Point of intersection of a General Pair of Straight Lines

Shifting a Homogenous Pair of Straight Lines

Given a homogeneous pair of straight lines \(ax^2 + 2hxy + by^2 = 0\), which intersect at the origin, consider shifting this pair of straight lines parallelly (without rotation, only upward or downward translation) to a new location such that the intersection point is \((x_1, y_1)\). The new equation of this pair of straight lines is found by substituting \(x - x_1\) for \(x\) and \(y - y_1\) for \(y\) in the original equation.

Thus, the new equation becomes:

\[ a(x - x_1)^2 + 2h(x - x_1)(y - y_1) + b(y - y_1)^2 = 0 \]

Invariance of Quadratic terms because of Shifting

Expanding this equation:

\[ a(x - x_1)^2 + 2h(x - x_1)(y - y_1) + b(y - y_1)^2 = 0 \]

\[ a(x^2 - 2x_1x + x_1^2) + 2h(xy - x_1y - y_1x + x_1y_1) + b(y^2 - 2y_1y + y_1^2) = 0 \]

Combining like terms, we get:

\[ ax^2 - 2ax_1x + ax_1^2 + 2hxy - 2hx_1y - 2hy_1x + 2hx_1y_1 + by^2 - 2by_1y + by_1^2 = 0 \]

Grouping terms, we have:

\[ ax^2 + 2hxy + by^2 - 2(ax_1 + hy_1)x - 2(hx_1 + by_1)y + (ax_1^2 + 2hx_1y_1 + by_1^2) = 0 \]

From this expanded form, we can observe that:

Quadratic terms: The first three terms \(ax^2 + 2hxy + by^2\) remain unchanged, preserving the original quadratic nature of the pair of lines.
Linear terms: There is an addition of two new linear terms:

\[ -2(ax_1 + hy_1)x \]

\[ -2(hx_1 + by_1)y \]
Constant term: There is an additional constant term:

\[ ax_1^2 + 2hx_1y_1 + by_1^2 \]

This leads us to an important conclusion: A pair of straight lines parallel to the lines given by \(ax^2 + 2hxy + by^2 = 0\), and having a point of intersection at \((x_1, y_1)\), can be written as:

\[ a(x - x_1)^2 + 2h(x - x_1)(y - y_1) + b(y - y_1)^2 = 0 \]

In this new equation, each line is parallel to its corresponding line in the original pair. This means that the orientation and the angle between the lines remain unchanged, but their position is shifted to the new intersection point \((x_1, y_1)\).

When this new equation is expanded, the quadratic terms remain the same, and we get additional linear and constant terms. Using the result from the previous expansion, we have:

\[ ax^2 + 2hxy + by^2 - 2(ax_1 + hy_1)x - 2(hx_1 + by_1)y + (ax_1^2 + 2hx_1y_1 + by_1^2) = 0 \]

This clearly shows that the quadratic terms \(ax^2 + 2hxy + by^2\) are unchanged, preserving the original structure of the pair of lines. However, there are new linear terms \(-2(ax_1 + hy_1)x\) and \(-2(hx_1 + by_1)y\), as well as a new constant term \(ax_1^2 + 2hx_1y_1 + by_1^2\).

Pair of Straight Lines perpendicular to a given pair

A pair of straight lines intersecting at \((x_1, y_1)\) and perpendicular to the pair of lines given by \(ax^2 + 2hxy + by^2 = 0\) is represented by:

\[ b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0 \]

Proof

Pair of Perpendicular Lines through the Origin:

The pair of straight lines passing through the origin and perpendicular to the lines given by \(ax^2 + 2hxy + by^2 = 0\) is:

\[ bx^2 - 2hxy + ay^2 = 0 \]

This follows because, for two lines to be perpendicular, the product of their slopes must be \(-1\). Therefore, swapping the coefficients of \(x^2\) and \(y^2\) and changing the sign of the mixed term \(2hxy\) gives us the equation of the perpendicular lines.
Translation to Intersect at \((x_1, y_1)\):

To find the required pair of lines intersecting at \((x_1, y_1)\) and perpendicular to the original lines, we translate the perpendicular lines through the origin to intersect at the new point \((x_1, y_1)\).

Substituting \(x - x_1\) for \(x\) and \(y - y_1\) for \(y\) in the perpendicular equation, we get:

\[ b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0 \]
Conclusion:

Therefore, the equation of the pair of straight lines intersecting at \((x_1, y_1)\) and perpendicular to the given pair of lines is:

\[ b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0 \]

This completes the proof.

Point of Intersection of a Pair of Straight Lines

Given a general pair of straight lines represented by the equation:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

We wish to find the point of intersection of the lines that this equation represents. To achieve this, we use the concept of shifting the origin.

Let the intersection point be \((x_1, y_1)\). If we shift the origin to \((x_1, y_1)\), then the equation of the pair of straight lines should transform to \(ax^2 + 2hxy + by^2 = 0\), with no linear terms and the constant term eliminated.

To do this, replace \(x\) with \((x + x_1)\) and \(y\) with \((y + y_1)\) in the original equation. Expanding this transformation, we get:

\[ a(x + x_1)^2 + 2h(x + x_1)(y + y_1) + b(y + y_1)^2 + 2g(x + x_1) + 2f(y + y_1) + c = 0 \]

Expanding and collecting like terms, we have:

\[ a(x + x_1)^2 + 2h(x + x_1)(y + y_1) + b(y + y_1)^2 + 2g(x + x_1) + 2f(y + y_1) + c = 0 \]

\[ a(x^2 + 2x_1x + x_1^2) + 2h(xy + x_1y + y_1x + x_1y_1) + b(y^2 + 2y_1y + y_1^2) + 2g(x + x_1) + 2f(y + y_1) + c = 0 \]

\[ ax^2 + 2ax_1x + ax_1^2 + 2hxy + 2hx_1y + 2hy_1x + 2hx_1y_1 + by^2 + 2by_1y + by_1^2 + 2gx + 2gx_1 + 2fy + 2fy_1 + c = 0 \]

Combining like terms:

\[ ax^2 + 2hxy + by^2 + (2ax_1 + 2hy_1 + 2g)x + (2hx_1 + 2by_1 + 2f)y + (ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c) = 0 \]

For the equation to be free of linear terms and the constant term, the coefficients of the linear terms must be zero. Thus, we set the coefficients of \(x\) and \(y\) to zero:

\[ 2ax_1 + 2hy_1 + 2g = 0 \]

\[ 2hx_1 + 2by_1 + 2f = 0 \]

Dividing by 2, we get the system of linear equations:

\[ ax_1 + hy_1 + g = 0 \]

\[ hx_1 + by_1 + f = 0 \]

The solution of these two linear equations gives us the point of intersection \((x_1, y_1)\).

Thus, the point of intersection of the pair of straight lines is:

\[ x_1 = \frac{hf - bg}{ab - h^2} \]

\[ y_1 = \frac{hg - af}{ab - h^2} \]

Now, you must be wondering what happened to the constant term \(ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c\) in the expansion. It turns out that it is equal to zero automatically as we set the coefficients of the linear terms to zero.

Let's write the constant term as follows:

\[ ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 + c \]

We can factor and group the terms involving \(x_1\) and \(y_1\):

\[ x_1(ax_1 + hy_1 + g) + y_1(by_1 + hx_1 + f) + gx_1 + fy_1 + c \]

Using the linear equations \(ax_1 + hy_1 + g = 0\) and \(hx_1 + by_1 + f = 0\), the first two terms become zero:

\[ x_1(0) + y_1(0) + gx_1 + fy_1 + c \]

Now, substituting \(x_1 = \frac{hf - bg}{ab - h^2}\) and \(y_1 = \frac{hg - af}{ab - h^2}\) into the remaining terms, we get:

\[ g \left( \frac{hf - bg}{ab - h^2} \right) + f \left( \frac{hg - af}{ab - h^2} \right) + c \]

Combining these terms, we have:

\[ \frac{g(hf - bg) + f(hg - af) + c(ab - h^2)}{ab - h^2} \]

Simplifying the numerator:

\[ g(hf - bg) + f(hg - af) + c(ab - h^2) = 2fgh - g^2b - f^2a - ch^2 + abc\]

Since \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) represents a pair of straight lines, the condition \(abc + 2fgh - af^2 - bg^2 - ch^2 = 0\) holds, making the numerator zero:

\[ 2fgh - g^2b + fhg - f^2a + abc - ch^2 = 0 \]

Thus, the constant term becomes:

\[ \frac{0}{ab - h^2} = 0 \]

Therefore, the constant term is zero, confirming that the equation transforms correctly with no constant term after shifting the origin.

Here is an interesting thing though. For the intersection point \((x_1, y_1)\), the following equations hold:

\[ ax_1 + hy_1 + g = 0 \]

\[ hx_1 + by_1 + f = 0 \]

\[ gx_1 + fy_1 + c = 0 \]

So, we can use any two of these equations to find \(x_1\) and \(y_1\).

We can easily remember these equations using the condition for a pair of straight lines:

\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

This condition can be represented in determinant form as:

\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]

The coefficients of the equations for the intersection point are nothing but the rows of this determinant. Therefore, the equations:

\[ ax_1 + hy_1 + g = 0 \]

\[ hx_1 + by_1 + f = 0 \]

\[ gx_1 + fy_1 + c = 0 \]

correspond to the rows of the determinant:

\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]

This makes it easier to remember and derive these equations for finding the intersection point \((x_1, y_1)\) of the pair of straight lines.

Alternative Method to Calculate Intersection Point

Consider a pair of straight lines represented by the equation:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \tag{1} \]

Let \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \) be the lines represented by (1).

The joint equation of \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \) is:

\[ (m_1x - y + c_1)(m_2x - y + c_2) = 0 \]

Expanding this, we get:

\[ m_1m_2x^2 - (m_1 + m_2)xy + y^2 + (m_1c_2 + m_2c_1)x - (c_1 + c_2)y + c_1c_2 = 0 \tag{2} \]

Equation (1) and Equation (2) represent the same pair of lines, so their coefficients must be proportional:

\[ \frac{m_1m_2}{a} = \frac{-(m_1 + m_2)}{2h} = \frac{1}{b} = \frac{m_1c_2 + m_2c_1}{2g} = \frac{-(c_1 + c_2)}{2f} = \frac{c_1c_2}{c} \]

From these proportions, we get:

\[ m_1 + m_2 = -\frac{2h}{b} \]

\[ m_1m_2 = \frac{a}{b} \]

\[ m_1c_2 + m_2c_1 = \frac{2g}{b} \]

\[ c_1 + c_2 = -\frac{2f}{b} \]

\[ c_1c_2 = \frac{c}{b} \]

To find the intersection point \((x_1, y_1)\), we solve the system of equations given by the lines \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \):

Setting \( y = m_1x + c_1 \) equal to \( y = m_2x + c_2 \):

\[ m_1x + c_1 = m_2x + c_2 \]

Solving for \(x\):

\[ x = \frac{c_2 - c_1}{m_1 - m_2} \]

Next, find \(y\)

\[ \frac{y - c_1}{m_1} = \frac{y - c_2}{m_2} \]

\[ m_2y - c_1m_2 = m_1y - m_1c_2 \]

\[ y = \frac{m_1c_2 - c_1m_2}{m_1 - m_2} \]

Thus, the intersection point \((x_1, y_1)\) is:

\[ x_1 = \frac{c_2 - c_1}{m_1 - m_2} \]

\[ y_1 = \frac{m_1c_2 - c_1m_2}{m_1 - m_2} \]

Consider:

\[ (c_2 - c_1)^2 = (c_1 + c_2)^2 - 4c_1c_2 \]

\[ = \left( \frac{4f^2}{b^2} \right) - \left( \frac{4c}{b} \right) \]

\[ = 4 \left( \frac{f^2 - bc}{b^2} \right) \]

Similarly:

\[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 \]

\[ = \left( \frac{4h^2}{b^2} \right) - \left( \frac{4a}{b} \right) \]

\[ = 4 \left( \frac{h^2 - ab}{b^2} \right) \]

For the expression involving \(y_1\):

\[ (m_1c_2 - m_2c_1)^2 = (m_1c_2 + m_2c_1)^2 - 4m_1m_2c_1c_2 \]

\[ = \left( \frac{4g^2}{b^2} \right) - \left( \frac{4ac}{b^2} \right) \]

\[ = 4 \left( \frac{g^2 - ac}{b^2} \right) \]

Therefore:

\[ x_1^2 = \left( \frac{c_2 - c_1}{m_1 - m_2} \right)^2 = \frac{f^2 - bc}{h^2 - ab} \]

\[ y_1^2 = \left( \frac{m_1c_2 - m_2c_1}{m_1 - m_2} \right)^2 = \frac{g^2 - ac}{h^2 - ab} \]

Thus, the intersection point \((x_1, y_1)\) is:

\[ \boxed{\left( \pm \sqrt{\frac{f^2 - bc}{h^2 - ab}}, \pm \sqrt{\frac{g^2 - ac}{h^2 - ab}} \right)}\]

This formula contains some ambiguity regarding signs but can be a useful formula when we are interested in \(x_1^2\) and \(y_1^2\).

Example

Like for example, the distance of the point of intersection from the origin can be calculated as follows:

Given the coordinates of the intersection point \((x_1, y_1)\), where:

\[ x_1^2 = \frac{f^2 - bc}{h^2 - ab} \]

\[ y_1^2 = \frac{g^2 - ac}{h^2 - ab} \]

The distance \(d\) from the origin to the intersection point \((x_1, y_1)\) is:

\[ d = \sqrt{x_1^2 + y_1^2} \]

Substituting the expressions for \(x_1^2\) and \(y_1^2\):

\[ d = \sqrt{\frac{f^2 - bc}{h^2 - ab} + \frac{g^2 - ac}{h^2 - ab}} \]

Combining the terms under the square root:

\[ d = \sqrt{\frac{f^2 - bc + g^2 - ac}{h^2 - ab}} \]

This simplifies to:

\[ d = \sqrt{\frac{f^2 + g^2 - bc - ac}{h^2 - ab}} \]

Since the numerator \(f^2 + g^2 - c(a + b)\) is a combination of terms involving \(f\), \(g\), \(a\), \(b\), and \(c\), we can write the distance as:

\[ d = \sqrt{\frac{f^2 + g^2 - c(a + b)}{h^2 - ab}} \]

Thus, the distance of the point of intersection from the origin is:

\[ d = \sqrt{\frac{f^2 + g^2 - c(a + b)}{h^2 - ab}} \]

This provides a succinct way to calculate the distance from the origin to the intersection point of the pair of straight lines represented by the equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\).