Angle between Lines MCQ Quiz - Objective Question with Answer for Angle between Lines - Download Free PDF

Explanation:
If θ be the angle between the lines y = m1x + c1 and y = m2x + c₂ then θ = \(\rm \pm \tan^{-1} \left(\frac{m_1 - m_2}{1+m_1m_2} \right )\)

Additional Information

When two lines are perpendicular, the product of their slopes = m₁m₂ = -1

When two lines are parallel then m₁ = m₂

Calculation: 

Given equation of the second degree is:

\(x^{2} + 2\sqrt{2}xy + 2y^{2} + 4x + 4\sqrt{2}y + 1 = 0\)

This equation can be represented as a pair of straight lines. To determine the distance between the lines, we need to simplify and rewrite the equation in a recognizable form.

First, rewrite the equation as:

\( (x^{2} + 2\sqrt{2}xy + 2y^{2}) + (4x + 4\sqrt{2}y + 1) = 0\)

Now, we need to complete the square for both \(x\) and \(y\) terms:

\( x^{2} + 2\sqrt{2}xy + 2y^{2} \) can be written as \( (\sqrt{2}x + y)^{2} \)

So, the equation becomes:

\( (\sqrt{2}x + y)^{2} + 4x + 4\sqrt{2}y + 1 = 0\)

Rewriting it:

\( (\sqrt{2}x + y)^{2} + 4(\sqrt{2}x + y) = 0\)

Let \(z = \sqrt{2}x + y\), then the equation reduces to:

\( z^{2} + 4z + 1 = 0\)

Solving for \(z\):

\( z = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}\)

So, we have two lines:

\( \sqrt{2}x + y = -2 + \sqrt{3}\)

\( \sqrt{2}x + y = -2 - \sqrt{3}\)

To find the distance \(d\) between these two lines, we use the formula:

\( d = \frac{ \left| c_{1} - c_{2} \right| }{\sqrt{a^{2} + b^{2}}} \)

Here, \( c_{1} = -2 + \sqrt{3} \), \( c_{2} = -2 - \sqrt{3} \), \(a = \sqrt{2}\), and \(b = 1\).

So,

\( d = \frac{ \left| (-2 + \sqrt{3}) - (-2 - \sqrt{3}) \right| }{\sqrt{ (\sqrt{2})^{2} + 1^{2} }} = \frac{ \left| 2\sqrt{3} \right| }{\sqrt{2 + 1}} = \frac{ 2\sqrt{3} }{\sqrt{3}} = 2 \)

∴ The correct answer is option 3.

Concept:

Angle Between Two Lines in 3D:

• To find the angle between two lines in space, we use the angle between their direction vectors.
• If direction vectors are a and b, then angle θ between them is given by:
• cosθ = (a · b) / (|a| × |b|)
• Here, a · b is the dot product of vectors, and |a| is the magnitude of vector a.

Dot Product:

• The dot product of vectors a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k is:
• a · b = a₁b₁ + a₂b₂ + a₃b₃

Calculation:

Given, direction vector of first line = 4i + 6j + 12k

Let a = 4i + 6j + 12k

Direction vector of second line = 5i + 8j − 4k

Let b = 5i + 8j − 4k

⇒ a · b = (4)(5) + (6)(8) + (12)(−4)

⇒ a · b = 20 + 48 − 48 = 20

⇒ |a| = √(4² + 6² + 12²) = √(16 + 36 + 144) = √196 = 14

⇒ |b| = √(5² + 8² + (−4)²) = √(25 + 64 + 16) = √105

⇒ cosθ = (a · b) / (|a| × |b|) = 20 / (14 × √105)

⇒ cosθ = 2 / (√105 / 7)

⇒ θ = cos⁻¹(20 / (14√105))

⇒ θ =  cos−1(10 / (7√105))

∴ Hence, Option 1 is the correct answer.

Concept: 

The slope of a line of the form y = mx + c is 

m = tan θ and θ = tan^-1(m)

The angle between the two lines with slopes m₁ and m₂ is,

tan θ = \(\Big|\frac{m_1-m_2}{1+m_1m_2}\Big|\)

Calculation: 

Given, the slope of a line L is 2.

The angle between the two lines with slopes m1 and m2 is,

tan θ = \(\Big|\frac{m_1-m_2}{1+m_1m_2}\Big|\)

Given that the lines with slopes m₁ and m₂ are inclined at an angle \(\frac{\pi}{6}\) with L.

∴ tan \(\frac{\pi}{6}\) = \(\left| \frac{ m_1- 2}{1+2m_1} \right|\)

\(\frac{1}{√{3}}\) = \(\left| \frac{m_1-2}{1+2m_1} \right|\)

\(\frac{m_1-2}{1+2m_1}=\frac{1}{\sqrt3}\) , \(\frac{m_1-2}{1+2m_1}=-\frac{1}{\sqrt3}\)

\(m_1=\frac{-2\sqrt3- 1}{2-\sqrt3}\), \(m_1=\frac{2\sqrt3- 1}{2+\sqrt3}\)

Similarly, \(m_2=\frac{-2\sqrt3- 1}{2-\sqrt3}\), \(m_2=\frac{2\sqrt3- 1}{2+\sqrt3}\)

Take \(m_1=\frac{-2\sqrt3- 1}{2-\sqrt3}\) and \(m_2=\frac{2\sqrt3- 1}{2+\sqrt3}\)

m₁ + m₂ = \(\frac{-2\sqrt3- 1}{2-\sqrt3}\) \(+\frac{2\sqrt3- 1}{2+\sqrt3}\)

\(m_1+m_2=\frac{-4\sqrt 3-6-2-\sqrt3+4\sqrt3-6-2+\sqrt3}{4-3}\)

∴ m1 + m2 = - 16

The correct answer is option (4).

Calculation

The given lines are:

y = (2 - √3)x + 5 and y = (2 + √3)x - 7

Therefore, slope of first line = m₁ = 2 - √3 and slope of second line = m₂ = 2 + √3

∴ \(\tan \theta=\left|\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|\)

= \(\frac{2 \sqrt{3}}{2} \left\lvert\,=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}\right.\)

Hence option 1 is correct

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = \(\rm \left|\frac{m_1 - m_2}{1+m_1m_2} \right |\)

Calculation:

Given lines are y = x + 4 and y =  2x - 3

Let slope of 1st and 2nd line are m₁ and m₂ respectively,

Therefore, m₁ = 1 and m₂ = 2

As we know, tan θ = \(\rm \left|\frac{m_1 - m_2}{1+m_1m_2} \right |\)

⇒ tan θ = \(\rm \left|\frac{1 - 2}{1+1 \times 2} \right | = \frac 1 3\)

∴ θ = \(\rm \tan^{-1} \left(\frac 1 3 \right)\)

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = \(\rm \left|\frac{m_1 - m_2}{1+m_1m_2} \right |\)

Calculation:

Let slope of 1st and 2nd line are m1 and m2 respectively,

Therefore, m1 = \(\sqrt 3\) and m2 = \(\frac {1}{\sqrt 3}\)

As we know, tan θ = \(\rm \left|\frac{\sqrt 3 - \frac {1}{\sqrt 3}}{1+\sqrt 3 \times \frac {1}{\sqrt 3}} \right |\)

⇒ tan θ = \(\rm \left|\frac{\frac{3 - 1}{\sqrt 3}}{1+1} \right | = \frac {1} {\sqrt3}\)

∴ θ = \(\rm \tan^{-1} \left(\frac 1 {\sqrt3} \right)\) = \(\rm \frac {\pi}{6}\)

Concept:

• The angle θ between the two lines y = m₁x + c₁ and y = m₂x + c₂, is given by:

  θ = \(\rm\tan^{-1}\left|m_2\ -\ m_1\over1\ +\ m_1m_2\right|\).

• The slope (m) of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by:

  m = \(\rm y_2\ -\ y_1\over x_2\ -\ x_1\).

Calculation:

Given that A = (1, 3), B = (2, 2) and C = (3, 4).

The angle B in the triangle ABC is the angle between the lines BA and BC.

Slope of BA = m₁ = \(\rm 2\ -\ 3\over 2\ -\ 1\) = -1.

Slope of BC = m₂ = \(\rm 4\ -\ 2\over 3\ -\ 2\) = 2.

Angle B = \(\rm\tan^{-1}\left|m_2\ -\ m_1\over1\ +\ m_1m_2\right|=\tan^{-1}\left|2\ -\ (-1) \over1\ +\ (2)(-1)\right|\ =\ \tan^{-1}\left|3\over-1\right|\) = tan^-1 3.

CONCEPT:

If α is the acute angle between two non-vertical and non-perpendicular lines L₁ and L₂ with slopes m₁ and m₂ respectively then \(\tan α = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot {m_2}}}} \right|\)

CALCULATION:

Here, we have to find the angle between the lines whose slopes are \(\sqrt 3 \ and \frac{1}{\sqrt 3}\)

Let \(m_1 = \sqrt 3 \ and \ m_2 = \frac{1}{\sqrt 3}\)

As we know that, \(\tan α = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot {m_2}}}} \right|\)

⇒ \(\tan α = \left| {\frac{{{\frac{1}{\sqrt 3}} - {\sqrt 3}}}{{1 + {\sqrt 3} \cdot {\frac{1}{\sqrt 3}}}}} \right| \)

⇒ \(tan \ α = \frac{1}{\sqrt3}\)

⇒ α = 30°

So, the angle between the lines whose slopes are \(\sqrt 3 \ and \frac{1}{\sqrt 3}\) is 45° 

Hence, option C is the correct answer.

CONCEPT:

If θ is the acute angle between two non-vertical and non-perpendicular lines L1 and L2 with slopes m1 and m2 respectively then \(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot {m_2}}}} \right|\)

CALCULATION:

Here, we have to find the angle between the lines whose slopes are 1/2 and 3

Let m₁ = 1/2 and m₂ = 3

As we know that, \(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot {m_2}}}} \right|\)

⇒ \(\tan θ = \left| {\frac{{{3} - {\frac{1}{2}}}}{{1 + {\frac{1}{2}} \cdot {3}}}} \right| = 1\)

⇒ θ = 45°

So, the angle between the lines whose slopes are 1/2 and 3 is 45° 

Hence, option D is the correct answer.

Concept:

Angle between two lines is given by, \(\rm \theta =tan^{-1}|\frac{m_2-m_1}{1+m_1m_2}|\), m₁ and m₂ are slopes of lines.

Equatin of straight line: y = mx + c, where m =slope

Calculation:

Here, the pair of straight lines are 2x + y = 1 and 3x - y = 2

2x + y = 1 ⇒y = -2x + 1 so, m₁ = -2

And, 3x - y = 2 ⇒y = 3x - 2 so, m₂ = 3

So, angle between given pair of straight lines = \(\rm \theta =tan^{-1}|\frac{m_2-m_1}{1+m_1m_2}|\)

 \(\rm =tan^{-1}|\frac{3-(-2)}{1+3(-2)}|\\ =tan^{-1} 1\)

Hence, option (1) is correct. 

Concept:

The angle between the lines whose slopes are m1 and m2 be given by,

\(\rm\tan\;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|\)

Calculations:

The angle between the lines whose slopes are m1 and m2 be given by,

\(\rm\tan\;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|\)

Given, m1 = 2 - √3 and m2 = 2 + √3

\(\rm\tan\;\theta = |\dfrac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}|\)

⇒ \(\rm\tan\;\theta = |\dfrac{-2\sqrt3}{2}|\)

⇒ \(\rm\tan\;\theta = |-\sqrt3|\)

⇒ \(\rm\tan\;\theta =\pm \sqrt3\)

⇒ \(\rm\theta = \dfrac{\pi}{3}, \dfrac {2\pi} {3}\)

Concept:

The angle between two lines 

If θ is the angle between two intersecting lines defined by y = m₁x + c₁ and y = m₂x + c₂, then, the angle θ is given by

tanθ = \(\rm \left | \frac{m_{2} - m_{1}}{ 1 + m_{1}m_{2} } \right |\)

\(\)Calculation:

Given lines are

2x - y = 3      ....(1)

and x - 2y = 3      ...(2)

From equation (1),

2x - y = 3 

⇒ y = 2x - 3

Here, m₁ = 2

From equation (2),

x - 2y = 3 

2y = x - 3

y = \(\rm \frac{x}{2} - \frac{3}{2}\)

Here. m₂ = \(\rm \frac{1}{2}\)

Now, 

tan θ = \(\rm \left | \frac{\frac{1}{2} - 2}{ 1 + \frac{1}{2}. 2} \right |\) = \(\rm \frac{3}{4}\)

θ = tan^-1(\(\rm \frac{3}{4}\))

The angle between the lines 2x - y = 3 and x - 2y = 3 is tan-1(\(\rm \frac{3}{4}\))

Concept:

Angle between two lines: The angle θ between the lines having slope m₁ and m₂ is given by

\(\tan {\rm{\theta }} = {\rm{\;}}\left| {\frac{{{{\rm{m}}_2} - {{\rm{m}}_1}}}{{1 + {\rm{\;}}{{\rm{m}}_1}{\rm{\;}}{{\rm{m}}_2}}}} \right|\)

Calculation:

Given: y - √3x – 5 = 0 & √3y – x + 6 = 0

y - √3x – 5 = 0

⇒ y = √3x + 5

So, slope of line, m₁ = √3

√3y – x + 6 = 0

\(\Rightarrow {\rm{y}} = \frac{{\rm{x}}}{{\sqrt 3 }} - \frac{6}{{\sqrt 3 }}\)

So, slope of the line, m₂ = \(\frac{1}{{\sqrt 3 }}\)

Let θ be the acute angle between the lines.

\(\tan {\rm{\theta }} = \left| {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}{{\rm{m}}_2}}}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\sqrt 3 - \frac{1}{{\sqrt 3 }}}}{{1 + \sqrt 3 \times \frac{1}{{\sqrt 3 }}}}} \right|\)

\( \Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\frac{2}{{\sqrt 3 }}}}{2}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \frac{1}{{\sqrt 3 }}\)

⇒ θ = 30° 

Concept :

The angle θ between the lines having slope m₁ and m₂ is given by, 

\(\rm \tan θ = \left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |\) 

Calculation :

We have given equation of lines are,  y - 3x + 2 = 0 and 9x = 3y + 7 

⇒ y = 3x - 2 

⇒ m₁ = 3            ____( i ) 

and , 3y = 9x - 7 

⇒ y = 3x - 7/3 

⇒ m₂ = 3           ____( ii ) 

We know that , \(\rm tanθ = \left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |\) 

⇒ tan θ = \(\rm\left | \frac{3-3}{1+9} \right |\)

⇒ tan θ = 0 

⇒ θ = 0 .

The correct option is 3.

Additional Information

The slopes of parallel lines are equal.

If lines are parallel then angle θ between the lines is zero.