# 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

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