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Straight lines is an extremely important topic of IIT JEE Mathematics. It often fetches some direct questions in various competitions like the IIT JEE. Since the topic is quite vast, students are advised to spend sufficient time on grasping the various concepts. Angle between pair of straight lines is an important head under straight lines. We begin with the concept of angle between pair of lines and then discuss some of the illustrations on the same:
Suppose we have two straight lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2}, then the angle between these two lines is given by tan θ = |(m_{1} – m_{2})/ (1 + m_{1}m_{2})|.
If m_{1}, m_{2} and m_{3} are the slopes of three lines L_{1} = 0, L_{2} = 0 and L_{3} = 0, where m_{1} > m_{2} > m_{3} then the interior angles of the triangle ABC formed by these lines are given by,
tan A = (m_{1 }- m_{2})/(1 + m_{1}m_{2}), tan B = (m_{2 }- m_{3})/(1 + m_{2}m_{3}) and tan C = (m_{3 }- m_{1})/(1 + m_{3}m_{1}).
If one of the line is parallel to y-axis then the angle between two straight lines is given by tan θ = ±1/m where ‘m’ is the slope of the other straight line.
If the two lines are a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, then the formula becomes tan θ = |(a_{1}b_{2 }- b_{1}a_{2})/(a_{1}a_{2} + b_{1}b_{2})|
Generally speaking, the angle between these two lines is assumed to be acute and hence, the value of tan θ is taken to be positive.
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Angle between pair of lines represented by ax^{2} + 2hxy + by^{2} = 0
Comparing the coefficients of x^{2}, y^{2} and xy, we get
b(y – m_{1}x) (y – m_{2}x) = ax^{2} + 2hxy + by^{2}
m_{1} + m_{2} = –2h/b and
m_{1} m_{2} = a/b
tan θ_{acute} = |(m_{2} – m_{1})/(1 + m_{1}m_{2})|
= |√(m_{1}–m_{2})^{2} – 4m_{1}m_{2}/(1 + m_{1}m_{2})|
= |2√(h^{2} – ab)/(a + b)|
If two lines through the origin are represented by y = m_{1}x and y = m_{2}x, we cannot write
⇒ (y – m_{1}x) (y – m_{2}x) ≡ ax^{2} + 2hxy + by^{2}
Because coefficient of y^{2} on left hand side is one on right hand side, it is b.
The given equation represents real lines only when h^{2} – ab > 0
If two lines are coincident then tan θ = 0 ⇒ h^{2} = ab
If two lines are perpendicular then m_{1}m_{2} = 1 ⇒ a + b = 0
i.e. x^{2} + 2hxy – y^{2} always represents pair of mutually perpendicular lines through origin.
Two lines are equally inclined to axes but are not parallel:
For such a case let us take a line l_{1} which is inclined at an angle θ, then l_{2} is inclined at (π – θ).
tan (π – θ) = – tan θ which is the condition for two lines inclined equally to axes.
m_{1} = –h + √(h^{2}–ab)/2 and m_{2} = –h – √(h^{2}–ab)/2
What is the equation of the pair of lines through origin and perpendicular to ax^{2} + 2hxy + by^{2} = 0?
Let ax^{2} + 2hxy + by^{2} = 0 represent the lines y = m_{1}x (i) and y = m_{2}x (ii)
Lines perpendicular to the lines (i) and (ii) are y = –1/m_{1} x and y = –1/m_{2} x respectively and passing through origin
i.e. m_{1}y + x = 0 and m_{2}y + x = 0
Their combined equation is given by
(m_{1}y + x) (m_{2}y + x) = 0
⇒ m_{1}m_{2} y^{2} + (m_{1} + m_{2}) xy + x^{2} = 0
⇒ a/2 y^{2} – 2h/b xy + x^{2} = 0
⇒ bx^{2} – 2hxy + ay^{2} = 0 is the equation of the pair of lines perpendicular to pair of lines ax^{2} + 2hxy + by^{2} = 0.
Find the angle between the lines represented by the equation 2x^{2} – 7xy + 3y^{2} = 0.
The given equation is 2x^{2} – 7xy + 3y^{2} = 0.
This can be written as 2x^{2} – 7xy + 3y^{2} = (2x – y)(x-3y)
Therefore, the given lines are (2x – y) = 0 and (x-3y) = 0.
Now, here, a = 2, h = -7/2 and b = 3.
Hence, the angle between the lines is given by tan θ = 2√{(-7/2)^{2} – 6}/5 = 1.
Hence, θ = 45°.
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