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 xaxis.
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 yaxis. Since the slope of the line parallel to yaxis 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°.
11 and 12 Grade Math
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