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
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