Concurrent Lines: Three or more lines passing through a single point in a plane are called concurrent lines. Whenever two non-parallel lines coincide with each other, they form a point of intersection. When another line also passes through the point of intersection made by the first two lines, these three lines are said to be concurrent lines.

The point at which all the three lines meet is called the “Point of Concurrency”. For example, we can see that three altitudes drawn on a triangle intersect at a point called the orthocentre. It is to be noted that only non-parallel lines can have a point of concurrence since they extend indefinitely and meet at a point somewhere.

Learn About Different Types of Lines

Concurrent Lines Definition

Lines (three or more) that pass through a single point on a Cartesian plane are called concurrent lines. The point where the concurrent lines intersect is called the point of concurrency.

A few examples include the diameter of a circle that is concurrent at the centre of a circle. In quadrilaterals, the line segments joining the midpoints of opposite sides and the diagonals are concurrent. In the figure below, the three lines intersect at point \({\rm{P}}.\) All the three lines are concurrent with each other.

Concurrent Lines Formula and Condition for Three Lines to be Concurrent

Method 1:

If three lines are said to be concurrent, then the point of intersection of two lines lies on the third line. Assume the equations of three lines as:
\({a_1}x + {b_1}y + {c_1} = 0\)……(i)
\({a_2}x + {b_2}y + {c_2} = 0\)……(ii)
\({a_3}x + {b_3}y + {c_3} = 0\)……(iii)
Thus, the condition for three lines concurrent to each other is given by:
\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

Concurrent Lines Example

Find the point where the set of lines \(ax + by + c = 0\) and \(5a + 6b + 7c = 0\) are concurrent. \(ax + by + c = 0 \Rightarrow \frac{{ax}}{{ – c}} + \frac{{by}}{{ – c}} = 1\)
\( \Rightarrow 5a + 6b + 7 = 0\)
\( \Rightarrow \frac{a}{{\left( {\frac{{ – 7}}{5}} \right)}} + \frac{b}{{\left( {\frac{{ – 7}}{6}} \right)}} = 1\)
Hence, the equation passes through \(\left( {\frac{5}{7},\,\frac{6}{7}} \right).\)

Method 2:

To check if three lines are concurrent, we first find the point of intersection of two lines and then check to see if the third line passes through the intersection point. It will ensure that all three lines are concurrent. Let us understand this better with an example. The equations of any three lines are as follows.
\(2x – y – 2 = 0\)…..(i)
\(y = x + 2\)…..(ii)
\(2x + 3y = 26\)……(iii)

Step 1: To find the point of intersection of line \(1\) and line \(2,\) solve the equations \(\left( 1 \right)\) and \(\left( 2 \right)\) by the substitution method.
Substituting the value of ? from equation \(\left( 2 \right)\) in equation \(\left( 1 \right),\) we get
\(2x – \left( {x + 2} \right) – 2 = 0\)
\( \Rightarrow 2x – x – 2 – 2 = 0\)
\( \Rightarrow x – 4 = 0\)
\( \Rightarrow x = 4.\)
Substituting the value of \(x = 4\) in equation \(\left( 2 \right),\) we get the value of \(y.\)
\(y = x + 2\)….(ii)
\( \Rightarrow y = 4 + 2\)
\( \Rightarrow y = 6\)
Therefore, line \(1\) and line \(2\) intersect at a point \(\left( {4,\,6} \right).\)

Step 2: Substitute the point of intersection of the first two lines in the equation of the third line.
The equation of the third line is \(2x + 3y = 26\)….(iii)
Substituting the values of \(\left( {4,\,6} \right)\) in equation (iii), we get
\( \Rightarrow 2\left( 4 \right) + 3\left( 6 \right) = 26\)
\( \Rightarrow 8 + 18 = 26\)
\( \Rightarrow 26 = 26\)
Therefore, the point of intersection goes right with the third line equation, which means the three lines intersect each other and are concurrent lines.

Concurrent Lines and Point of Concurrency

Three or more lines in a plane intersecting each other at a single common point are called concurrent lines. Two lines in a plane intersecting one another at one common point are called intersecting lines. The common point where all the concurrent lines meet each other is called the point of concurrency. In the figure given below, point \({\rm{P}}\) is the point of concurrency.

Concurrent Lines Geometry

Triangle

In a triangle, \(4\) basic types of sets of concurrent lines are altitudes, angle bisectors, medians, and perpendicular bisectors.

Quadrilateral

The two segments joining the midpoints of opposite sides and the line segment joining the midpoints of the diagonals are concurrent. They are all bisected by their point of intersection.

In a cyclic quadrilateral, \(4\) line segments, each perpendicular to one side and passing through the opposite side’s midpoint, are concurrent.

Polygon

Suppose a regular polygon has an even number of sides. In that case, the diagonals joining opposite vertices are concurrent at the centre of the polygon.

Circle

The perpendicular bisectors of all the chords of a circle are concurrent at the centre of the circle.
All perimeter bisectors and area bisectors of a circle are diameters, and they are concurrent at the circle’s centre.
The lines perpendicular to the tangents to a circle at the points of tangency are concurrent at the centre.

Concurrent Lines in a Triangle

A triangle is a two-dimensional shape that has three sides and three angles. Concurrent lines can be seen inside triangles when some particular types of line segments are drawn inside them. We can locate four different points of concurrency in a triangle. 

(i) Incenter: The point of intersection of three angular bisectors inside a triangle is called the incenter of a triangle.

(ii) Circumcenter: The point of intersection of three perpendicular bisectors inside a triangle is called the circumcenter of a triangle.

(iii) Centroid: The point of intersection of the three medians of a triangle is called the centroid of a triangle.

(iv) Orthocenter: The point of intersection of three altitudes of a triangle is called the orthocenter of a triangle.

Solved Examples – Concurrent Lines

Q.1. Show that the lines \(4x – 6y + 10 = 0,\,6x + 8y – 14 = 0\) and \(18x – 10y + 16 = 0\) are concurrent.
Ans: We know that if the equations of three straight lines \({a_1}x + {b_1}y + {c_1} = 0,\,{a_2}x + {b_2}y + {c_2} = 0\) and \({a_3}x + {b_3}y + {c_3} = 0\) are concurrent, then
\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)
The given lines are \(4x – 6y + 10 = 0,\,6x + 8y – 14 = 0\) and \(18x – 10y + 16 = 0\)
We have
\(\left| {\begin{array}{*{20}{c}} 4&{ – 6}&{10}\\ 6&8&{ – 14}\\ {18}&{ – 10}&{16} \end{array}} \right| = 0\)
\( \Rightarrow 4\left( {128 – 140} \right) + 6\left( {96 + 252} \right) + 10\left( { – 60 – 144} \right)\)
\( = \, – 48 + 2088 – 2040\)
\( = 2088 – 2088\)
\( = 0\)
Therefore, the three straight lines given are concurrent.

Q.2. From the figure given below, find out the concurrent lines and the point of concurrency.

Ans: The straight lines \(AE,\,BF,\,CG\) and \(DH\) are concurrent lines because these lines are passing through a single point \(‘O’.\)
Therefore, \(O\) is the point of concurrency.

Q.3. Verify whether the following lines are concurrent or not. The line equations are, \(x + 2y – 4 = 0,\,x – y – 1 = 0,\,4x + 5y – 13 = 0.\)
Ans: To check if three lines are concurrent, the following condition should be satisfied.
\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)
Comparing the given three line equations to \({a_1}x + {b_1}y + {c_1} = 0,\,{a_2}x + {b_2}y + {c_2} = 0\) and \({a_3}x + {b_3}y + {c_3} = 0,\) let us find the values of \({a_1},\,{a_2},\,{a_3},\,{b_1},\,{b_2},\,{b_3},\,{c_1},\,{c_2}\) and \({c_3}\)
\({a_1} = 1,\,{b_1} = 2,\,{c_1} = \, – 4\)
\({a_2} = 1,\,{b_2} = \, – 1,\,{c_2} = \, – 1\)
\({a_3} = 4,\,{b_3} = \,5,\,{c_3} = \, – 13\)
Arranging them in the determinants form, we get
\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)
On solving this, we get
\( \Rightarrow 1\left( {13 + 5} \right) – 2\left( { – 13 + 4} \right) – 4\left( {5 + 4} \right)\)
\( = 18 + 18 – 36\)
\( = 36 – 36\)
\( = 0\)
The above condition holds good for the three lines. Therefore, the three lines are concurrent.

Q.4. If the lines \(2x + y – 3 = 0,\,5x + ky – 3 = 0\) and \(3x – y – 2 = 0\) are concurrent, find the value of \(k.\)
Ans: the condition, if the three lines are concurrent to each other, is;
\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)
Substituting the values in the condition to find \(k\)
\(\left| {\begin{array}{*{20}{c}} 2&1&{ – 3}\\ 5&k&{ – 3}\\ 3&{ – 1}&{ – 2} \end{array}} \right| = 0\)
\(2\left[ {k \times \left( { – 2} \right) – 3} \right] – 1\left[ {\left( {5 \times – 2} \right) – \left( {3 \times – 3} \right)} \right] – 3\left[ {\left( {5 \times – 1} \right) – 3 \times k} \right] = 0\)
\( \Rightarrow \, – 4k – 6 + 1 + 15 + 9k = 0\)
\( \Rightarrow 5k + 10 = 0\)
\( \Rightarrow k = \, – 2\)

Q.5. Show that the three lines \(3p – 4q + 5 = 0,\,7p – 8q + 5 = 0\) and \(4p + 5q = 45\) are concurrent.
Ans: Let
\(3p – 4q + 5 = 0\)….(i)
\(7p – 8q + 5 = 0\)…..(ii)
\(4p + 5q = 45\)….(iii)
Let us use the substitution method and solve equations \(1\) and \(2\) given above
\(3p – 4q + 5 = 0\)….(i)
\(7p – 8q + 5 = 0\) or \(7p – 2\left( {4q} \right) + 5 = 0\)
Now substituting \(4q = 3p + 5\)….(iii)
\( \Rightarrow 7p – 2\left( {3p + 5} \right) + 5 = 0\)
\( \Rightarrow 7p – 6p – 10 + 5 = 0\)
\( \Rightarrow p – 5 = 0\)
\( \Rightarrow p = 5\)
From equation (i), we get \(3 \times 5 – 4q + 5 = 0\)
\( \Rightarrow 4q = 20\)
\( \Rightarrow q = 5\)
Hence, the point of intersection of lines \(\left( 1 \right)\) and \(\left( 2 \right)\) is \(\left( {5,\,5} \right).\)
Substituting the values \(\left( {5,\,5} \right).\) in equation (iii), we get
\(4p + 5q = 45\)
\( \Rightarrow 4 \times 5 + 5 \times 5 = 45\)
\( \Rightarrow 20 + 25 = 45\)
\( \Rightarrow 45 = 45\)
Hence, the given three lines are concurrent and pass through the point of concurrency \(\left( {5,\,5} \right).\)

Summary

In this article, we defined concurrent lines, listed the difference between concurrent lines and intersecting lines. We also learnt the condition for three lines to be concurrent.

Also, we studied concurrent lines in geometry, concurrent lines in the triangle formed by the point of intersection of three angular bisectors called the incenter, the point of intersection of three perpendicular bisectors called the circumcenter, the point of intersection of three medians called the centroid, and lastly, the point of intersection of three altitudes called the orthocenter of a triangle. Also, solved examples that are related to concurrent lines are discussed.

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Frequently Asked Questions (FAQs) – Concurrent Lines

Q.1. What are concurrent lines?
Ans: When three or more line segments intersect each other at a single point, then they are said to be concurrent lines.

Q.2. What is the difference between intersecting lines and concurrent lines?
Ans:

Intersecting Lines	Concurrent Lines
Only two lines intersect each other.	Three or more lines pass through a common point.
The point where two lines intersect is called the intersection point or the point of intersection.	The single point at which these lines intersect each other is called a concurrency point or a point of concurrency.
The figure shows intersecting lines.	The figure shows concurrent lines.

Q.3. Are medians of triangle concurrent?
Ans: Medians of a triangle intersect each other at a single point. Thus, they are referred to as concurrent, and the common point where they intersect is the centroid of the triangle.

Q.4. How to prove that two lines are concurrent?
Ans: Two lines in a plane that intersect each other at one common point are termed intersecting lines. And, for the lines to be concurrent, there must be a minimum of three lines intersecting at a single point.

Q.5. How to check the concurrency of three lines? Or How to find if the given lines are concurrent?
Ans: Steps to check concurrency of three lines are as follows:
(i) Solve two equations from the given three equations of the straight lines and obtain their point of intersection.
(ii) Plug the coordinates of the point of intersection in the third equation.
(iii) Check whether the third equation is satisfied.
(iv) If it is satisfied, the point lies on the third line, and hence the three straight lines are concurrent.

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