The relationship between two lines can be determined by the angle generated by the two lines. It's the angle formed by the two lines that are being calculated. The acute angle and the obtuse angle are two angles formed by two intersecting lines. We consider the acute angle between two lines while calculating the angle between two lines.
The angle formed by two lines can be used to calculate the angle formed by two sides of a closed polygon. Let's look at the formulas and examples for the angle formed by two lines in a coordinate plane, as well as three-dimensional space.
The topic of straight lines is particularly significant in IIT JEE Mathematics. In competitions such as the IIT JEE, it frequently yields some straight questions. Because the subject is so broad, students should devote enough time to comprehending the numerous concepts. Under straight lines, the angle between two straight lines is a significant head.
When two straight lines intersect, two sets of angles are formed. A pair of acute and another pair of obtuse angles are formed by the intersection. Angles' absolute values are determined by the slopes of crossing lines.
It's also worth noting that if one of the lines intersects with the y-axis, the angle generated by the intersection cannot be determined because the slope of a line parallel to the y-axis is ambiguous.
Angle Between Two Lines Coordinate Geometry
If you know the coordinates of the three points A, B, and C, you can use analytic geometry to calculate the angle between the lines AB and BC.
When the slope of each line is known from the equation, the angle between the two lines may be calculated by calculating the slope of each line and then using it in the formula to get the angle between two lines.
Calculating Angle Between Two Lines in Coordinate Geometry
1. How to Find Angle Between Two Lines
Let us consider three points are given on the x-axis and y-axis whose coordinates are given.
Consider a line whose endpoints have coordinates (x1 y1) and (x2 y2).
The equation of the slope will be
m = y₂ - y₁/x₂ - x₁
m1 and m2 can be calculated by substituting this in the above formula then the values of m1 and m2 can be substituted in the formula given.
tan θ = ± (m₁ - m₂ ) / (1- m₁*m₂)
2. If the Lines are Parallel
(image will be uploaded soon)
The two lines are parallel means, the angle between them is zero Ɵ = 0°
i.e tan Ɵ =0
m₁ - m₂/1+m₁m₂=0
m₁ - m₂=0
m₁ = m₂
The slopes are equal. It shows that the lines are parallel
Note: The value of tan Ɵ will always be positive.
3. If the Lines are Perpendicular
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When the two lines are perpendicular angle between them will be 90° i.e Ɵ=90°
1/tan Ɵ = 0
1+m₁m₂/m₁-m₂ = 0
1+m₁m₂ =0
m₁m₂ = -1
The product of their slope is -1. It shows that the lines are perpendicular.
Formula to Find the Angle Between Two Lines
Consider two nonparallel lines which have slopes m1 and m2 and Ɵ is the angle between the lines, then the formula for finding the angle between the two lines would be:
Derivation of the Formula
Considering the below figure here, we can see that we have a plane with an x-axis and a y-axis.
Two lines L1 and L2 are intersecting each other. Forming one acute and the other obtuse angle. Let us consider the acute angle as Ɵ. (image will be uploaded soon)
Step 1: At first, We have to show that
θ = θ₂ - θ₁
Step 2: For proving above, Let us consider ∆ABC
By angle sum property we can state that
θ + θ₂ + x = 180 ……..(1)
x + θ₂ = 180…………. (2) (as angle x and θ2 forms a linear pair)
Step 3: From equations 1 and 2, we can equate them.
θ + θ₁ + x = x + θ₂ = 180
Subtracting x from both sides
θ + θ₁ + x - x = x + θ₂ - x
we get
θ + θ₁ = θ₂
Step 4: Subtracting θ₁ from both sides
θ + θ₁ - θ₁ = θ₂ - θ₁
We get,
θ = θ₂ - θ₁
Step 5: Applying tangent on both sides
tan θ = tan ( θ₂ - θ₁)
Using the formula of tangents we get
tan θ = tan θ₂ - tan θ₁/1+tan θ₁tanθ₂
From the inclination of a line, we know that tan θ = m
Hence we can substitute tan θ₁ = m₁ and tan θ₂ = m2 we get,
tan Ɵ= |(m₂ - m₁)/(1+m₁m₂)| |
Solved Examples:
Example 1:
If P (2, -1), Q (5, 3), and R (-2, 6) are three points, find the angle between the straight lines PQ and QR.
Solution:
The slope of PQ is given by
m = ( y2 – y1 ) / (x2 – x1)
m = ( 3 – (- 1) ) / (5 – 2 ))
m = 4/3
Therefore, m₁ = 4/3
The slope of QR is given by
m= (6 - 3 ) / (−2−5)
m= 3/-7
Therefore, m₂ = 3/-7
Substituting the values of m2 and m1 in the formula for the angle between two lines we get,
tan θ = ± (m₂ – m₁ ) / (1- m₁ m₂)
tan θ = ± (3/-7) – (4/3) ) / (1- (3/-7)(4/3))
tan θ = ± (37/33)
Therefore, θ = tan -1 (37 / 33)
Example 2:
Find the angle between the following two lines.
Line 1: 4x -3y = 8
Line 2: 2x + 5y = 4
Solution:
Put 4x -3y = 8 into slope-intercept form so you can clearly identify the slope.
4x -3y = 8
3y = 4x - 8
y = 4x / 3 - 8/3
y = (4/3)x - 8/3
Put 2x + 5y = 4 into slope-intercept form so you can clearly identify the slope.
2x + 5y = 4
5y = -2x + 4
y = -2x/5 + 4/5
y = (-2/5)x + 4/5
The slopes are 4/3 and -2/5 or 1.33 and -0.4. It does not matter which one is m₁ or m₂. You will get the same answer.
Let m₁ = 1.33 and m₂ = -0.4
tan θ = ± (m1 – m₂ ) / (1+ m₁*m₂)
tan θ = ± (1.33 - (- 0.4)) / (1- (1.33)*(-0.4))
tan θ = ± (1.73) / (1- 0.532)
tan θ = ± (1.73 ) / (0.468)
tan θ= 3.696
θ = tan⁻¹ (3.69)
Example 3:
Find the acute angle between y = 3x+1 and y = -4x+3
Solution:
m₁= 3 and m₂ = -4
tan θ = ± (m1 – m2 ) / (1+ m₁*m₂)
tan θ = ± (3-(-4) ) / (1+ 3*-4)
tan θ = ± (7 ) / (1+(-12))
tan θ = ± (7 ) / (-11)
tan θ = ± (7/11)
tan θ = 0.636
θ = tan⁻¹ (0.636)
This is all about angles formed between two lines and how to measure them. Focus on the derivation of the formulas to understand the concepts and grab hold of the idea of measuring angles between two lines.
Here is a way to do it with the cosine rule. It's very long but if you are not getting the right answer you could try this method.
First, find the intercept of the two equations:
$4x+2=−x+3, x=0.2, y=4(0.2)+2, y=2.8, (0.2, 2.8)$
Now, sub in any y value into both the equations and find the coordinates. I will be using 10 as an example.
Eqn 1: $10=4x+2, x=2, (2,10)$
Eqn 2: $10=-x+3, x=-7, (-7,10)$
Then, find the distance between all three of the points. These three distances will be used subbed in the cosine rule so make sure that the distance that does not connect to the intercept point is $a$
$a$: $2-(-7)=9$
$b$: $\sqrt{(10-2.8)^2+(2-0.2)^2}=7.42$
$c$: $\sqrt{(10-2.8)^2+(-7-0.2)^2}=10.18$
Sub the values into the formula $cos(\theta)=\frac{b^2+c^2-a^2}{2bc}$
$\theta=cos^{-1}(\frac{(7.42)^2+(10.18)^2-(9)^2}{2(7.42)(10.18)})=59°$(approximate answer)