Here we are going to see how to find the angle between two straight lines.

**θ = tan-¹ |(m₁ - m₂)/(1 + m₁ m₂)|**

Here m₁ is slope of the first line and m₂ is the slope of the second line.To find angle between two lines, first we need to find slope of both lines separately and then we have to apply their values in the above formula.

Here you can find two example problems to understand this topic clearly.

**Example 1:**

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

**Solution :**

To find the angle between two lines we have to find the slopes of the two lines.

Slope of a line = - coefficient of x/coefficient of y

Slope of the fist line x + 2y -1 = 0

m₁ = -1/2

Slope of the second line 3x - 2y +5=0

m₂ = -3/(-2)

m₂ = 3/2

Angle between the lines

θ = tan-¹ |(m₁ - m₂)/(1 + m₁ m₂)|

θ = tan-¹ |(-1/2 - 3/2) /(1+ (-1/2) (3/2))|

θ = tan-¹ |[(-1 - 3)/2] /[1 + (-3/4)]|

θ = tan-¹ |[(-4)/2] /[4 + (-3)/4)]|

θ = tan-¹ |[(-2) /[1/4)]|

θ = tan-¹ |[(-2) x[4/1]|

θ = tan-¹ |-8|

θ = tan-¹ (-8)

**Example 2:**

Find the angle between the lines 2x + y = 4 and x + 3y = 5

**Solution:**

To find the angle between two lines, we have to find the slopes of the two lines.

Slope of line = - coefficient of x/coefficient of y

Slope of 1^{st} line 2x + y = 4 = 0

m₁ = -2/1

m₁ = -2

Slope of 2^{nd} line x + 3y = 5

m₂ = -1/3

Angle between the lines

** θ = tan-¹ |(m₁ - m₂)/(1 + m₁ m₂)|**

θ = tan-¹ |(-2 -(-1/3) /(1+ (-2) (-1/3))|

θ = tan-¹ |[(-2 + 1/3)] /[1 + (2/3)]|

θ = tan-¹ |[(-6+1)/3] /[(3 + 2)/3)]|

θ = tan-¹ |[(-5/3) /[5/3)]|

θ = tan-¹ |[(-5/3) x[3/5]|

θ = tan-¹ |-1|

θ = tan-¹ (1)

θ = 45°

After having gone through the stuff given above, we hope that the students would have understood "Angle between two straight lines".

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