Interaction between Lines
Intersection of two Lines
Consider two lines given by the equations:
There are three possible cases for the relationship between these two lines:
Case 1: Parallel Lines
Two lines are parallel if:
In this case, the lines do not intersect.
Case 2: Coincident Lines
Two lines are coincident if:
In this case, the lines overlap completely and have infinitely many points of intersection.
Case 3: Intersecting Lines
Two lines intersect at a unique point if:
In this case, the lines intersect at a single point. The intersection point \((x, y)\) is given by:
From this, the coordinates of the intersection point can be found as:
Proof
To prove that two lines \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) intersect at a unique point when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), we can use the elimination method to solve the system of linear equations formed by these two lines.
Given:
Eliminate \(y\) First

Multiply Each Equation to Align Coefficients of \(y\):
Multiply equation (1) by \(b_2\) and equation (2) by \(b_1\) to make the coefficients of \(y\) align:
\[ b_2(a_1x + b_1y + c_1) = 0 \quad \Rightarrow \quad a_1b_2x + b_1b_2y + b_2c_1 = 0 \quad \text{(3)} \]\[ b_1(a_2x + b_2y + c_2) = 0 \quad \Rightarrow \quad a_2b_1x + b_2b_1y + b_1c_2 = 0 \quad \text{(4)} \] 
Subtract the Equations to Eliminate \(y\):
Subtract equation (4) from equation (3):
\[ (a_1b_2x + b_1b_2y + b_2c_1)  (a_2b_1x + b_2b_1y + b_1c_2) = 0 \]Simplify the equation:
\[ a_1b_2x  a_2b_1x + b_2c_1  b_1c_2 = 0 \]\[ (a_1b_2  a_2b_1)x = b_1c_2  b_2c_1 \] Solve for \(x\): Since \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), we know that \(a_1b_2  a_2b_1 \ne 0\). Therefore, we can solve for \(x\):
\[ x = \frac{b_1c_2  b_2c_1}{a_1b_2  a_2b_1} \]** Eliminate \(x\) Next**

Multiply Each Equation to Align Coefficients of \(x\):
Multiply equation (1) by \(a_2\) and equation (2) by \(a_1\) to make the coefficients of \(x\) align:
\[ a_2(a_1x + b_1y + c_1) = 0 \quad \Rightarrow \quad a_1a_2x + a_2b_1y + a_2c_1 = 0 \quad \text{(5)} \]\[ a_1(a_2x + b_2y + c_2) = 0 \quad \Rightarrow \quad a_2a_1x + a_1b_2y + a_1c_2 = 0 \quad \text{(6)} \] 
Subtract the Equations to Eliminate \(x\):
Subtract equation (6) from equation (5):
\[ (a_1a_2x + a_2b_1y + a_2c_1)  (a_2a_1x + a_1b_2y + a_1c_2) = 0 \]Simplify the equation:
\[ a_2b_1y  a_1b_2y + a_2c_1  a_1c_2 = 0 \]\[ (a_2b_1  a_1b_2)y = a_1c_2  a_2c_1 \] 
Solve for \(y\):
Since \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\), we know that \(a_2b_1  a_1b_2 \ne 0\). Therefore, we can solve for \(y\):
\[ y = \frac{a_1c_2  a_2c_1}{a_2b_1  a_1b_2} \]Conclusion
Therefore, the intersection point \((x, y)\) is given by:
\[ x = \frac{b_1c_2  b_2c_1}{a_1b_2  a_2b_1} \]\[ y = \frac{a_1c_2  a_2c_1}{a_2b_1  a_1b_2} \]This completes the proof that two lines intersect at a unique point when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\).
Example
Given the two lines:
Let's determine the intersection point.

Check the ratios to ensure the lines are neither parallel nor coincident:
\[ \frac{2}{4} \neq \frac{3}{1} \]Since the ratios are not equal, the lines intersect at a unique point.

Calculate the intersection point:
\[ x = \frac{(3)(2)  (1)(5)}{(2)(1)  (4)(3)} = \frac{6 + 5}{2  12} = \frac{1}{14} = \frac{1}{14} \]\[ y = \frac{(4)(5)  (2)(2)}{(2)(1)  (4)(3)} = \frac{20 + 4}{2  12} = \frac{24}{14} = \frac{12}{7} \]Thus, the intersection point is:
\[ \left( \frac{1}{14}, \frac{12}{7} \right) \]By following these steps, you can determine whether two lines are parallel, coincident, or intersecting, and if they intersect, you can find the intersection point.
Intersection of Three Lines at a Common Point
Given three lines in 2D space:
These three lines intersect at a common point if and only if the determinant of the coefficients is zero. The determinant is formed by placing the coefficients of \(x\), \(y\), and the constants from the three equations into a matrix. The condition for the lines to intersect at a common point is:
Proof
To prove that three lines \( L_1 \), \( L_2 \), and \( L_3 \) intersect at a common point if and only if the determinant of their coefficients is zero, we can first find the intersection point of \( L_1 \) and \( L_2 \) using the previously derived formula. Then, we will substitute this intersection point into the equation of \( L_3 \) and derive the determinant condition.
Given the equations of the three lines:
Step 1: Find the Intersection of \( L_1 \) and \( L_2 \)
Using the formula for the intersection point of two lines:
Let's denote this intersection point as \( (x_0, y_0) \):
Step 2: Substitute \( (x_0, y_0) \) into the Equation of \( L_3 \)
Substitute \( x_0 \) and \( y_0 \) into the equation of \( L_3 \):
Substitute the expressions for \( x_0 \) and \( y_0 \):
Step 3: Simplify the Expression
Multiply through by \( a_1b_2  a_2b_1 \) to clear the denominators:
Step 4: Expand and Rearrange
Expand and rearrange the terms:
Step 5: Form the Determinant
This equation can be written in the form of a determinant:
Conclusion
Thus, the three lines \( L_1 \), \( L_2 \), and \( L_3 \) intersect at a common point if and only if the determinant of the coefficients is zero:
This condition ensures that there exists a single point \((x, y)\) that satisfies all three equations, indicating that the lines intersect at that point.
Parallel Lines

Condition for Parallel Lines
Two lines \( L_1: a_1x + b_1y + c_1 = 0 \) and \( L_2: a_2x + b_2y + c_2 = 0 \) are parallel if:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \]This condition means that the ratios of the coefficients of \(x\) and \(y\) are equal, indicating that the lines have the same slope and will never intersect.

Equation of a Line Parallel to a Given Line
The equation of a line parallel to \(ax + by + c = 0\) can be written as \(ax + by + k = 0\), where \(k\) is a constant. The only difference between the two equations is the constant term, which translates the line parallelly without changing its slope.
For example, if the given line is:
\[ 3x + 4y + 5 = 0 \]A parallel line would have the form:
\[ 3x + 4y + k = 0 \]where \(k\) is a different constant.

Equation of a Parallel Line Passing Through a Given Point
To find the equation of a line parallel to \(ax + by + c = 0\) and passing through a specific point \((x_1, y_1)\), we use the following steps:
 Start with the general form of the parallel line: \(ax + by + k = 0\).
 Substitute the coordinates of the given point \((x_1, y_1)\) into the equation to find \(k\).
The equation of the parallel line passing through \((x_1, y_1)\) is:
\[ ax + by + k = 0 \]Substitute \((x_1, y_1)\):
\[ a(x_1) + b(y_1) + k = 0 \]Solving for \(k\), we get:
\[ k =  (ax_1 + by_1) \]Thus, the equation of the line parallel to \(ax + by + c = 0\) and passing through \((x_1, y_1)\) is:
\[ a(x  x_1) + b(y  y_1) = 0 \]This can also be written as:
\[ ax + by  (ax_1 + by_1) = 0 \]Example
Given the line:
\[ 2x + 3y + 6 = 0 \]To find a parallel line passing through the point \((1, 2)\), we use the formula:
\[ 2(x  1) + 3(y + 2) = 0 \]Simplifying:
\[ 2x  2 + 3y + 6 = 0 \]\[ 2x + 3y + 4 = 0 \] 
Distance Between Two Parallel Lines
The distance between two parallel lines \(L_1: ax + by + c_1 = 0\) and \(L_2: ax + by + c_2 = 0\) is given by:
\[ d = \frac{c_1  c_2}{\sqrt{a^2 + b^2}} \]Proof:

Assume a Point on \(L_1\):
Let \(P(x_1, y_1)\) be a point on the line \(L_1\). Therefore, the coordinates \((x_1, y_1)\) satisfy the equation of \(L_1\):
\[ ax_1 + by_1 + c_1 = 0 \] 
Distance from Point to Line:
The distance from point \(P(x_1, y_1)\) to the line \(L_2: ax + by + c_2 = 0\) is given by the distance formula for a point to a line:
\[ d = \frac{ax_1 + by_1 + c_2}{\sqrt{a^2 + b^2}} \] 
Substitute \(x_1\) and \(y_1\) into \(L_1\):
Since \(P(x_1, y_1)\) lies on \(L_1\), we have:
\[ ax_1 + by_1 = c_1 \] 
Replace \(ax_1 + by_1\) in the Distance Formula:
Substitute \(ax_1 + by_1 = c_1\) into the distance formula:
\[ d = \frac{ c_1 + c_2 }{\sqrt{a^2 + b^2}} \]Simplify the absolute value expression:
\[ d = \frac{c_2  c_1}{\sqrt{a^2 + b^2}} \] 
Conclusion:
Therefore, the distance between the two parallel lines \(L_1\) and \(L_2\) is:
\[ d = \frac{c_1  c_2}{\sqrt{a^2 + b^2}} \]


Equation of a Line Parallel to \( ax + by + c = 0 \) at a Distance \( d \)
The equations of lines parallel to \( ax + by + c = 0 \) at a distance \( d \) from it are given by:
\[ ax + by + c \pm d \sqrt{a^2 + b^2} = 0 \]Proof:
To find the equation of a line parallel to \( ax + by + c = 0 \) at a distance \( d \), we start by assuming that the equation of the parallel line is \( ax + by + k = 0 \).
The distance \( d \) between the two parallel lines \( ax + by + c = 0 \) and \( ax + by + k = 0 \) is given by the formula:
\[ d = \frac{k  c}{\sqrt{a^2 + b^2}} \]First, we multiply both sides of the equation by \( \sqrt{a^2 + b^2} \) to clear the denominator:
\[ k  c = d \sqrt{a^2 + b^2} \]The absolute value equation \( k  c = d \sqrt{a^2 + b^2} \) can be split into two linear equations:
\[ k  c = d \sqrt{a^2 + b^2} \quad \text{or} \quad k  c = d \sqrt{a^2 + b^2} \]Solving these equations for \( k \), we get:
\[ k = c + d \sqrt{a^2 + b^2} \]\[ k = c  d \sqrt{a^2 + b^2} \]Therefore, the equations of the lines parallel to \( ax + by + c = 0 \) at a distance \( d \) are:
\[ ax + by + \left(c + d \sqrt{a^2 + b^2}\right) = 0 \]\[ ax + by + \left(c  d \sqrt{a^2 + b^2}\right) = 0 \]We can combine these results into a single expression:
\[ ax + by + c \pm d \sqrt{a^2 + b^2} = 0 \]Hence, the equations of the lines parallel to \( ax + by + c = 0 \) at a distance \( d \) from it are:
\[ ax + by + c \pm d \sqrt{a^2 + b^2} = 0 \] 
Equation of a Line Parallel to Two Given Parallel Lines and Dividing the Region in a Given Ratio
Given two parallel lines \( L_1: ax + by + c_1 = 0 \) and \( L_2: ax + by + c_2 = 0 \), a line \( L \) parallel to both \( L_1 \) and \( L_2 \) that divides the region between \( L_1 \) and \( L_2 \) in the ratio \( m:n \) is given by:
\[ ax + by + \frac{mc_2 + nc_1}{m + n} = 0 \]Proof:
Consider the two parallel lines \( L_1: ax + by + c_1 = 0 \) and \( L_2: ax + by + c_2 = 0 \). Let \( P(x_1, y_1) \) be a point on \( L_1 \) and \( Q(x_2, y_2) \) be a point on \( L_2 \). Assume that the line \( L \), which is parallel to both \( L_1 \) and \( L_2 \), intersects the line segment \( PQ \) at point \( R \), dividing \( PQ \) in the ratio \( m:n \).
The coordinates of point \( R \) can be determined using the section formula. Thus, \( R \) has coordinates:
\[ R = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \]Since \( L \) is parallel to \( L_1 \) and \( L_2 \), its equation can be written in the form \( ax + by + k = 0 \). To determine the value of \( k \), we use the fact that the distance between the parallel lines \( L_1 \) and \( L_2 \) is divided in the ratio \( m:n \) by the line \( L \).
Consider the equation of line \( L \) in the form \( ax + by + k = 0 \) passing through point \( R \). Substituting the coordinates of \( R \) into the equation of \( L \) gives:
\[ a \left( \frac{mx_2 + nx_1}{m + n} \right) + b \left( \frac{my_2 + ny_1}{m + n} \right) + k = 0 \]Simplifying the above equation, we get:
\[ \frac{a(mx_2 + nx_1) + b(my_2 + ny_1)}{m + n} + k = 0 \]Since \( x_1, y_1 \) and \( x_2, y_2 \) satisfy the equations of \( L_1 \) and \( L_2 \), respectively, we can write:
\[ ax_1 + by_1 = c_1 \]\[ ax_2 + by_2 = c_2 \]Substituting these into our equation, we have:
\[ \frac{a(mx_2 + nx_1) + b(my_2 + ny_1)}{m + n} + k = 0 \]\[ \frac{m(ax_2 + by_2) + n(ax_1 + by_1)}{m + n} + k = 0 \]\[ \frac{m(c_2) + n(c_1)}{m + n} + k = 0 \]\[ \frac{mc_2  nc_1}{m + n} + k = 0 \]Solving for \( k \), we get:
\[ k = \frac{mc_2 + nc_1}{m + n} \]Therefore, the equation of the line \( L \) parallel to \( L_1 \) and \( L_2 \) and dividing the region between them in the ratio \( m:n \) is:
\[ ax + by + \frac{mc_2 + nc_1}{m + n} = 0 \]This completes the proof that the equation of the parallel line dividing the region in the given ratio is given by the above formula.
Example
Given two parallel lines:
\[ L_1: 2x + 3y + 6 = 0 \]\[ L_2: 2x + 3y  4 = 0 \]We want to find the equation of a line parallel to both \( L_1 \) and \( L_2 \) that bisects the region between them. This means the ratio \( m:n \) is 1:1.
Using the formula:
\[ ax + by + \frac{mc_2 + nc_1}{m + n} = 0 \]For \( L_1 \) and \( L_2 \):
 \( a = 2 \)
 \( b = 3 \)
 \( c_1 = 6 \)
 \( c_2 = 4 \)
 \( m = 1 \)
 \( n = 1 \)
Substituting these values:
\[ 2x + 3y + \frac{1 \cdot (4) + 1 \cdot 6}{1 + 1} = 0 \]\[ 2x + 3y + \frac{4 + 6}{2} = 0 \]\[ 2x + 3y + 1 = 0 \]The equation of the bisecting line is:
\[ 2x + 3y + 1 = 0 \]This line bisects the region between \( L_1 \) and \( L_2 \).
Ceva's Theorem
If the lines joining any point \( O \) to the vertices \( A \), \( B \), and \( C \) of a triangle meet the opposite sides in \( D \), \( E \), and \( F \) respectively, then:
Proof
Let's place \(\triangle ABC\) in a Cartesian coordinate system such that \( A \), \( B \), and \( C \) have coordinates \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) respectively, and let point \( O \) be \((0, 0)\).
The equation of \( AD \) can be written as:
The line \( AD \) divides \( BC \) in the ratio:
Similarly, for the line \( BE \), the ratio \(\frac{CE}{EA}\) is:
And for the line \( CF \), the ratio \(\frac{AF}{FB}\) is:
By multiplying these ratios together, we get:
The negative signs cancel out, and we have:
Thus, we have proven Ceva's theorem.
Let \( AD \), \( BE \), and \( CF \) be concurrent at point \( X \).
From the diagram, we can start by considering the ratio \(\frac{BD}{DC}\):
Using the areas of triangles:
Here, we used the fact that \(\frac{a}{b} = \frac{c}{d} = \frac{a  c}{b  d}\) to simplify.
Similarly, for the ratio \(\frac{CE}{EA}\):
And for the ratio \(\frac{AF}{FB}\):
Now, combining these ratios, we get:
Thus, we have proven Ceva's theorem.
Example
Problem: Let \( \triangle ABC \) be a triangle with a line \( AD \) dividing \( BC \) in the ratio \( 1:2 \) and a line \( BE \) dividing \( CA \) in the ratio \( 2:3 \). The lines \( AD \) and \( BE \) intersect at point \( X \). Determine the ratio in which the line \( CX \) cuts \( AB \).
Solution:
Given the ratios \(\frac{BD}{DC} = \frac{1}{2}\) and \(\frac{CE}{EA} = \frac{2}{3}\), we will use Ceva's theorem to find the ratio \(\frac{AF}{FB}\), where \( F \) is the intersection of \( CX \) with \( AB \).
Ceva's theorem states that if the lines \( AD \), \( BE \), and \( CF \) are concurrent, then:
Substituting the given ratios into Ceva's theorem:
Let \(\frac{AF}{FB} = k\). Then, the equation becomes:
Simplifying the equation:
Therefore, \(\frac{AF}{FB} = 3\). Thus, the line \( CX \) cuts \( AB \) in the ratio \( 3:1 \).
Menelau's Theorem
Consider a triangle \( \Delta ABC \). If a transversal cuts the sides \( BC \), \( CA \), and \( AB \) at \( D \), \( E \), and \( F \) respectively, then
Proof:
Let the coordinates of vertices \( A \), \( B \), and \( C \) be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \). Assume the equation of the transversal \( T \) is given by \( ax + by + c = 0 \).
The ratio in which the transversal \( T \) divides the side \( BC \),
Similarly, for \( E \) on \( CA \),
and for \( F \) on \( AB \),
Multiplying these ratios,
Each term in the numerator and the denominator cancels out, leading to
Hence, Menelaus' Theorem is proved.
The product is negative, why?
In Menelaus' theorem, the transversal either cuts all three sides of the triangle externally or cuts two sides internally and one side externally. It is not possible for the transversal to cut all three sides internally.
To understand why, consider the following:

All sides cut externally: When the transversal cuts all three sides externally, it intersects the extensions of the sides of the triangle. This configuration is depicted by the negative product of the ratios, consistent with the theorem.

Two sides cut internally, one externally: In this case, the transversal intersects two sides of the triangle within the segments connecting the vertices, while intersecting the third side externally. This configuration is also valid under Menelaus' theorem, as the product of the ratios still results in \(1\), indicating the correct relationship.

All sides cut internally (impossible): If the transversal were to cut all three sides internally, it would imply that the transversal lies completely within the triangle, forming three intersection points strictly within the sides. However, geometrically, this configuration is not feasible while maintaining the conditions set by Menelaus' theorem. The intersection points would not align correctly to satisfy the required ratio product of \(1\).