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

  • Last Updated : 28 Mar, 2021

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 formulam =\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 is2\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 angle2\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} andm_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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