If three are more straight lines intersect at a common point, then the lines are said to be concurrent lines.
The common point at where the lines are intersecting is called point of concurrency.
In the figure above, the lines AB, CD and EF are concurrent lines, as they all intersect at a common point P. And, P is the point of concurrency.
Method 1 :
(i) Solve any two of the givem equations and obtain point of intersection.
(ii) Substitute the point into the third equation and check, if the point satisfies the third equation.
(iv) If the point satisfies the third equation, then the three lines are concuurent.
Method 2 :
Consider the equations of three lines in general form as shown below.
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
a_{3}x + b_{3}y + c_{3} = 0
If the above three lines are concurrent, then the coefficients of x, y and the constant terms must satisfy the following condition.
To get the point of concurrency, solve any two of the given three equations.
Example 1 :
Show that the following straight lines are concurrent and find the point of concurrency.
2x - 3y + 4 = 0
9x + 5y = 19
2x -7y + 12 = 0
Solution :
2x - 3y + 4 = 0 ----(1)
9x + 5y - 19 = 0 ----(2)
2x -7y + 12 = 0 ----(3)
5(1) - 3(2) :
37x = 37
x = 1
Substitute x = 1 into (1).
2(1) - 3y + 4 = 0
2 - 3y + 4 = 0
-3y + 6 = 0
-3y = -6
y = 2
The point of interesection of (1) and (2) is (1, 2).
Substitute the point (1, 2) into (3).
2(1) - 7(2) + 12 = 0
2 -14 +12 = 0
-12 + 12 = 0
0 = 0 (true)
The point (1, 2) satisfies (3). So, the point (1, 2) lies on the third line.
Therefore, the given three lines are concurrent and the point of concurrency is (1, 2).
Example 2 :
Show that the following straight lines are concurrent and find the point of concurrency.
3x + 4y = 13
2x - 7y = -1
5x - y = 14
Solution :
Write the given equations in general form.
3x + 4y - 13 = 0 ----(1)
2x - 7y + 1 = 0 ----(2)
5x - y - 14 = 0 ----(3)
Find the value of the determinant with the coefficients of x, y and constant terms as explained in Method 2 above.
= 3(98 + 1) - 4(-28 - 5) - 13(-2 + 35)
= 3(99) - 4(-33) - 13(33)
= 297 + 132 - 429
= 429 - 429
= 0
Since the value of the determinant is zero, the given lines are concurrent.
Solve any two of (1), (2) and (3) to find the point of concurrency.
2(1) - 3(2) :
29y = 29
y = 1
Substitute y = 1 into (1).
3x + 4(1) - 13 = 0
3x + 4 - 13 = 0
3x - 9 = 0
3x = 9
x = 3
The point of concurrency is (3, 1).
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