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 = $\pm {\tan }^{-1}\left(\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right)$

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:

$d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Here, $c_1=-2+\sqrt{3}$, $c_2=-2-\sqrt{3}$, $a=\sqrt{2}$, and $b=1$.

So,

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

∴ 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 = $|\frac{m_1-m_2}{1+m_1{m}_2}|$

Calculation: 

Given, the slope of a line L is 2.

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

tan θ = $|\frac{m_1-m_2}{1+m_1{m}_2}|$

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

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

$\frac{1}{\surd 3}$ = $\left|\frac{m_1-2}{1+2m_1}\right|$

$\frac{m_1-2}{1+2m_1}=\frac{1}{\sqrt{3}}$ , $\frac{m_1-2}{1+2m_1}=-\frac{1}{\sqrt{3}}$

$m_1=\frac{-2\sqrt{3}-1}{2-\sqrt{3}}$, $m_1=\frac{2\sqrt{3}-1}{2+\sqrt{3}}$

Similarly, $m_2=\frac{-2\sqrt{3}-1}{2-\sqrt{3}}$, $m_2=\frac{2\sqrt{3}-1}{2+\sqrt{3}}$

Take $m_1=\frac{-2\sqrt{3}-1}{2-\sqrt{3}}$ and $m_2=\frac{2\sqrt{3}-1}{2+\sqrt{3}}$

m₁ + m₂ = $\frac{-2\sqrt{3}-1}{2-\sqrt{3}}$ $+\frac{2\sqrt{3}-1}{2+\sqrt{3}}$

$m_1+m_2=\frac{-4\sqrt{3}-6-2-\sqrt{3}+4\sqrt{3}-6-2+\sqrt{3}}{4-3}$

∴ 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

∴ $\tan \theta =\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|$

= $\frac{2\sqrt{3}}{2}|=\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{3}={60}^{\circ }$

Hence option 1 is correct

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

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 = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

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

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

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

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 = $\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}\right|$

⇒ tan θ = $\left|\frac{\frac{3-1}{\sqrt{3}}}{1+1}\right|=\frac{1}{\sqrt{3}}$

∴ θ = ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ = $\frac{\pi }{6}$

Concept:

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

  θ = ${\tan }^{-1}\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$.

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

  m = $\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$.

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 = $\frac{2-3}{2-1}$ = -1.

Slope of BC = m₂ = $\frac{4-2}{3-2}$ = 2.

Angle B = ${\tan }^{-1}\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|={\tan }^{-1}\left|\frac{2-(-1)}{1+(2)(-1)}\right|={\tan }^{-1}\left|\frac{3}{-1}\right|$ = 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 $\tan \alpha =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

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, $\tan \alpha =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

⇒ $\tan \alpha =\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}\cdot \frac{1}{\sqrt{3}}}\right|$

⇒ $tan\alpha =\frac{1}{\sqrt{3}}$

⇒ 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 $\tan \theta =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

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, $\tan \theta =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

⇒ $\tan \theta =\left|\frac{3-\frac{1}{2}}{1+\frac{1}{2}\cdot 3}\right|=1$

⇒ 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, $\theta =\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}|$, 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 = $\theta =\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}|$

 $$\begin{gathered} =\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}|\frac{3-(-2)}{1+3(-2)}| \\ =\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}1 \end{gathered}$$

Hence, option (1) is correct. 

Concept:

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

$\tan \theta =|{\displaystyle \frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}}|$

Calculations:

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

$\tan \theta =|{\displaystyle \frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}}|$

Given, m1 = 2 - √3 and m2 = 2 + √3

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

⇒ $\tan \theta =|{\displaystyle \frac{-2\sqrt{3}}{2}}|$

⇒ $\tan \theta =|-\sqrt{3}|$

⇒ $\tan \theta =\pm \sqrt{3}$

⇒ $\theta ={\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}$

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 = $\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

$$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 = $\frac{\mathrm{x}}{2}-\frac{3}{2}$

Here. m₂ = $\frac{1}{2}$

Now, 

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

θ = tan^-1($\frac{3}{4}$)

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

Concept:

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

$\tan \theta =\left|\frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

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

$\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{\sqrt{3}}-\frac{6}{\sqrt{3}}$

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

Let θ be the acute angle between the lines.

$\tan \theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

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

$\Rightarrow \tan \theta =\left|\frac{\frac{2}{\sqrt{3}}}{2}\right|$

$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

⇒ θ = 30° 

Concept :

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

$\tan \theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$ 

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 , $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$ 

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

⇒ 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.