Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.13

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Question 1: Find the angles between each of the following pairs of straight lines.

(i) 3x+y+12=0 and x+2y-1=0

Solution:

Given equations of lines are,3x + y + 12 = 0, x + 2y -1 = 0

Letm₁ andm₂ be the slopes of these lines respectively.

By y = mx +c, we getm₁=-3 and m₂=-1/2

Let θ be the angle between the two lines,

By using formula $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

$\tan\theta=|\frac{(-3+\frac{1}{2})}{1+\frac{3}{2}}|$

⇒ $|\frac{-5/2}{5/2}|$

⇒ 1

Therefore, $\theta=\tan^{-1}(1)$

The angles between the two lines is 45°.

(ii) 3x-y+5 = 0 and x-3y+1 = 0

Solution:

Given equations of lines are 3x – y + 5 = 0, x – 3y +1 = 0

Let m₁ and m₂ be the slopes of these lines respectively.

By y = mx +c, we get m₁=3 and m₂=1/3

Let θ be the angle between the two lines,

We know that, $\tan\theta=|\frac{m_1-m_2}{1+m_1m_2}|$

$\tan\theta=|\frac{(3-\frac{1}{3})}{1+1}|$

⇒ $\frac{4}{3}$

Therefore, $\theta=\tan^{-1}(\frac{4}{3})$

The angle between the two lines is $\tan^{-1}(\frac{4}{3})$

(iii) 3x+4y -7 = 0 and 4x-3y+5 = 0

Solution:

Given equations of lines are 3x + 4y – 7 = 0, 4x – 3y+5 = 0

Letm₁ andm₂ be the slopes of these lines respectively.

By y= mx +c, we get m₁= $\frac{-3}{4}$ and m₂= $\frac{4}{3}$

Here, if we carefully observe, m₁m₂= -1, which means

From the formula, $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$ denominator will become 0,

Therefore, $\tan\theta=\infty$ , $\theta=\tan^{-1}(\infty)$

The angle between the two lines is 90°.

(iv) x-4y = 3 and 6x-y = 11

Solution:

Given equations of lines are x – 4y =3, 6x – y =11

Letm₁ andm₂ be the slopes of these lines respectively.

By y = mx +c, we get, m₁=1/4 and m₂=6

Let θ be the angle between the two lines,

We know that, $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{\frac{1}{4}-6}{1+\frac{3}{2}}|$

⇒ $|\frac{\frac{-23}{4}}{\frac{5}{2}}|$

⇒ $\frac{23}{10}$

Therefore, $\theta=\tan^{-1}(\frac{23}{10})$

The angle between the two lines is $\tan^{-1}(\frac{23}{10})$

(v) (m²-mn)y = (mn+n²)x + n³ and (mn+m²)y = (mn-n²)x + m³

Solution:

Given two lines, letm₁ andm₂ be the slopes of these lines.

By y = mx +c, we getm₁= $\frac{mn+m^2}{m^2-mn}$ and m₂= $\frac{mn-n^2}{m^2+mn}$

Let θ be the angle between two lines,

We know that $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{\frac{mn+n^2}{m^2-mn}-\frac{mn-n^2}{m^2+mn}}{1+\frac{(mn+n^2)(mn-n^2)}{{(m^2-nm)(m^2+mn)}}}|$

⇒ $|\frac{(m^3n+m^2n^2+m^2n^2+m^3n-mn^3+m^2n^2+m^2n^2-mn^3)}{(m^4-n^4)}|$

⇒ $\frac{4m^2n^2}{m^4-n^4}$

Therefore, Angle between two lines is $\tan^{-1}(\frac{4m^2n^2}{m^4-n^4})$ .

Question 2: Find the acute angle between the lines 2x-y+3 = 0 and x+y+2 = 0

Solution:

Letm₁ andm₂ be the slopes of these two lines

By y = mx +c, we get m₁=2 and m₂=-1

Let θ be the angle between the two lines,

We know that, $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{2-(-1)}{(1-2)}|$

⇒ $|\frac{3}{(-1)}|$

Here we need acute angle, $\tan\theta$ is positive if the angle is acute and negative if obtuse.

Therefore, $\theta=\tan^{-1}(3)$ .

The acute angle between the two lines is $\tan^{-1}(3)$ .

Question 3: Prove that points (2, -1), (0, 2), (2, 3), and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Solution:

Let the given points are A = (2,-1), B = (0, 2), C = (2, 3) and D = (4, 0) are coordinates of a parallelogram.

For these points to form a parallelogram, it is must that any pair of two lines formed by these points are parallel to each other.

So, Now lets find the slopes of lines AB, BC, CD, DA using formula $m =\frac{y_2 - y_1}{x_2-x_1}$

Slope of line $AB = \frac{-3}{2}$

Slope of line $BC =\frac{1}{2}$

Slope of line $CD = \frac{-3}{2}$

Slope of line $DA = \frac{1}{2}$

Since the lines AB parallel to CD and BC parallel to DA, the points form a parallelogram.

Now, Angle between the diagonals of parallelogram = Angle between the lines AC and BD.

Letm₁ andm₂ be the slopes of these lines,

From the figure, the angle between diagonals $∅ = \tan^{-1}(\frac{-1}{2})-90°$

We know that $\tan^{-1}\theta+\tan^{-1}-\theta=\pi$

⇒ $∅=180°-\tan^{-1}(\frac{1}{2})-90°=90°-\tan^{-1}(\frac{1}{2})$

Therefore, The angle between the diagonals is $\frac{\pi}{2}-\tan^{-1}(\frac{1}{2})$

Question 4: Find the angles between the line joining the points (2, 0), (0, 3), and the line x + y = 1.

Solution:

Let slope of line joining the points (2, 0) and (0, 3) is m₁= -3/2

slope of line m₂ =-1

Let θ be the angle between the two lines,

We know that, $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{\frac{-3}{2}+1}{1+\frac{3}{2}}|$

⇒ $\frac{1}{5}$

Therefore, the acute angle between the line and the line joining the points is $\tan^{-1}(\frac{1}{5})$

Question 5: If θ is the angle which the straight line joining the points (x1,y1)and (x2,y2) subtends at origin, prove that $\tan\theta=\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}$ and $\cos\theta=\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^1)(x_2^2+y_2^2)}}$

Solution:

Let the points A = (x₁, y₁), B = (x₂, y₂) and origin O = (0, 0)

Slopes of lines joining OA and OB are m₁= y₁/x₁ and m₂ = y₂/x₂

Let θ be the angle between the lines OA and OB.

We know that, $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{(\frac{y_1}{x_1}-\frac{y_2}{x_2})}{(1+\frac{y_1y_2}{x_1x_2}}|$

Therefore, $\tan\theta=\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}$

By the formula $\sec^2\theta-\tan^2\theta=1$ ,we get $\cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}$

Substituting Tanθ from above equation, we get,

$\cos\theta=\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^1)(x_2^2+y_2^2)}}$

Therefore, hence proved.

Question 6: Prove that the straight lines(a+ b)x+(a – b)y=2ab, (a – b)x+(a + b)y=2ab and x + y=0 form an isosceles triangle whose vertical angle is $2\tan^{-1}(\frac{a}{b})$

Solution:

Let m1, m2, m3 be the slopes of given lines respectively.

m₁= $\frac{-(a+b)}{a-b}$ , m₂= $\frac{-(a-b)}{a+b}$ and m₃= -1

Let θ₁, θ₂, θ₃ be the angles between the lines

Now, $\tan\theta_1$ = $|\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{(a-b)^2-(a+b)^2}{2(a^2-b^2)}|$

⇒ $|\frac{2ab}{a^2-b^2}|$ ⇒ $|\frac{2\frac{a}{b}}{1-(\frac{a}{b})^2}|$

We know that $\tan2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$ , using this above equation

⇒ $\theta_1=2\tan^{-1}(\frac{a}{b})$

$\tan\theta_2$ = $|\frac{m_2 - m_3}{1 + m_2m_3}|$

⇒ $\frac{(-a + b +a + b)}{2a}⇒\frac{a}{b}$

⇒ $\theta_2=\tan^{-1}(\frac{b}{a})$

$\tan\theta_3$ = $|\frac{m_3 - m_1}{1 + m_3m_1}|$

⇒ $\frac{(-a + b +a + b)}{2a}⇒\frac{a}{b}$

⇒ $\theta_3=\tan^{-1}(\frac{b}{a})$

Here, angle θ₂ and θ₃ are equal, and θ₁ is the vertical angle

Therefore, the given lines forms Isosceles triangle with vertical angle $2\tan^{-1}(\frac{a}{b})$

Question 7: Find the angle between the lines x = a, by + c = 0

Solution:

The given lines are in the form of x = constant and y=constant respectively

where x=c and y= -c/b

x = c line is parallel to y-axis as there is no y-coefficient

and $y=\frac{-c}{b}$ is parallel to x-axis as there is no x-coefficient

Therefore, the Angle between the two lines is 90°

Question 8: Find the tangent of the angle between the lines which have intercepts 3, 4, and 1, 8 on the axes respectively.

Solution:

Equation of line which have intercepts a, b on x and y-axis is $\frac{x}{a}+\frac{y}{b}=1$

Therefore, the line with intercepts 3,4 is $\frac{x}{3}+\frac{y}{4}=1$

and the line with intercepts 1, 8 is $\frac{x}{1}+\frac{y}{8}=1$

letm₁ and m₂ be the slopes of these lines,

m₁= $\frac{-4}{3}$ and m₂= -8

Now, let θ be the angle between the lines,

$\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{\frac{-4}{3}+8}{1+\frac{32}{3}}|$

⇒ $\frac{20}{35}$ ⇒ $\frac{4}{7}$

Therefore, The tangent of angle between the lines is 4/7

Question 9: Show that line a²x+ay+1 = 0 is perpendicular to line x-ay = 1

Solution:

Letm₁ and m₂ be the slopes of given lines,m₁=-a andm₂=1/a

Here, if we carefully observe,m₁m₂=-1 ,which means

From the formula $\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$ , denominator will become 0,

Therefore, $\tan\theta = \infty$ , $\theta = \tan^{-1}(\infty)\ ⇒ 90°$ .

The angle between the two lines is 90°, and they are perpendicular to each other.

Therefore, Hence proved.

Question 10: Show that tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is $\frac{2ab}{a^2-b^2}$ .

Solution:

Given lines, $\frac{x}{a} + \frac{y}{b} = 1$ ⇒ $bx + ay = ab$

$\frac{x}{a} - \frac{y}{b} = 1$ ⇒ $bx-ay = ab$

Let slopes of these lines arem₁ andm₂ respectively.

$m_1=\frac{-b}{a}$ and $m_2=\frac{b}{a}$

Now, let θ be the angle between the lines,

$\tan\theta = |\frac{m_1-m_2}{1+m_1m_2}|$

⇒ $|\frac{\frac{-b}{a}-\frac{b}{a}}{1 + \frac{b^2}{a^2}}|$

⇒ $| \frac{2ab}{{a^2}-{b^2}}|$

Therefore, Tangent of angle between the lines is $\frac{2ab}{a^2-b^2}$

Hence, proved.

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Last Updated : 28 Mar, 2021

Class 11 RD Sharma Solutions - Chapter 23 The Straight Lines- Exercise 23.12 | Set 3

Class 11 RD Sharma Solutions - Chapter 23 The Straight Lines- Exercise 23.14

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.13

Question 1: Find the angles between each of the following pairs of straight lines.

(i) 3x+y+12=0 and x+2y-1=0

(ii) 3x-y+5 = 0 and x-3y+1 = 0

(iii) 3x+4y -7 = 0 and 4x-3y+5 = 0

(iv) x-4y = 3 and 6x-y = 11

(v) (m2-mn)y = (mn+n2)x + n3 and (mn+m2)y = (mn-n2)x + m3

Question 2: Find the acute angle between the lines 2x-y+3 = 0 and x+y+2 = 0

Question 3: Prove that points (2, -1), (0, 2), (2, 3), and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Question 4: Find the angles between the line joining the points (2, 0), (0, 3), and the line x + y = 1.

Question 5: If θ is the angle which the straight line joining the points (x1,y1)and (x2,y2) subtends at origin, prove thatand

Question 6: Prove that the straight lines(a+ b)x+(a – b)y=2ab, (a – b)x+(a + b)y=2ab and x + y=0 form an isosceles triangle whose vertical angle is

Question 7: Find the angle between the lines x = a, by + c = 0

Question 8: Find the tangent of the angle between the lines which have intercepts 3, 4, and 1, 8 on the axes respectively.

Question 9: Show that line a2x+ay+1 = 0 is perpendicular to line x-ay = 1

Question 10: Show that tangent of an angle between the linesand is.

(v) (m²-mn)y = (mn+n²)x + n³ and (mn+m²)y = (mn-n²)x + m³

Question 5: If θ is the angle which the straight line joining the points (x1,y1)and (x2,y2) subtends at origin, prove that $\tan\theta=\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}$ and $\cos\theta=\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^1)(x_2^2+y_2^2)}}$

Question 6: Prove that the straight lines(a+ b)x+(a – b)y=2ab, (a – b)x+(a + b)y=2ab and x + y=0 form an isosceles triangle whose vertical angle is $2\tan^{-1}(\frac{a}{b})$

Question 9: Show that line a²x+ay+1 = 0 is perpendicular to line x-ay = 1

Question 10: Show that tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is $\frac{2ab}{a^2-b^2}$ .