Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

I : Choose correct alternatives.

Question 1.

If the equation 4x^{2} + hxy + y^{2} = 0 represents two coincident lines, then h = _________.

(A) ± 2

(B) ± 3

(C) ± 4

(D) ± 5

Solution:

(C) ± 4

Question 2.

If the lines represented by kx^{2} – 3xy + 6y^{2} = 0 are perpendicular to each other then _________.

(A) k = 6

(B) k = -6

(C) k = 3

(D) k = -3

Solution:

(B) k = -6

Question 3.

Auxiliary equation of 2x^{2} + 3xy – 9y^{2} = 0 is _________.

(A) 2m^{2} + 3m – 9 = 0

(B) 9m^{2} – 3m – 2 = 0

(C) 2m^{2} – 3m + 9 = 0

(D) -9m^{2} – 3m + 2 = 0

Solution:

(B) 9m^{2} – 3m – 2 = 0

Question 4.

The difference between the slopes of the lines represented by 3x^{2} – 4xy + y^{2} = 0 is _________.

(A) 2

(B) 1

(C) 3

(D) 4

Solution:

(A) 2

Question 5.

If the two lines ax^{2} +2hxy+ by^{2} = 0 make angles α and β with X-axis, then tan (α + β) = _____.

(A) \(\frac{h}{a+b}\)

(B) \(\frac{h}{a-b}\)

(C) \(\frac{2 h}{a+b}\)

(D) \(\frac{2 h}{a-b}\)

Solution:

(D) \(\frac{2 h}{a-b}\)

Question 6.

If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h^{2} = ___.

(A) 1 : 2

(B) 2 : 1

(C) 8 : 9

(D) 9 : 8

Solution:

(D) 9 : 8

Question 7.

The joint equation of the lines through the origin and perpendicular to the pair of lines 3x^{2} + 4xy – 5y^{2} = 0 is _________.

(A) 5x^{2} + 4xy – 3y^{2} = 0

(B) 3x^{2} + 4xy – 5y^{2} = 0

(C) 3x^{2} – 4xy + 5y^{2} = 0

(D) 5x^{2} + 4xy + 3y^{2} = 0

Solution:

(A) 5x^{2} + 4xy – 3y^{2} = 0

Question 8.

If acute angle between lines ax^{2} + 2hxy + by^{2} = 0 is, \(\frac{\pi}{4}\) then 4h^{2} = _________.

(A) a^{2} + 4ab + b^{2}

(B) a^{2} + 6ab + b^{2}

(C) (a + 2b)(a + 3b)

(D) (a – 2b)(2a + b)

Solution:

(B) a^{2} + 6ab + b^{2}

Question 9.

If the equation 3x^{2} – 8xy + qy^{2} + 2x + 14y + p = 1 represents a pair of perpendicular lines then

the values of p and q are respectively _________.

(A) -3 and -7

(B) -7 and -3

(C) 3 and 7

(D) -7 and 3

Solution:

(B) -7 and -3

Question 10.

The area of triangle formed by the lines x^{2} + 4xy + y^{2} = 0 and x – y – 4 = 0 is _________.

(A) \(\frac{4}{\sqrt{3}}\) Sq. units

(B) \(\frac{8}{\sqrt{3}}\) Sq. units

(C) \(\frac{16}{\sqrt{3}}\) Sq. units

(D)\(\frac{15}{\sqrt{3}}\) Sq. units

Solution:

(B) \(\frac{8}{\sqrt{3}}\) Sq. units

[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x – y – 4 = 0]

Question 11.

The combined equation of the co-ordinate axes is _________.

(A) x + y = 0

(B) x y = k

(C) xy = 0

(D) x – y = k

Solution:

(C) xy = 0

Question 12.

If h^{2} = ab, then slope of lines ax^{2} + 2hxy + by^{2} = 0 are in the ratio _________.

(A) 1 : 2

(B) 2 : 1

(C) 2 : 3

(D) 1 : 1

Solution:

(D) 1 : 1

[Hint: If h^{2} = ab, then lines are coincident. Therefore slopes of the lines are equal.]

Question 13.

If slope of one of the lines ax^{2} + 2hxy + by^{2} = 0 is 5 times the slope of the other, then 5h^{2} = _________.

(A) ab

(B) 2 ab

(C) 7 ab

(D) 9 ab

Solution:

(D) 9 ab

Question 14.

If distance between lines (x – 2y)^{2} + k(x – 2y) = 0 is 3 units, then k =

(A) ± 3

(B) ± 5\(\sqrt {5}\)

(C) 0

(D) ± 3\(\sqrt {5}\)

Solution:

(D) ± 3\(\sqrt {5}\)

[Hint: (x – 2y)^{2} + k(x – 2y) = 0

∴ (x – 2y)(x – 2y + k) = 0

∴ equations of the lines are x – 2y = 0 and x – 2y + k = 0 which are parallel to each other.

∴ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3

∴ k = ± 3\(\sqrt {5}\)

II. Solve the following.

Question 1.

Find the joint equation of lines:

(i) x – y = 0 and x + y = 0

Solution:

The joint equation of the lines x – y = 0 and

x + y = 0 is

(x – y)(x + y) = 0

∴ x^{2} – y^{2} = 0.

(ii) x + y – 3 = 0 and 2x + y – 1 = 0

Solution:

The joint equation of the lines x + y – 3 = 0 and 2x + y – 1 = 0 is

(x + y – 3)(2x + y – 1) = 0

∴ 2x^{2} + xy – x + 2xy + y^{2} – y – 6x – 3y + 3 = 0

∴ 2x^{2} + 3xy + y^{2} – 7x – 4y + 3 = 0.

(iii) Passing through the origin and having slopes 2 and 3.

Solution:

We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.

i.e. their equations are

2x – y = 0 and 3x – y = 0 respectively.

∴ their joint equation is (2x – y)(3x – y) = 0

∴ 6x^{2} – 2xy – 3xy + y^{2} = 0

∴ 6x^{2} – 5xy + y^{2} = 0.

(iv) Passing through the origin and having inclinations 60° and 120°.

Solution:

Slope of the line having inclination θ is tan θ .

Inclinations of the given lines are 60° and 120°

∴ their slopes are m_{1} = tan60° = \(\sqrt {3}\) and

m_{2} = tan 120° = tan (180° – 60°)

= -tan 60° = –\(\sqrt {3}\)

Since the lines pass through the origin, their equa-tions are

y = \(\sqrt {3}\)x and y= –\(\sqrt {3}\)x

i.e., \(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0

∴ the joint equation of these lines is

(\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0

∴ 3x^{2} – y^{2} = 0.

(v) Passing through (1, 2) amd parallel to the co-ordinate axes.

Solution:

Equations of the coordinate axes are x = 0 and y = 0

∴ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1

i.e. x – 1 = 0 and y – 2 0

∴ their combined equation is

(x – 1)(y – 2) = 0

∴ x(y – 2) – 1(y – 2) = 0

∴ xy – 2x – y + 2 = 0

(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.

Solution:

Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.

i.e. x – 3 = 0 and y – 2 = 0

∴ their joint equation is

(x – 3)(y – 2) = 0

∴ xy – 2x – 3y + 6 = 0.

(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.

Solution:

Let L_{1} and L_{2} be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 respectively.

Slopes of the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.

∴ slopes of the lines L_{1}and L_{2} are 2 and \(\frac{-4}{3}\) respectively.

Since the lines L_{1} and L_{2} pass through the point (-1, 2), their equations are

∴ (y – y_{1}) = m(x – x_{1})

∴ (y – 2) = 2(x + 1)

⇒ y – 1 = 2x + 2

⇒ 2x – y + 4 = 0 and

∴ (y – 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)

⇒ 3y – 6 = (-4)(x + 1)

⇒ 3y – 6 = -4x + 4

⇒ 4x + 3y – 6 + 4 = 0

⇒ 4x + 3y – 2 = 0

their combined equation is

∴ (2x – y + 4)(4x + 3y – 2) = 0

∴ 8x^{2} + 6xy – 4x – 4xy – 3y^{2} + 2y + 16x + 12y – 8 = 0

∴ 8x^{2} + 2xy + 12x – 3y^{2} + 14y – 8 = 0

(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 – \(\sqrt {3}\)

Solution:

Let l_{1} and l_{2} be the two lines. Slopes of l_{1} is 1 + \(\sqrt {3}\) and that of l_{2} is 1 – \(\sqrt {3}\)

Therefore the equation of a line (l_{1}) passing through the origin and having slope is

y = (1 + \(\sqrt {3}\))x

∴ (1 + \(\sqrt {3}\))x – y = 0 ..(1)

Similarly, the equation of the line (l_{2}) passing through the origin and having slope is

y = (1 – \(\sqrt {3}\))x

∴ (1 – \(\sqrt {3}\))x – y = 0 …(2)

From (1) and (2) the required combined equation is

∴ (1 – 3)x^{2} – 2xy + y^{2} = 0

∴ -2x^{2} – 2xy + y^{2} = 0

∴ 2x^{2} + 2xy – y^{2} = 0

This is the required combined equation.

(ix) Which are at a distance of 9 units from the Y – axis.

Solution:

Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9

i.e. x – 9 = 0 and x + 9 = 0

∴ their combined equation is

(x – 9)(x + 9) = 0

∴ x^{2} – 81 = 0.

(x) Passing through the point (3, 2), one of which is parallel to the line x – 2y = 2 and other is perpendicular to the line y = 3.

Solution:

Let L_{1} be the line passes through (3, 2) and parallel to the line x – 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)

∴ slope of the line L_{1} is \(\frac{1}{2}\).

∴ equation of the line L_{1} is

y – 2 = \(\frac{1}{2}\)(x – 3)

∴ 2y – 4 = x – 3 ∴ x – 2y + 1 = 0

Let L_{2} be the line passes through (3, 2) and perpendicular to the line y = 3.

∴ equation of the line L_{2} is of the form x = a.

Since L_{2} passes through (3, 2), 3 = a

∴ equation of the line L_{2} is x = 3, i.e. x – 3 = 0

Hence, the equations of the required lines are

x – 2y + 1 = 0 and x – 3 = 0

∴ their joint equation is

(x – 2y + 1)(x – 3) = 0

∴ x^{2} – 2xy + x – 3x + 6y – 3 = 0

∴ x^{2} – 2xy – 2x + 6y – 3 = 0.

(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.

Solution:

Let L_{1} and L_{2} be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.

Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.

Since the lines L_{1} and L_{2} pass through the origin, their equations are

y = 2x and y = \(\frac{1}{3}\)x

i.e. 2x – y = 0 and x – 3y = 0

∴ their combined equation is

(2x – y)(x – 3y) = 0

∴ 2x^{2} – 6xy – xy + 3y^{2} = 0

∴ 2x^{2} – 7xy + 3y^{2} = 0.

Question 2.

Show that each of the following equation represents a pair of lines.

(i) x^{2} + 2xy – y^{2} = 0

Solution:

Comparing the equation x^{2} + 2xy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = 2, i.e. h = 1 and b = -1

∴ h^{2} – ab = (1)^{2} – 1(-1) = 1 + 1=2 > 0

Since the equation x^{2} + 2xy – y^{2} = 0 is a homogeneous equation of second degree and h^{2} – ab > 0, the given equation represents a pair of lines which are real and distinct.

(ii) 4x^{2} + 4xy + y^{2} = 0

Solution:

Comparing the equation 4x^{2} + 4xy + y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 4, 2h = 4, i.e. h = 2 and b = 1

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

Since the equation 4x^{2} + 4xy + y^{2} = 0 is a homogeneous equation of second degree and h^{2} – ab = 0, the given equation represents a pair of lines which are real and coincident.

(iii) x^{2} – y^{2} = 0

Solution:

Comparing the equation x^{2} – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = 0, i.e. h = 0 and b = -1

∴ h^{2} – ab = (0)^{2} – 1(-1) = 0 + 1 = 1 > 0

Since the equation x^{2} – y^{2} = 0 is a homogeneous equation of second degree and h^{2} – ab > 0, the given equation represents a pair of lines which are real and distinct.

(iv) x^{2} + 7xy – 2y^{2} = 0

Solution:

Comparing the equation x^{2} + 7xy – 2y^{2} = 0

a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2

∴ h^{2} – ab = \(\left(\frac{7}{2}\right)^{2}\) – 1(-2)

= \(\frac{49}{4}\) + 2

= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0

Since the equation x^{2} + 7xy – 2y^{2} = 0 is a homogeneous equation of second degree and h^{2} – ab > 0, the given equation represents a pair of lines which are real and distinct.

(v) x^{2} – 2\(\sqrt {3}\) xy – y^{2} = 0

Solution:

Comparing the equation x^{2} – 2\(\sqrt {3}\) xy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h= -2\(\sqrt {3}\), i.e. h = –\(\sqrt {3}\) and b = 1

∴ h^{2} – ab = (-\(\sqrt {3}\))^{2} – 1(1) = 3 – 1 = 2 > 0

Since the equation x^{2} – 2\(\sqrt {3}\)xy – y^{2} = 0 is a homo¬geneous equation of second degree and h^{2} – ab > 0, the given equation represents a pair of lines which are real and distinct.

Question 3.

Find the separate equations of lines represented by the following equations:

(i) 6x^{2} – 5xy – 6y^{2} = 0

Solution:

6x^{2} – 5xy – 6y^{2} = 0

∴ 6x^{2} – 9xy + 4xy – 6y^{2} = 0

∴ 3x(2x – 3y) + 2y(2x – 3y) = 0

∴ (2x – 3y)(3x + 2y) = 0

∴ the separate equations of the lines are

2x – 3y = 0 and 3x + 2y = 0.

(ii) x^{2} – 4y^{2} = 0

Solution:

x^{2} – 4y^{2} = 0

∴ x^{2} – (2y)^{2} = 0

∴(x – 2y)(x + 2y) = 0

∴ the separate equations of the lines are

x – 2y = 0 and x + 2y = 0.

(iii) 3x^{2} – y^{2} = 0

Solution:

3x^{2} – y^{2} = 0

∴ (\(\sqrt {3}\) x)^{2} – y^{2} = 0

∴ (\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0

∴ the separate equations of the lines are

\(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0.

(iv) 2x^{2} + 2xy – y^{2} = 0

Solution:

2x^{2} + 2xy – y^{2} = 0

∴ The auxiliary equation is -m^{2} + 2m + 2 = 0

∴ m^{2} – 2m – 2 = 0

m_{1} = 1 + \(\sqrt {3}\) and m_{2} = 1 – \(\sqrt {3}\) are the slopes of the lines.

∴ their separate equations are

y = m_{1}x and y = m_{2}x

i.e. y = (1 + \(\sqrt {3}\))x and y = (1 – \(\sqrt {3}\))x

i.e. (\(\sqrt {3}\) + 1)x – y = 0 and (\(\sqrt {3}\) – 1)x + y = 0.

Question 4.

Find the joint equation of the pair of lines through the origin and perpendicular to the lines

given by :

(i) x^{2} + 4xy – 5y^{2} = 0

Solution:

Comparing the equation x^{2} + 4xy – 5y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = 4, b= -5

Let m_{1} and m_{2} be the slopes of the lines represented by x^{2} + 4xy – 5y^{2} = 0.

Now, required lines are perpendicular to these lines

∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)

Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x

i.e. m_{1}y = -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + \(\frac{4}{5}\)xy – \(\frac{1}{5}\)y^{2} = 0 …[By (1)]

∴ 5x^{2} + 4xy – y^{2} = 0

(ii) 2x^{2} – 3xy – 9y^{2} = 0

Solution:

Comparing the equation 2x^{2} – 3xy – 9y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 2, 2h = -3, b = -9

Let m_{1} and m_{2} be the slopes of the lines represented by 2x^{2} – 3xy – 9y^{2} = 0

∴ m_{1} + m_{2} =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m_{1}m_{2} = \(\frac{a}{b}=-\frac{2}{9}\) …(1)

Now, required lines are perpendicular to these lines

∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)

Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x

i.e. m_{1}y = -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}y)(x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y^{2} = 0 …[By (1)]

∴ 9x^{2} – 3xy – 2y^{2} = 0

(iii) x^{2} + xy – y^{2} = 0

Solution:

Comparing the equation x^{2}+ xy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

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

Let m_{1} and m_{2} be the slopes of the lines represented by x^{2} + xy – y^{2} = 0

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m_{1}m_{2} = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)

Now, required lines are perpendicular to these lines

∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)

Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x

i.e. m_{1}y = -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}y)(x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2}) + m_{1}m_{2}y^{2} = 0

∴ x^{2} + 1xy + (-1)y^{2} = 0 …[By (1)]

∴ x^{2} + xy – y^{2} = 0

Question 5.

Find k if

(i) The sum of the slopes of the lines given by 3x^{2} + kxy – y^{2} = 0 is zero.

Solution:

Comparing the equation 3x^{2} + kxy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 3, 2h = k, b = -1

Let m_{1} and m_{2} be the slopes of the lines represented by 3x^{2} + kxy – y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k

Now, m_{1} + m_{2} = 0 … (Given)

∴ k = 0.

(ii) The sum of slopes of the lines given by 2x^{2} + kxy – 3y^{2} = 0 is equal to their product.

Question is modified.

The sum of slopes of the lines given by x^{2} + kxy – 3y^{2} = 0 is equal to their product.

Solution:

Comparing the equation x^{2} + kxy – 3y^{2} = 0, with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = k, b = -3

Let m_{1} and m_{2} be the slopes of the lines represented by x^{2} + kxy – 3y^{2} = 0.

∴ m_{1} + m_{2} = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)

and m_{1}m_{2} = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)

Now, m_{1} + m_{2} = m_{1}m_{2} … (Given)

∴ \(\frac{k}{3}=\frac{-1}{3}\)

∴ k = -1.

(iii) The slope of one of the lines given by 3x^{2} – 4xy + ky^{2} = 0 is 1.

Solution:

The auxiliary equation of the lines given by 3x^{2} – 4xy + ky^{2} = 0 is km^{2} – 4m + 3 = 0.

Given, slope of one of the lines is 1.

∴ m = 1 is the root of the auxiliary equation km^{2} – 4m + 3 = 0.

∴ k(1)^{2} – 4(1) + 3 = 0

∴ k – 4 + 3 = 0

∴ k = 1.

(iv) One of the lines given by 3x^{2} – kxy + 5y^{2} = 0 is perpendicular to the 5x + 3y = 0.

Solution:

The auxiliary equation of the lines represented by 3x^{2} – kxy + 5y^{2} = 0 is 5m^{2} – km + 3 = 0.

Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).

∴ slope of that line = m = \(\frac{3}{5}\)

∴ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5

5m^{2} – km + 3 = 0.

∴ 5\(\left(\frac{3}{5}\right)^{2}\) – k\(\left(\frac{3}{5}\right)\) + 3 = 0

∴ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0

∴ 9 – 3k + 15 = 0

∴ 3k = 24

∴ k = 8.

(v) The slope of one of the lines given by 3x^{2} + 4xy + ky^{2} = 0 is three times the other.

Solution:

3x^{2} + 4xy + ky^{2} = 0

∴ divide by x^{2}

∴ y = mx

∴ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m

put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)

Comparing the equation km^{2} + 4m + 3 = 0 with ax^{2} + 2hxy+ by^{2} = 0, we get,

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

m_{1} = 3m_{2} ..(given condition)

m_{1} + m_{2} = \(\frac{-2 h}{k}=-\frac{4}{k}\)

m_{1}m_{2} = \(\frac{a}{b}=\frac{3}{k}\)

m_{1} + m_{2} = \(-\frac{4}{\mathrm{k}}\)

4m_{2} = \(-\frac{4}{\mathrm{k}}\) …(m_{1} = 3m_{2})

m_{2} = \(-\frac{1}{\mathrm{k}}\)

m_{1}m_{2} = \(\frac{3}{k}\)

\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) …(m_{1} = 3m_{2})

\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) …(m_{2} = \(-\frac{1}{k}\))

\(\frac{1}{k^{2}}=\frac{1}{k}\)

k^{2} = k

k = 1 or k = 0

(vi) The slopes of lines given by kx^{2} + 5xy + y^{2} = 0 differ by 1.

Solution:

Comparing the equation kx^{2} + 5xy +y^{2} = 0 with ax^{2} + 2hxy + by^{2}

a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)

m_{1} + m_{2} = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5

and m_{1}m_{2} = \(\frac{a}{b}=\frac{k}{1}\) = k

the slope of the line differ by (m_{1} – m_{2}) = 1 …(1)

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

(m_{1} – m_{2})^{2} = (-5)^{2} – 4(k)

(m_{1} – m_{2})^{2} = 25 – 4k

1 = 25 – 4k ..[By (1)]

4k = 24

k = 6

(vii) One of the lines given by 6x^{2} + kxy + y^{2} = 0 is 2x + y = 0.

Solution:

The auxiliary equation of the lines represented by 6x^{2} + kxy + y^{2} = 0 is

m^{2} + km + 6 = 0.

Since one of the line is 2x + y = 0 whose slope is m = -2.

∴ m = -2 is the root of the auxiliary equation m^{2} + km + 6 = 0.

∴ (-2)^{2} + k(-2) + 6 = 0

∴ 4 – 2k + 6 = 0

∴ 2k = 10 ∴ k = 5

Question 6.

Find the joint equation of the pair of lines which bisect angle between the lines given by x^{2} + 3xy + 2y^{2} = 0

Solution:

x^{2} + 3xy + 2y^{2} = 0

∴ x^{2} + 2xy + xy + 2y^{2} = 0

∴ x(x + 2y) + y(x + 2y) = 0

∴ (x + 2y)(x + y) = 0

∴ separate equations of the lines represented by x^{2} + 3xy + 2y^{2} = 0 are x + 2y = 0 and x + y = 0.

Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P (x, y) from the line x + 2y = 0

= the distance of P(x, y) from the line x + y = 0

∴ 2(x + 2y)^{2} = 5(x + y)^{2}

∴ 2(x^{2} + 4xy + 4y^{2}) = 5(x^{2} + 2xy + y^{2})

∴ 2x^{2} + 8xy + 8y^{2} = 5x^{2} + 10xy + 5y^{2}

∴ 3x^{2} + 2xy – 3y^{2} = 0.

This is the required joint equation of the lines which bisect the angles between the lines represented by x^{2} + 3xy + 2y^{2} = 0.

Question 7.

Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.

Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the line x = 3.

∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis

∴ slope of OA = tan 30° = 1/\(\sqrt {3}\)

∴ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x

∴ \(\sqrt {3}\)y = x ∴ x – \(\sqrt {3}\)y = 0

Slope of OB = tan 150° = tan (180° – 30°)

= – tan 30°= -1/\(\sqrt {3}\)

∴ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x

∴ \(\sqrt {3}\)y = -x ∴ x + \(\sqrt {3}\)y = 0

∴ required combined equation of the lines is

(x – \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0

i.e. x^{2} – 3y^{2} = 0.

Question 8.

Show that the lines x^{2} – 4xy + y^{2} = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.

Solution:

We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is -1.

Let OA (or OB) has slope m.

∴ its equation is y = mx … (1)

Also, tan 60° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)

∴ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)

Squaring both sides, we get,

3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)

∴ 3(1 – 2m + m^{2}) = m^{2} + 2m + 1

∴ 3 – 6m + 3m^{2} = m^{2} + 2m + 1

∴ 2m^{2} – 8m + 2 = 0

∴ m^{2} – 4m + 1 = 0

∴ \(\left(\frac{y}{x}\right)^{2}\) – 4\(\left(\frac{y}{x}\right)\) + 1 = 0 …[By (1)]

∴ y^{2} – 4xy + x^{2} = 0

∴ x^{2} – 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10

∴ x^{2} – 4xy + y^{2} = 0 and x + y = 10 form a triangle OAB which is equilateral.

Let seg OM ⊥^{r} line AB whose question is x + y = 10

Question 9.

If the slope of one of the lines represented by ax^{2} + 2hxy + by^{2} = 0 is three times the other then prove that 3h^{2} = 4ab.

Solution:

Let m_{1} and m_{2} be the slopes of the lines represented by ax^{2} + 2hxy + by^{2} = 0.

∴ m_{1} + m_{2} = \(-\frac{2 h}{b}\) and m_{1}m_{2} = \(\frac{a}{b}\)

We are given that m_{2} = 3m_{1}

∴ m_{1} + 3m_{1} = \(-\frac{2 h}{b}\) 4m_{1} = \(-\frac{2 h}{b}\)

∴ m_{1} = \(-\frac{h}{2 b}\) …(1)

Also, m_{1}(3m_{1}) = \(\frac{a}{b}\) ∴ 3m_{1}^{2} = \(\frac{a}{b}\)

∴ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) ….[By (1)]

∴ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)

∴ 3h^{2} = 4ab, as b ≠0.

Question 10.

Find the combined equation of the bisectors of the angles between the lines represented by 5x^{2} + 6xy – y^{2} = 0.

Solution:

Comparing the equation 5x^{2} + 6xy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 5, 2h = 6, b = -1

Let m_{1} and m_{2} be the slopes of the lines represented by 5x^{2} + 6xy – y^{2} = 0.

The separate equations of the lines are

y = m_{1}x and y = m_{2}x, where m1 ≠ m2

i.e. m_{1}x – y = 0 and m_{1}x – y = 0.

Let P (x, y) be any point on one of the bisector of the angles between the lines.

∴ the distance of P from the line m_{1}x – y = 0 is equal to the distance of P from the line m_{2}x – y = 0.

∴ (m_{2}^{2} + 1)(m_{1}x – y)^{2} = (m_{1}^{2} + 1)(m_{2}x – y)^{2}

∴ (m_{2}^{2} + 1)(m_{1}^{2}x^{2} – 2m_{1}xy + y^{2}) = (m_{1}^{2} + 1)(m_{2}^{2}x^{2} – 2m_{2}xy + y^{2})

∴ m_{1}^{2}m_{2}^{2}x^{2} – 2m_{1}m_{1}^{2}y^{2}xy + m_{2}^{2}y^{2} + m_{1}^{2}x^{2} – 2m_{1}^{2}xy + y^{2}

= m_{1}^{2}m_{2}^{2}x^{2} – 2m_{1}^{2}m_{2}xy + m_{1}^{2}y^{2} + m_{2}^{2}x^{2} – 2m_{2}xy + y^{2}

∴ (m_{1}^{2} – m_{2}^{2})x^{2} + 2m_{1}m_{2}(m_{1} – m_{2})xy – 2(m_{1} – m_{2})xy – (m_{1}^{2} – m_{2}^{2})y^{2} = 0

Dividing throughout by m_{1} – m_{2} (≠0), we get,

(m_{1} + m_{2})x^{2} + 2m_{1}m_{2}xy – 2xy – (m_{1} + m_{2})y^{2} = 0

∴ 6x^{2} – 10xy – 2xy – 6y^{2} = 0 …[By (1)]

∴ 6x^{2} – 12xy – 6y^{2} = 0

∴ x^{2} – 2xy – y^{2} = 0

This is the joint equation of the bisectors of the angles between the lines represented by 5x^{2} + 6xy – y^{2} = 0.

Question 11.

Find a, if the sum of the slopes of the lines represented by ax^{2} + 8xy + 5y^{2} = 0 is twice their product.

Solution :

Comparing the equation ax^{2} + 8xy + 5y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0,

we get, a = a, 2h = 8, b = 5

Let m_{1} and m_{2} be the slopes of the lines represented by ax^{2} + 8xy + 5y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=-\frac{8}{5}\)

and m_{1}m_{2} = \(\frac{a}{b}=\frac{a}{5}\)

Now, (m_{1} + m_{2}) = 2(m_{1}m_{2})

\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)

a = -4

Question 12.

If the line 4x – 5y = 0 coincides with one of the lines given by ax^{2} + 2hxy + by^{2} = 0, then show that 25a + 40h +16b = 0.

Solution :

The auxiliary equation of the lines represented by ax^{2} + 2hxy + by^{2} = 0 is bm^{2} + 2hm + a = 0

Given that 4x – 5y = 0 is one of the lines represented by ax^{2} + 2hxy + by^{2} = 0.

The slope of the line 4x – 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)

∴ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm^{2} + 2hm + a = 0.

∴ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0

∴ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0

∴ 16b + 40h + 25a = 0 i.e.

∴ 25a + 40h + 16b = 0

Question 13.

Show that the following equations represent a pair of lines. Find the acute angle between them :

(i) 9x^{2} – 6xy + y^{2} + 18x – 6y + 8 = 0

Solution:

Comparing this equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, we get,

a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.

∴ D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|=\left|\begin{array}{rrr}

9 & -3 & 9 \\

-3 & 1 & -3 \\

9 & -3 & 8

\end{array}\right|\)

= 9(8 – 9) + 3(-24 + 27) + 9(9 – 9)

= 9(-1) + 3(3) + 9(0)

= -9 + 9 + 0 = 0

and h^{2} – ab = (-3)^{2} – 9(1) = 9 – 9 = 0

∴ the given equation represents a pair of lines.

Let θ be the acute angle between the lines.

∴ tan θ = tan0°

∴ θ = 0°.

(ii) 2x^{2} + xy – y^{2} + x + 4y – 3 = 0

Solution:

Comparing this equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy+ c = 0, we get,

a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3

= -2 + 1 + 1

= -2 + 2= 0

∴ the given equation represents a pair of lines.

Let θ be the acute angle between the lines.

∴ tan θ = tan 3

∴ θ = tan^{-1}(3)

(iii) (x – 3)^{2} + (x – 3)(y – 4) – 2(y – 4)^{2} = 0.

Solution :

Put x – 3 = X and y – 4 = Y in the given equation, we get,

X^{2} + XY – 2Y^{2} = 0

Comparing this equation with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, h = \(\frac{1}{2}\), b = -2

This is the homogeneous equation of second degreeand h^{2} – ab = \(\left(\frac{1}{2}\right)^{2}\) – 1(-2)

= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0

Hence, it represents a pair of lines passing through the new origin (3, 4).

Let θ be the acute angle between the lines.

∴ tanθ = 3 ∴ θ = tan^{-1}(3)

Question 14.

Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.

Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.

Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.

∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)

∴ equation of the line OA is

y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0

Slope of OB = tan 150° = tan (180° – 30°)

= tan 30° = \(-\frac{1}{\sqrt{3}}\)

∴ equation of the line OB is

y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0

∴ required combined equation is

(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0

i.e. x^{2} – 3y^{2} = 0.

Question 15.

If lines representedby ax^{2} + 2hxy + by^{2} = 0 make angles of equal measures with the co-ordinate

axes then show that a = ± b.

OR

Show that, one of the lines represented by ax^{2} + 2hxy + by^{2} = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.

Solution:

Let OA and OB be the two lines through the origin represented by ax^{2} + 2hxy + by^{2} = 0.

Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and \(\frac{\pi}{2}\) – ∝ with the positive direction of X-axis or ∝ and \(\frac{\pi}{2}\) + ∝ with thepositive direction of X-axis.

∴ slope of the line OA = m_{1} = tan ∝

and slope of the line OB = m_{2}

= tan(\(\frac{\pi}{2}\) – ∝) or tan(\(\frac{\pi}{2}\) + ∝)

i.e. m_{2} = cot ∝ or m_{2} = -cot ∝

∴ m_{1}m_{2} – tan ∝ x cot ∝ = 1

OR m_{1}m_{2} = tan ∝ (-cot ∝) = -1

i.e. m_{1}m_{2} = ± 1

But m_{1}m_{2} = \(\frac{a}{b}\)

∴ \(\frac{a}{b}\)= ±1 ∴ a = ±b

This is the required condition.

Question 16.

Show that the combined equation of a pair of lines through the origin and each making an angle of ∝ with the line x + y = 0 is x^{2} + 2(sec 2∝) xy + y^{2} = 0.

Solution:

Let OA and OB be the required lines.

Let OA (or OB) has slope m.

∴ its equation is y = mx … (1)

It makes an angle ∝ with x + y = 0 whose slope is -1. m +1

∴ tan ∝ = \(\left|\frac{m+1}{1+m(-1)}\right|\)

Squaring both sides, we get,

tan^{2}∝ = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)

∴ tan^{2}∝(1 – 2m + m^{2}) = m^{2} + 2m + 1

∴ tan^{2}∝ – 2m tan^{2}∝ + m^{2}tan^{2}∝ = m^{2} + 2m + 1

∴ (tan^{2}∝ – 1)m^{2} – 2(1 + tan^{2}∝)m + (tan^{2}∝ – 1) = 0

∴ y^{2} + 2xysec2∝ + x^{2} = 0

∴ x^{2} + 2(sec2∝)xy + y^{2} = 0 is the required equation.

Question 17.

Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)^{2} – 3(4x – 3y)^{2} =0 form an equilateral triangle.

Solution:

The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)

Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m_{1} is 60°.

On squaring both sides, we get,

3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)

∴ 3 (4 – 3m)^{2} = (4m + 3)^{2}

∴ 3(16 – 24m + 9m^{2}) = 16m^{2} + 24m + 9

∴ 48 – 72m + 27m^{2} = 16m^{2} + 24m + 9

∴ 11m^{2} – 96m + 39 = 0

This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).

∴ the combined equation of the two lines is

11\(\left(\frac{y}{x}\right)^{2}\) – 96\(\left(\frac{y}{x}\right)\) + 39 = 0

∴ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0

∴ 11y^{2} – 96xy + 39x^{2} = 0

∴ 39x^{2} – 96xy + 11y^{2} = 0.

∴ 39x^{2} – 96xy + 11y^{2} = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.

The equation 39x^{2} – 96xy + 11y^{2} = 0 can be written as :

-39x^{2} + 96xy – 11y^{2} = 0

i.e., (9x^{2} – 48x^{2}) + (24xy + 72xy) + (16y^{2} – 27y^{2}) = 0

i.e. (9x^{2} + 24xy + 16y^{2}) – (48x^{2} – 72xy + 27y^{2}) = 0

i.e. (9x^{2} + 24xy + 16y^{2}) – 3(16x^{2} – 24xy + 9y^{2}) = 0

i.e. (3x + 4y)^{2} – 3(4x – 3y)^{2} = 0

Hence, the line 3x + 4y + 5 = 0 and the lines

(3x + 4y)^{2} – 3(4x – 3y)^{2} form the sides of an equilateral triangle.

Question 18.

Show that lines x^{2} – 4xy + y^{2} = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.

Solution:

x^{2} – 4xy + y^{2} = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.

Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)

In right angled triangle OAM,

sin 60° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) ∴ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)

∴ OA = 2

∴ length of the each side of the equilateral triangle OAB = 2 units.

∴ perimeter of ∆ OAB = 3 × length of each side

= 3 × 2 = 6 units.

Question 19.

If the slope of one of the lines given by ax^{2} + 2hxy + by^{2} = 0 is square of the other then show that a^{2}b + ab^{2} + 8h^{3} = 6abh.

Solution:

Let m be the slope of one of the lines given by ax^{2} + 2hxy + by^{2} = 0.

Then the other line has slope m^{2}

Multiplying by b^{3}, we get,

-8h^{3} = ab^{2} + a^{2}b – 6abh

∴ a^{2}b + ab^{2} + 8h^{3} = 6abh

This is the required condition.

Question 20.

Prove that the product of lengths of perpendiculars drawn from P (x_{1}, y_{1}) to the lines repersented by ax^{2} + 2hxy + by^{2} = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)

Solution:

Let m_{1} and m_{2} be the slopes of the lines represented by ax^{2} + 2hxy + by^{2} = 0.

∴ m^{1} + m^{2} = \(-\frac{2 h}{b}\) and m^{1}m^{2} = \(\frac{a}{b}\) …(1)

The separate equations of the lines represented by

ax^{2} + 2hxy + by^{2} = 0 are

y = m_{1}x and y = m_{2}x

i.e. m_{1}x – y = 0 and m_{2}x – y = 0

Length of perpendicular from P(x_{1}, _{1}) on

Question 21.

Show that the difference between the slopes of lines given by (tan^{2}θ + cos^{2}θ )x^{2} – 2xytanθ + (sin^{2}θ )y^{2} = 0 is two.

Solution:

Comparing the equation (tan^{2}θ + cos^{2}θ)x^{2} – 2xy tan θ + (sin^{2}θ) y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = tan^{2}θ + cos^{2}θ, 2h = -2 tan θ and b = sin2θ

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

Question 22.

Find the condition that the equation ay^{2} + bxy + ex + dy = 0 may represent a pair of lines.

Solution:

Comparing the equation

ay^{2} + bxy + ex + dy = 0 with

Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0, we get,

A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0

The given equation represents a pair of lines,

i.e. if bed – ae^{2} = 0

i.e. if e(bd – ae) = 0

i.e. e = 0 or bd – ae = 0

i.e. e = 0 or bd = ae

This is the required condition.

Question 23.

If the lines given by ax^{2} + 2hxy + by^{2} = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h^{2}.

Solution:

Since the lines ax^{2} + 2hxy + by^{2} = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax^{2} + 2hxy + by^{2} = 0 is 60°.

∴ 3(a + b)^{2} = 4(h^{2} – ab)

∴ 3(a^{2} + 2ab + b^{2}) = 4h^{2} – 4ab

∴ 3a^{2} + 6ab + 3b^{2} + 4ab = 4h^{2}

∴ 3a^{2} + 10ab + 3b^{2} = 4h^{2}

∴ 3a^{2} + 9ab + ab + 3b^{2} = 4h^{2}

∴ 3a(a + 3b) + b(a + 3b) = 4h^{2}

∴ (3a + b)(a + 3b) = 4h^{2}

This is the required condition.

Question 24.

If line x + 2 = 0 coincides with one of the lines represented by the equation x^{2} + 2xy + 4y + k = 0 then show that k = -4.

Solution:

One of the lines represented by

x^{2} + 2xy + 4y + k = 0 … (1)

is x + 2 = 0.

Let the other line represented by (1) be ax + by + c = 0.

∴ their combined equation is (x + 2)(ax + by + c) = 0

∴ ax^{2} + bxy + cx + 2ax + 2by + 2c = 0

∴ ax^{2} + bxy + (2a + c)x + 2by + 2c — 0 … (2)

As the equations (1) and (2) are the combined equations of the same two lines, they are identical.

∴ by comparing their corresponding coefficients, we get,

∴ 1 = \(\frac{-4}{k}\)

∴ k = -4.

Question 25.

Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0

Solution:

Let m_{1} and m_{2} be the slopes of the lines represented by ax^{2} + 2hxy + by^{2} = 0.

Now, required lines are perpendicular to these lines.

∴ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)

Since these lines are passing through the origin, their separate equations are

y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x

i.e. m_{1}y= -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}y)(x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2}\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y^{2} = 0

∴ bx^{2} – 2hxy + ay^{2} = 0.

Question 26.

If equation ax^{2} – y^{2} + 2y + c = 1 represents a pair of perpendicular lines then find a and c.

Solution:

The given equation represents a pair of lines perpendicular to each other.

∴ coefficient of x^{2} + coefficient of y^{2} = 0

∴ a – 1 = 0 ∴ a = 1

With this value of a, the given equation is

x^{2} – y^{2} + 2y + c – 1 = 0

Comparing this equation with

Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0, we get,

A = 1, H = 0, B = -1, G = 0, F = 1, C = c – 1

Since the given equation represents a pair of lines,

D = \(\left|\begin{array}{ccc}

A & H & G \\

H & B & F \\

G & F & C

\end{array}\right|\) = 0

∴ \(\left|\begin{array}{rrr}

1 & 0 & 0 \\

0 & -1 & 1 \\

0 & 1 & c-1

\end{array}\right|\) = 0

∴ 1(-c + 1 – 1) – 0 + 0 = 0

∴ -c = 0

∴ c = 0.

Hence, a = 1, c = 0.