1 + m2 = – 2h/b

∴  km2 + m2 =- 2h/b

∴ (k + 1)m2 =- 2h/b

∴  m2 = – 2h/b(k + 1)

Now, m1 . m= a/b

∴  km2 . m= a/b

∴  k(m2)2 = a/b

∴  (m2)2 = a/bk

4kh2 = ab(1+ k)2 (proved as required)

#### Example – 19:

• If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is 5 times the slope of the other, prove that 5h2 = 9ab.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

Given m1 = 5m2

Now, m1 + m2 = – 2h/b

∴  5m2 + m2 =- 2h/b

∴    6m2 =- 2h/b

∴  m2 = – h/3b

Now, m1 . m= a/b

∴  5m2 . m= a/b

∴  5(m2)2 = a/b

5h2 = 9ab  (proved as required)

#### Example – 20:

• If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is 4 times the slope of the other, prove that 16h2 = 25ab.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

Given m1 = 4m2

Now, m1 + m2 = – 2h/b

∴  4m2 + m2 =- 2h/b

∴    5m2 =- 2h/b

∴  m2 = – 2h/5b

Now, m1 . m= a/b

∴  4m2 . m= a/b

∴  4(m2)2 = a/b

16h2 = 25ab  (proved as required)

#### Example – 21:

• If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is 3 times the slope of the other, prove that 3h2 = 4ab.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

Given m1 = 3m2

Now, m1 + m2 = – 2h/b

∴  3m2 + m2 =- 2h/b

∴    4m2 =- 2h/b

∴  m2 = – h/2b

Now, m1 . m= a/b

∴  3m2 . m= a/b

∴  3(m2)2 = a/b

3h2 = 4ab  (proved as required)

#### Example – 22:

• Find the condition that slope of one of the lines represented by ax2 + 2hxy  + by2 = 0 is square of the slope of another line.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

Given m1 = (m2)2

Now, m1 . m= a/b

∴  (m2)2 . m= a/b

∴  (m2)3 = a/b

Now, m1 + m2 = – 2h/b

∴  (m2)2 + m2 = – 2h/b

Making cube of both sides

((m2)2 + m2)3 = (- 2h/b)3

Multiplying both sides by b3

a2b – 6abh + ab2 = -8h3

a2b  + ab2 + 8h= 6abh

This is the required condition

#### Example – 23:

• Show that one of the straight line  given by ax2 + 2hxy + by2 = 0 bisects an angle between the co-ordinate axes if and only if  (a + b)2 = 4h2.
• Solution:

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= 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 m2 = ± 1

#### Example – 24:

• If the straight lines represented by a joint equation  ax2 + 2hxy + by2 = 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

m1 = tan α

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

m2 = tan (π/2 ± α) = ± cot α

The given joint equation is ax2 + 2hxy + by2 = 0

Let m1 and m2 be the slopes of the two lines represented by given joint equation

(m1 + m2) = – 2h/b, and  m1 . m= a/b

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