There are two conditions of concurrency of lines which are given below :

(a) Three lines are said to be concurrent if they pass through a common point i.e. they meet at a point.

Thus, Three lines \(a_1x + b_1y + c_1\) = 0 and \(a_2x + b_2y + c_2\) = 0 and \(a_3x + b_3y + c_3\) = 0 are concurrent, if

\(\begin{vmatrix}

a_1 & b_1 & c_1 \\

a_2 & b_2 & c_2 \\

a_3 & b_3 & c_3 \\

\end{vmatrix}\) = 0

This is the required condition of concurrency of lines.

Example : Prove that the lines 3x + y – 14 = 0, x – 2y = 0 and 3x – 8y + 4 = 0.

Solution : Given lines are 3x + y – 14 = 0, x – 2y = 0 and 3x – 8y + 4 = 0

We have, \(\begin{vmatrix}
3 & 1 & -14 \\
1 & -2 & 0 \\
3 & -8 & 4 \\
\end{vmatrix}\) = 3(-8 + 0) – 1(4 – 0) – 14(-8 + 6)

= -24 – 4 + 28 = 0

So, the given lines are concurrent.

## Another Condition of Concurrency of Lines :

(b) To test the concurrency of lines, first find out the point of intersection of the three lines. If this point lies on third line ( coordinates of the point satisfy the equation of third line) then the three lines are concurrent otherwise not concurrent.

Example : Show that the lines x – y – 6 = 0, 4x – 3y – 20 = 0 and 6x + 5y + 8 = 0.

Solution : The Given lines are x – y – 6 = 0 ….(i), 4x – 3y – 20 = 0 ….(ii) and 6x + 5y + 8 = 0 ….(iii)

from equation (i) and (ii), we get

x = 2 and y = -4

Thus, the first two lines intersect at the point (2, -4). Putting x = 2 and y = -4 in equation (iii), we get

6\(\times\)2 + 5\(\times\)(-4) + 8 = 0

So, Point (2, -4) lies on third line

Hence, the given lines are concurrent and common point of intersection is (2, -4).

Hope you learnt condition of concurrency of lines, learn more concepts of straight lines and practice more questions to get ahead in the competition. Good luck!