There are two conditions ofย 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 โ 14 = 0, x โ 2y = 0 and 3x โ 8y + 4 = 0.
Solution : Given lines are 3x + y โ 14 = 0, x โ 2y = 0 and 3x โ 8y + 4 = 0
We have, \(\begin{vmatrix}
3 & 1 & -14 \\
1 & -2 & 0 \\
3 & -8 & 4 \\
\end{vmatrix}\) = 3(-8 + 0) โ 1(4 โ 0) โ 14(-8 + 6)
= -24 โ 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 โ y โ 6 = 0, 4x โ 3y โ 20 = 0 and 6x + 5y + 8 = 0.
Solution : The Given lines are x โ y โ 6 = 0 โฆ.(i), 4x โ 3y โ 20 = 0 โฆ.(ii) and 6x + 5y + 8 = 0 โฆ.(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!
