When two lines intersect each other, they create a small angle between them.  In addition, you don't even need another line, if the line intersects the axis, it still creates an angle from the axis. The angle between two lines is the smaller of the angles formed by the intersection of the two lines The angle can be obtained from:

Derivation of the Formula to Measure Angle

Angle formed by intersection of two lines and x-axis

Since it forms a triangle, we know that the sum of { \theta }_{ 1 } + \theta + ( 180 - { \theta }_{ 2 }) = 180.

{ \theta }_{ 1 } + \theta + 180 - { \theta }_{ 2 } = 180

{ \theta }_{ 1 } + \theta - { \theta }_{ 2 } = 0

{ \theta }_{ 1 } + \theta = { \theta }_{ 2 }

Subtracting { \theta }_{ 1 } from both sides:

{ \theta }_{ 1 } + \theta - { \theta }_{ 1 }= { \theta }_{ 2 } - { \theta }_{ 1 }

\theta = { \theta }_{ 2 } - { \theta }_{ 1 }

Using \tan { } on both sides:

\tan { \theta } = \tan { { \theta }_{ 2 } - { \theta }_{ 1 } }

\tan { \theta } = \frac { \tan { { \theta }_{ 2 } } -\tan { { \theta }_{ 1 } } }{ 1+\tan { { \theta }_{ 1 } } .\tan { { \theta }_{ 2 } } }

Since you all know that \tan { \theta } = m, that means \tan { { \theta }_{ 2 } = { m }_{ 2 } and \tan { { \theta }_{ 1 } = { m }_{ 1 }:

\tan { \theta } = \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }.{ m }_{ 2 } }

Hence, the final formula will be:

\tan { \theta } =\left| \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }.{ m }_{ 2 } } \right|

We used the modulus so that the \tan { \theta } is always equal to a positive number.

Formulas

To find the slope, you need to use this formula:

\tan { \theta } =\left| \frac { { m }_{ 2 }-{ m }_{ 1 } }{ 1+{ m }_{ 1 }.{ m }_{ 2 } } \right|

To find their direction vectors:\cos { \theta } =\frac { \left| { u }_{ 1 } . v_{ 1 }+{ u }_{ 2 } . { v }_{ 2 } \right| }{ \sqrt { { { u }_{ 1 } }^{ 2 }+{ { u }_{ 2 } }^{ 2 } } \sqrt { { { v }_{ 1 } }^{ 2 }+{ { v }_{ 2 } }^{ 2 } } }

 

Examples

Find the angle between the lines r and s, if their directional vectors are: \overrightarrow { u } = (−2, 1) and \overrightarrow { v } = (2, −3)

 

\cos { \theta } =\frac { \left| -2 . 1 + 2 . -3 \right| }{ \sqrt { 5 } . \sqrt { 13} }

\cos { \theta } =\frac { \left| -7 \right| }{ \sqrt { 65} }

\cos { \theta } = 0.868

\theta = 29.7729°

 

Given the lines r \equiv 3x + y - 1 = 0 and s \equiv 2x + my - 8 = 0, calculate the value of m so that the two lines form an angle of 45°.

\theta = 45°

\cos { 45 } = \frac { sqrt { 2 } }{ 2 }

\overrightarrow { { v }_{ r } } = (-1, 3)

\overrightarrow { { v }_{ s } } = (-m, 2)

 

\frac { \sqrt { 2 } }{ 2 } = \frac { -1 . (-m) + 3 . 2 }{ \sqrt { 1 + 9 } . \sqrt { { m }^{ 2 } + 4 }}

Taking square on both sides:

{ (\frac { \sqrt { 2 } }{ 2 }) }^{ 2 } = { (\frac { -1 . (-m) + 3 . 2 }{ \sqrt { 1 + 9 } . \sqrt { { m }^{ 2 } + 4 }}) }^{ 2 }

\frac { 36 + 12m + { m }^{ 2 } }{ 10( 4 + { m }^{ 2 } ) } = \frac { 1 }{ 2 }

{ m }^{ 2 } - 3m - 4 = 0

{ m }^{ 2 } - m (4-1) - 4 = 0

{ m }^{ 2 } - 4m + m - 4 = 0

m(m - 4) + 1(m - 4) = 0

(m - 4)(m + 1) = 0

m -4 = 0 \qquad m + 1 = 0

m = 4 \qquad m = -1

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.