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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 m1 and m2 is given by tan θ = ± m2m11+m1m2

Let the equations of the straight lines AB and CD are y = m1 x + c1 and y = m2  x + c2  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 m1  and m2  respectively.

Then, m1  = tan θ1  and m2  = tan θ2

Now, from the above figure we get, θ2  = θ + θ1  

⇒ θ = θ2  - θ1

Now taking tangent on both sides, we get,

tan θ = tan (θ2  - θ1)

⇒ 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 

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