Concurrent Lines

Have you ever been at a crossroads where three or more roads meet at the same spot during a journey? Something similar happens in geometry; instead of roads, we refer to lines. In geometry, this intersection is referred to as concurrent lines. These concurrent lines are not random; in fact, these lines follow some special rules, especially in triangles. Here, we will deeply understand some important facts related to such lines. 

1.0Concurrent Lines Definition 

In geometry, concurrent lines are three or more lines that meet at a single point. That point is called the point of concurrency. The most important part of this definition is that all the lines must meet at one shared point. Concurrency is different from the intersection of lines because when only two lines meet, we just say they "intersect" or "cross", – while concurrency has three or more lines that interact with each other. 

Mathematically, if lines L₁, L₂, and L₃ all pass through point P(x,y), then the lines are said to be concurrent, and P is the point of concurrency. 

2.0Point of Concurrency

As discussed earlier, the point of concurrency is the specific point at which concurrent lines cross. This point is specific to each particular set of lines. This point also possesses some special properties, which include: 

• When the context of triangles is considered, it can be on the inside, outside, or within the triangle, depending on the lines and the triangle.
• In coordinate geometry, it is obtained by solving the system of equations of lines.

3.0How to Prove That the Lines Are Concurrent Lines? 

To prove that the lines are concurrent lines, there are two particularly important methods. These methods are: 

Algebraic Method 

This method involves solving the equations of lines using the standard methods of algebra. This technique is highly useful in cases when the equations of all three lines are given. This method includes the following steps: 

• Solve two equations together to determine their point of intersection.
• Plug the point of intersection into the equation of the third line.
• If the point works in the third equation, then all three lines intersect at the same point → the lines are concurrent. 

Method of Determinants 

The concurrency of lines can also be proved with the method of determinants. To check the concurrency, the lines must be in standard form like this:  

Let the three equations of lines be: 

• a₁​x + b₁​y + c₁​ = 0
• a₂x + b₂y + c₂ = 0
• a₃x + b₃y + c₃ = 0

And satisfy the following conditions: 

$\Delta =\left|\begin{array}{lll}{a}_1& b_1& c_3\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

This approach verifies if the system of three linear equations shares a common solution — i.e., one point common to all three lines. A zero determinant indicates that the system is dependent and hence crosses at a common point.

4.0Concurrent Lines of a Triangle

In triangle geometry, concurrency appears in several interesting ways. Some special lines that are constructed from the sides or vertices of a triangle always concur and are always concurrent, meaning that they intersect at a point, and each of these points is named and possesses special properties. These include: 

Medians → Centroid A median joins a vertex to the opposite side's midpoint. All medians of a triangle intersect at a point referred to as the centroid. The centroid bisects each median in a 2:1 proportion (vertex to midpoint). It's also called the centre of mass or balance point of the triangle.
Altitudes → Orthocenter An altitude is a perpendicular drawn from a vertex to the opposite side (or its extension). The three altitudes meet at the orthocenter. The orthocenter's location varies with the triangle: Inside for acute triangles, On the triangle for right triangles, Outside for obtuse triangles.
Angle Bisectors → Incenter An angle bisector cuts a triangle's angle into two equal parts. The three angle bisectors converge at the incenter. The incenter is equidistant from all three sides of the triangle and is the centre of the inscribed circle.
Perpendicular Bisectors → Circumcenter A perpendicular bisector divides a side into two equal parts at right angles.  Perpendicular bisectors of the three sides meet at the circumcenter.  The circumcenter is at the same distance from all three vertices and is the centre of the circumscribed circle.

5.0Non-Concurrent and Concurrent Lines 

In geometry, understanding the difference between non-concurrent lines and concurrent lines is extremely important, as this can help identify the concurrent lines much more easily. Some of the differences between these lines are: 

Concurrent Lines	Non-Concurrent Lines
Three or more lines intersect at a single point.	Lines that do not all intersect at the same point.
They share a single point of intersection.	They can intersect in pairs or be entirely distinct.
Example: Medians of a triangle.	Example: Two parallel lines intersected by a transversal.

6.0Solved Examples of Concurrent Lines 

Problem 1: Find the point of concurrency for the following concurrent lines: 

L₁ = x + y = 6
L₂ = 2x − y = 3

L₃ = x + 2y = 9

Solution: Solve the first two equations of the line; for this, add both equations: 

x + y = 6
2x − y = 3

x + y +2x – y = 6 + 3 

3x = 9 

x = 3 

For y, put the value of x in the first equation: 

9 + y = 6 

y = 3

Verify the points (3,3) by substituting the values in the third equation: 

3 + 2(3) = 9 

3 + 6 = 9 

9 = 9

Hence, the point of concurrency for the equations is (3,3). 

Problem 2: Show if the following lines are concurrent or not with the algebraic method: 

L₁​: x + y = 5 ….. (1)
L₂: 2x − y = 4 …..(2)
L₃: 3x + y = 9 …..(3)

Solution: Solve the equations 1 and 2 for x and y.For this, add equations 1 and 2

x + y + 2x – y = 5 + 4 

3x = 9 

x = 3

Now find y: 

y = 5−3 = 2

Point of intersection = (3, 2)

To check if the lines are concurrent, put the value of the point of intersection in equation 3: 

3(3) + 2 = 9 

9 + 2 = 9

$\text{ since }11\ne 9\text{, the given lines are not concurrent. }$

Problem 3: Prove that the following lines are concurrent using determinants:

2x + y − 5 = 0
x − y + 1 = 0
3x − 4 = 0

Solution: For the solution, write the coefficients of the equation in a 3×3 determinant:

$\Delta =\left|\begin{array}{ccc}2& 1& -5\\ 1& -1& 1\\ 3& 0& -4\end{array}\right|$

Now, expand the determinant: 

$\begin{array}{rl}& \Delta =2\left|\begin{array}{rr}-1& 1\\ 0& -4\end{array}\right|-1\left|\begin{array}{rr}1& 1\\ 3& -4\end{array}\right|+(-5)\left|\begin{array}{rr}1& -1\\ 3& 0\end{array}\right|\\ & \Delta =2(4-0)-1(-4-3)-5(0-(-3))\\ & \Delta =8+7-15=15-15\\ & \Delta =0\end{array}$

The determinant is zero; hence, the lines are concurrent.