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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 = ± m2−m11+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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