Angle between Two Straight Lines

We will learn how to find the angle between two straight lines.

The angle θ between the lines having slope m\(_{1}\) and m\(_{2}\) is given by tan θ = ± \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)

Let the equations of the straight lines AB and CD are y = m\(_{1}\) x + c\(_{1}\) and y = m\(_{2}\)  x + c\(_{2}\)  respectively intersect at a point P and make angles θ1 and θ2 respectively with the positive direction of x-axis.

Let ∠APC = θ is angle between the given lines AB and CD.

Clearly, the slope of the line AB and CD are m\(_{1}\)  and m\(_{2}\)  respectively.

Then, m\(_{1}\)  = tan θ\(_{1}\)  and m\(_{2}\)  = tan θ\(_{2}\)

Now, from the above figure we get, θ\(_{2}\)  = θ + θ\(_{1}\)  

⇒ θ = θ\(_{2}\)  - θ\(_{1}\)

Now taking tangent on both sides, we get,

tan θ = tan (θ\(_{2}\)  - θ\(_{1}\))

⇒ tan θ = \(\frac{tan θ_{2} - tan θ_{1}}{1 + tan θ_{1} tan θ_{2}}\), [Using the formula, tan (A + B) = \(\frac{tan A - tan B}{1 + tan A tan B}\)

⇒ tan θ = \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\), [Since, m\(_{1}\)  = tan θ\(_{1}\)  and m\(_{2}\)  = tan θ\(_{2}\)]

Therefore, θ = tan\(^{-1}\)\(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)

Again, the angle between the lines AB and CD be ∠APD = π - θ since ∠APC = θ

Therefore, tan ∠APD = tan (π - θ) = - tan θ = - \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)

Therefore, the angle θ between the lines AB and CD is given by,

tan θ = ± \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)

⇒ θ =  tan\(^{-1}\)(±\(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)) 


Notes:    

(i) The angle between the lines AB and CD is acute or obtuse according as the value of \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\) is positive or negative.

(ii) The angle between two intersecting straight lines means the measure of the acute angle between the lines.

(iii) The formula tan θ = ± \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\) cannot be used to find the angle between the lines AB and CD, if AB or CD is parallel to y-axis. Since the slope of the line parallel to y-axis is indeterminate.

 

Solved examples to find the angle between two given straight lines:

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

Solution:

Let the slope of the line AB and BC are m\(_{1}\) and m\(_{2}\) respectively.

Then,

m\(_{1}\) = \(\frac{3 - 1}{2 - (-2)}\) = \(\frac{2}{4}\)= ½ and

m\(_{2}\) = \(\frac{-4 - 3}{-2 - 2}\)= \(\frac{7}{4}\)

Let θ be the angle between AB and BC. Then,

tan θ = |\(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)| = |\(\frac{\frac{7}{4} - \frac{1}{2}}{1 + \frac{7}{4}\cdot \frac{1}{2}}\)| = |\(\frac{\frac{10}{8}}{\frac{15}{8}}\)|= ±\(\frac{2}{3}\).

⇒ θ = tan\(^{-1}\)(\(\frac{2}{3}\)), which is the required angle.

 

2. Find the acute angle between the lines 7x - 4y = 0 and 3x - 11y + 5 = 0.

Solution:  

First we need to find the slope of both the lines.

7x - 4y = 0     

⇒ y = \(\frac{7}{4}\)x

Therefore, the slope of the line 7x - 4y = 0 is \(\frac{7}{4}\)

Again, 3x - 11y + 5 = 0     

⇒ y = \(\frac{3}{11}\)x + \(\frac{5}{11}\)

Therefore, the slope of the line 3x - 11y + 5 = 0 is = \(\frac{3}{11}\)

Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is θ

Now,

tan θ = | \(\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}\)| = ±\(\frac{\frac{7}{4} - \frac{3}{11}}{1 + \frac{7}{4}\cdot \frac{3}{11}}\) = ± 1

Since θ is acute, hence we take, tan θ = 1 = tan 45°

Therefore, θ = 45°

Therefore, the required acute angle between the given lines is 45°.

 The Straight Line




11 and 12 Grade Math 

From Angle between Two Straight Lines to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Perimeter of a Rectangle | How to Find the Perimeter of a Rectangle?

    Apr 25, 24 02:01 PM

    Perimeter of a Rectangle
    We will discuss here how to find the perimeter of a rectangle. We know perimeter of a rectangle is the total length (distance) of the boundary of a rectangle. ABCD is a rectangle. We know that the opp…

    Read More

  2. Perimeter of a Square | How to Find the Perimeter of Square? |Examples

    Apr 25, 24 12:54 PM

    Perimeter of a Square
    We will discuss here how to find the perimeter of a square. Perimeter of a square is the total length (distance) of the boundary of a square. We know that all the sides of a square are equal. Perimete…

    Read More

  3. Perimeter of a Triangle | Perimeter of a Triangle Formula | Examples

    Apr 25, 24 12:53 PM

    Perimeter of a Triangle
    We will discuss here how to find the perimeter of a triangle. We know perimeter of a triangle is the total length (distance) of the boundary of a triangle. Perimeter of a triangle is the sum of length…

    Read More

  4. Dividing 3-Digit by 1-Digit Number | Long Division |Worksheet Answer

    Apr 24, 24 03:46 PM

    Dividing 3-Digit by 1-Digit Number
    Dividing 3-Digit by 1-Digit Numbers are discussed here step-by-step. How to divide 3-digit numbers by single-digit numbers? Let us follow the examples to learn to divide 3-digit number by one-digit nu…

    Read More

  5. Symmetrical Shapes | One, Two, Three, Four & Many-line Symmetry

    Apr 24, 24 03:45 PM

    Symmetrical Figures
    Symmetrical shapes are discussed here in this topic. Any object or shape which can be cut in two equal halves in such a way that both the parts are exactly the same is called symmetrical. The line whi…

    Read More