Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2

Question 1.

Show that lines represented by 3x^{2} – 4xy – 3y^{2} = 0 are perpendicular to each other.

Solution:

Comparing the equation 3x^{2} – 4 xy – 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x^{2} – 4xy – 3y^{2} = 0 are perpendicular to each other.

Question 2.

Show that lines represented by x^{2} + 6xy + gy^{2}= 0 are coincident.

The question is modified.

Show that lines represented by x^{2} + 6xy + 9y^{2}= 0 are coincident.

Solution:

Comparing the equation x^{2} + 6xy + 9y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = 6, i.e. h = 3 and b = 9

Since h^{2} – ab = (3)^{2} – 1(9)

= 9 – 9 = 0, .

the lines represented by x^{2} + 6xy + 9y^{2} = 0 are coincident.

Question 3.

Find the value of k if lines represented by kx^{2} + 4xy – 4y^{2} = 0 are perpendicular to each other.

Solution:

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

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

Since lines represented by kx^{2} + 4xy – 4y^{2} = 0 are perpendicular to each other,

a + b = 0

∴ k – 4 = 0 ∴ k = 4.

Question 4.

Find the measure of the acute angle between the lines represented by:

(i) 3x^{2} – 4\(\sqrt {3}\)xy + 3y^{2} = 0

Solution:

Comparing the equation 3x^{2} – 4\(\sqrt {3}\)xy + 3y^{2} = 0 with

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

a = 3, 2h = -4\(\sqrt {3}\), i.e. h = -24\(\sqrt {3}\) and b = 3

Let θ be the acute angle between the lines.

∴ θ = 30°.

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

Solution:

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

a = 4, 2h = 5, i.e. h = \(\frac{5}{2}\) and b = 1.

Let θ be the acute angle between the lines.

(iii) 2x^{2} + 7xy + 3y^{2} = 0

Solution:

Comparing the equation

2x^{2} + 7xy + 3y^{2} = 0 with

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

a = 2, 2h = 7 i.e. h = \(\frac{7}{2}\) and b = 3

Let θ be the acute angle between the lines.

tanθ = 1

∴ θ = tan 1 = 45°

∴ θ = 45°

(iv) (a^{2} – 3b^{2})x^{2} + 8abxy + (b^{2} – 3a^{2})y^{2} = 0

Solution:

Comparing the equation

(a^{2} – 3b^{2})x^{2} + 8abxy + (b^{2} – 3a^{2})y^{2} = 0, with

Ax^{2} + 2Hxy + By^{2} = 0, we have,

A = a^{2} – 3b^{2}, H = 4ab, B = b^{2} – 3a^{2}.

∴ H^{2} – AB = 16a^{2}b^{2} – (a^{2} – 3b^{2})(b^{2} – 3a^{2})

= 16a^{2}b^{2} + (a^{2} – 3b^{2})(3a^{2} – b^{2})

= 16a^{2}b^{2} + 3a^{4} – 10a^{2}b^{2} + 3b^{4}

= 3a^{4} + 6a^{2}b^{2} + 3b^{4}

= 3(a^{4} + 2a^{2}b^{2} + b^{4})

= 3 (a^{2} + b^{2})^{2}

∴ \(\sqrt{H^{2}-A B}\) = \(\sqrt {3}\) (a^{2} + b^{2})

Also, A + B = (a^{2} – 3b^{2}) + (b^{2} – 3a^{2})

= -2 (a^{2} + b^{2})

If θ is the acute angle between the lines, then

tan θ = \(\left|\frac{2 \sqrt{H^{2}-A B}}{A+B}\right|=\left|\frac{2 \sqrt{3}\left(a^{2}+b^{2}\right)}{-2\left(a^{2}+b^{2}\right)}\right|\)

= \(\sqrt {3}\) = tan 60°

∴ θ = 60°

Question 5.

Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0

Solution:

The slope of the line 3x + 2y – 11 = 0 is m_{1} = \(-\frac{3}{2}\) .

Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.

The angle between the lines having slopes m and m1 is 30°.

On squaring both sides, we get,

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

∴ (2 – 3m)^{2} = 3 (2m + 3)^{2}

∴ 4 – 12m + 9m^{2} = 3(4m^{2} + 12m + 9)

∴ 4 – 12m + 9m^{2} = 12m^{2} + 36m + 27

3m^{2} + 48m + 23 = 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

3\(\left(\frac{y}{x}\right)^{2}\) + 48\(\left(\frac{y}{x}\right)\) + 23 = 0

∴ \(\frac{3 y^{2}}{x^{2}}+\frac{48 y}{x}\) + 23 = 0

∴ 3y^{2} + 48xy + 23x^{2} = 0

∴ 23x^{2} + 48xy + 3y^{2} = 0.

Question 6.

If the angle between lines represented by ax^{2} + 2hxy + by^{2} = 0 is equal to the angle between lines represented by 2x^{2} – 5xy + 3y^{2} = 0 then show that 100(h^{2} – ab) = (a + b)^{2}.

Solution:

The acute angle θ between the lines ax^{2} + 2hxy + by^{2} = 0 is given by

tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\) ..(1)

Comparing the equation 2x^{2} – 5xy + 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 2, 2h= -5, i.e. h = \(-\frac{5}{2}\) and b = 3

Let ∝ be the acute angle between the lines 2x^{2} – 5xy + 3y^{2} = 0.

This is the required condition.

Question 7.

Find the combined equation of lines passing through the origin and each of which making angle 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.