Angle between Two Straight Lines

Angle between Two Straight Lines – Definition, Formula, Derivation, and Examples

The angle is nothing but the figure formed by two rays. If two straight lines meet, then they form two sets of angles. The intersection forms a pair of acute angles and another pair of obtuse angles. The angle values will be based on the slopes of the intersecting lines. Check out the formula to calculate the angle between two straight lines, derivation, example questions with answers in the following sections of this page.

Angle between Two Lines – Definition

In a plane when two straight and non-parallel lines meet at a point, then it forms two opposite vertical angles. In the formed angles, one is lesser than 90 degrees and the other is greater than 90 degrees. We will find the angle between two straight and perpendicular lines is 90 degrees and parallel lines is zero degrees.

Angle between Two Straight Lines Formula and Derivation

Let us consider θ as the angle between two intersecting straight lines. And those straight lines be y = mx + c, Y = MX + C, then the angle θ is given by

tan θ = ± \(\frac { (M – m) }{ (1 + mM)} \)

Derivation

Angle Between Two Straight Lines

Two straight lines L₁, L₂ are intersecting each other to form acute and obtuse angles.

Let us take the slope measurement can be taken as

tan θ₁ = m₁ and tan θ₂ = m₂

From the figure, we can say that θ = θ₂ – θ₁

Now, tan θ = tan(θ₂ – θ₁)

tan θ = \(\frac { (tan θ₂ – tan θ₁) }{ (1 + tan θ₁ tan θ₂) } \)

Substitute tan θ₁ = m₁, tan θ₂ = m₂

tan θ = \(\frac { (m₂ – m₁) }{ (1 + m₁ m₂) } \)

How to find Angle Between Two Straight Lines?

If three points on a coordinate plane are given, then endpoints of a line are (x₁, y₁) and (x₂, y₂)

The equation of the slope is m = \(\frac { (y₂ – y₁) }{ (x₂ – x₁) } \)

m₁ and m₂ can be calculated by substituting in the formula, then the angle between two lines is given by

tan θ = ± \(\frac { m₂ – m₁) }{ (1 + m₁ m₂) } \)

Also Check:

Angle between Two Straight Lines Examples

Example 1:

If A (-2, 1), B (2, 3), and C (-2, -4) are three points, find the angle between two straight lines AB, BC.

Solution:

Given that,

Three points are A (-2, 1), B (2, 3), and C (-2, -4)

The slope of line AB is m = \(\frac { (y₂ – y₁) }{ (x₂ – x₁) } \)

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

= \(\frac { (2) }{ (2 + 2) } \)

= \(\frac { 2 }{ 4 } \)

= \(\frac { 1 }{ 2 } \)

Therefore, m₁ = \(\frac { 1 }{ 2 } \)

The slope of line BC is given by

m = \(\frac { (y₂ – y₁) }{ (x₂ – x₁) } \)

m = \(\frac { (-4 – 3) }{ (-2 – 2) } \)

= \(\frac { -7 }{ -4 } \)

= \(\frac { 7 }{ 4 } \)

Therefore, m₂ = \(\frac { 7 }{ 4 } \)

Substituting the values of m2 and m1 in the formula for the angle between two lines when we know the slopes of two sides, we have,

tan θ = ± \(\frac { (m₂ – m₁) }{ (1 + m₁ m₂) } \)

= ± \(\frac { (\frac { 7 }{ 4 } – \frac { 1 }{ 2 } ) }{ (1 + \frac { 1 }{ 2 }  * \frac { 7 }{ 4 }) } \)

= ± \(\frac { 2 }{ 3 } \)

Therefore,  θ = tan -1 (⅔)

So, the angle between the lines AB, BC is tan -1 (⅔).

Example 2:

Find the angle between the following lines 4x – 3y = 8, 2x + 5y = 4.

Solution:

Given two straight lines are 4x – 3y = 8, 2x + 5y = 4

Converting the given lines into slope intercept form

4x – 3y = 8

4x = 8 + 3y

3y = 4x – 8

y = \(\frac { 4x – 8 }{ 3 } \)

y = \(\frac { 4x }{ 3 } – \frac { 8 }{ 3 } \)

Therefore, the slope of line 4x – 3y = 8 is \(\frac { 4 }{ 3 } \)

2x + 5y = 4

5y = 4 – 2x

y = \(\frac { 4 – 2x }{ 5 } \)

y = \(\frac { -2x }{ 5 } + \frac { 4 }{ 5 } \)

Therefore, the slope of the line 2x + 5y = 4 is –\(\frac { 2 }{ 5 } \)

The angle between lines is tan θ = ± \(\frac { (m₂ – m₁) }{ (1 + m₁ m₂) } \)

= ± \(\frac { (\frac { -2 }{ 5 } – \frac { 4 }{ 3 } ) }{ (1 + \frac { 4 }{ 3 }  * \frac { (-2) }{ 5 }) } \)

= \(\frac { -26 }{ 7 } \)

θ = tan -1 (\(\frac { -26 }{ 7 } \))

Example 3:

Find the angle between two lines x + y = 4, x + 2y = 3.

Solution:

The given two lines are x + y = 4, x + 2y = 3.

The slope-intercept form of the first line is

x + y = 4

y = 4 – x

Therefore, slope of x + y = 4 is m₁ = -1

The slope-intercept form of the second line is

x + 2y = 3

2y = 3 – x

y = \(\frac { (3 – x) }{ 2 } \)

y = \(\frac { 3 }{ 2 } – \frac { x }{ 2 } \)

Therefore, slope of x + 2y = 3 is m₂ = \(\frac { -1 }{ 2 } \)

The angle between lines is tan θ = ± \(\frac { (m₂ – m₁) }{ (1 + m₁ m₂) } \)

= ± \(\frac { (\frac { -1 }{ 2 } – (-1) ) }{ (1 + \frac { -1 }{ 2 }  * (-1)) } \)

= \(\frac { 1 }{ 3 } \)

θ = tan-1(\(\frac { 1 }{ 3 } \))

Example 4:

Find the angle between two straight lines x + 2y – 1 = 0 and 3x – 2y + 5 = 0

Solution:

The given lines are x + 2y – 1 = 0 and 3x – 2y + 5 = 0

The slope intercept form of first line is

x + 2y – 1 = 0

2y = 1 – x

y = \(\frac { 1 – x }{ 2 } \)

y = \(\frac { 1 }{ 2 } – \frac { x }{ 2 } \)

Therefore, the slope of line x + 2y – 1 = 0 is m₁ = \(\frac { -1 }{ 2 } \)

The slope-intercept form of the second line is

3x – 2y + 5 = 0

3x + 5 = 2y

y = \(\frac { 3x + 5 }{ 2 } \)

y = \(\frac { 3x }{ 2 } + \frac { 5 }{ 2 } \)

Therefore, the slope of line 3x – 2y + 5 = 0 is m₂ = \(\frac { 3 }{ 2 } \)

The angle between lines is tan θ = ± \(\frac { (m₂ – m₁) }{ (1 + m₁ m₂) } \)

= ± \(\frac { (\frac { 3 }{ 2 } – \frac { (-1 }{ 2 )} ) }{ (1 – \frac { 1 }{ 2 }  * \frac { (3) }{ 2 }) } \)

= 2

θ = tan-1(2)

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