Angle between pair of Straight Lines
Angle Between a Pair of Straight Lines
As a first example of the above theory, let us calculate the angle between a pair of straight lines. Consider the joint equation:
Let us assume that this equation represents two straight lines \( y = m_1x \) and \( y = m_2x \). Now, we know that:
Suppose the acute angle between these lines is \( \theta \). Then:
Using the identity \( (a  b)^2 = (a+b)^2  4ab \), we have:
Therefore:
Substituting the values of \( m_1 + m_2 \) and \( m_1m_2 \):
Hence:
Therefore:
So, the tangent of the acute angle \( \theta \) between the two straight lines is:
We can calculate \(\cos \theta\) as follows:
Thus:
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can find \(\sin \theta\):
Hence, we have:
Example
Prove that if the slope of one of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is twice the other, then \(9ab = 8h^2\).
Proof:
Let the slopes of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) be \(m\) and \(2m\). Then,
From the sum of slopes:
From the product of slopes:
Substitute \( m = \frac{2h}{3b} \) into the product of slopes:
Multiplying both sides by \(9b^2\):
Example
Prove that if the slope of one of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is \(\lambda\) times the other, then \((\lambda + 1)^2 ab = 4h^2 \lambda\).
Proof:
Let the slopes of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) be \(m\) and \(\lambda m\).
From the sum of slopes:
From the product of slopes:
Substitute \( m = \frac{2h}{b(\lambda + 1)} \) into the product of slopes:
Eliminate \( m \):
Therefore:
So, we have proved that if the slope of one of the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is \(\lambda\) times the other, then:
Perpendicular Lines
The lines represented by \( ax^2 + 2hxy + by^2 = 0 \) are perpendicular if and only if \( a + b = 0 \).
Proof:
Let the equation \( ax^2 + 2hxy + by^2 = 0 \) represent two straight lines. These lines can be expressed in the form \( y = m_1x \) and \( y = m_2x \), where \( m_1 \) and \( m_2 \) are the slopes of the lines.
From the given quadratic equation, we know the following relationships:

Sum of Slopes: [ m_1 + m_2 = \frac{2h}{b} ]

Product of Slopes: [ m_1m_2 = \frac{a}{b} ]
For the lines to be perpendicular, the product of their slopes must be \(1\). Thus, we have:
Using the given relationship for the product of slopes, we substitute:
Solving for \( a \) and \( b \):
Therefore, we can conclude:
This establishes that for the lines represented by \( ax^2 + 2hxy + by^2 = 0 \) to be perpendicular, the condition \( a + b = 0 \) must be satisfied. Conversely, if \( a + b = 0 \), the product of the slopes \( m_1 \) and \( m_2 \) will be \(1\), indicating that the lines are indeed perpendicular.
Hence, the theorem is proved.
This completes the proof.
Equation of perpendicular pair of straight lines to a given pair
The equation of a pair of straight lines, each perpendicular to a given pair of straight lines represented by \(ax^2 + 2hxy + by^2 = 0\), is given by \(bx^2  2hxy + ay^2 = 0\).
Proof:
Consider the equation of a pair of straight lines passing through the origin:
Let this equation represent two lines \(y = m_1x\) and \(y = m_2x\).
We know the following relationships between the slopes \(m_1\) and \(m_2\):
To find the equation of the lines perpendicular to these, we need the slopes of the perpendicular lines. If a line has a slope \(m\), then a line perpendicular to it has a slope \(\frac{1}{m}\). Therefore, the lines perpendicular to \(y = m_1x\) and \(y = m_2x\) will have equations:
These can be rewritten as:
To find the combined equation representing both these lines, we multiply the individual line equations:
Expanding this product, we get:
Substituting the known values of \(m_1 + m_2\) and \(m_1 m_2\) from above, we have:
Multiplying through by \(b\) to clear the denominators, the equation becomes:
Thus, the equation of the pair of straight lines perpendicular to the given pair of straight lines \(ax^2 + 2hxy + by^2 = 0\) is:
This completes the proof.
Equation of a Pair of Straight Lines Parallel to a given homogenous Pair
The joint equation of a pair of straight lines intersecting at \((x_1, y_1)\) and parallel to a homogeneous pair of straight lines given by \(ax^2 + 2hxy + by^2 = 0\) is:
Proof:
We derive this equation using the concept of shifting the origin. When the origin is shifted to \((h, k)\), a curve whose equation is \(f(x, y) = 0\) becomes \(f(x + h, y + k) = 0\), obtained by substituting \(x \to x + h\) and \(y \to y + k\)
Consider the pair of straight lines given by the homogeneous equation \(ax^2 + 2hxy + by^2 = 0\), passing through the origin. If we shift the origin to \((x_1, y_1)\), this pair of straight lines is now intersecting at \((x_1, y_1)\) and remains parallel to the original pair of lines described by \(ax^2 + 2hxy + by^2 = 0\).
The new equation, accounting for the shift in origin, is obtained by substituting \(x \to x  x_1\) and \(y \to y  y_1\) into the original equation:
Thus, the joint equation of the pair of straight lines intersecting at \((x_1, y_1)\) and parallel to the given homogeneous pair of lines is:
This completes the proof.
Equation of a Pair of Straight Lines Perpendicular to a Given Pair and Intersecting at a Given Point
Given the pair of straight lines represented by \(ax^2 + 2hxy + by^2 = 0\), the equation of a pair of straight lines perpendicular to this given pair and intersecting at the point \((x_1, y_1)\) is:
Common Line to Pair of Lines
When one line in a pair is parallel to a line in another pair
Suppose we wish to check whether two given pairs of straight lines:
have a line in common. We can determine this without actually factorizing the equations by the following condition:
If the pairs of straight lines have a line in common, then:
Proof:
Let \( y = mx \) be the common line between the two pairs of straight lines.
For the first pair of straight lines, substituting \( y = mx \) into the equation \( a_1x^2 + 2h_1xy + b_1y^2 = 0 \), we get:
Since \( x \neq 0 \), we have:
Similarly, for the second pair of straight lines, substituting \( y = mx \) into the equation \( a_2x^2 + 2h_2xy + b_2y^2 = 0 \), we get:
Since \( x \neq 0 \), we have:
From equations (1) and (2), we have:
Therefore,
Thus,
Therefore,
We did some rearragement of columns in the last step using properties of determinants.
This completes the proof.
Both Lines are common
If \( a_1x^2 + 2h_1xy + b_1y^2 = 0 \) and \( a_2x^2 + 2h_2xy + b_2y^2 = 0 \) are two given pairs of straight lines, then both lines are common if and only if:
When one line in one pair is perpendicular to a line in another pair
If one of the lines represented by \(a_1x^2 + 2h_1xy + b_1y^2 = 0\) is perpendicular to one of the lines represented by \(a_2x^2 + 2h_2xy + b_2y^2 = 0\), then:
Proof:
Let \( y = mx \) be the line in \( a_1x^2 + 2h_1xy + b_1y^2 = 0 \), which is perpendicular to one of the lines in \( a_2x^2 + 2h_2xy + b_2y^2 = 0 \).
Its equation is:
Substitute \( y = mx \) in \( a_1x^2 + 2h_1xy + b_1y^2 = 0 \):
Since \( x \neq 0 \), we have:
Substitute \( y = \frac{1}{m}x \) in \( a_2x^2 + 2h_2xy + b_2y^2 = 0 \):
Multiplying through by \( m^2 \), we get:
From equations (1) and (2), we have:
Therefore,
Thus,
Therefore,
Rearranging, we get the required result.