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When 3 lines are concurring, we can find the solution of the any two among the 3 equations and substitute the solution in the 3rd equation to get the condition.
Also when 3 lines are generally intersecting they intersect at 3 different points which form a triangle. If the 3 lines are concurrent, then the area formed by the three lines is zero.
We go by the first method.
Let a1x+b1y+c1 = 0 , a2x+b2y+c2 = 0 and a3x+b3y+c3 = 0 be the 3 different equations.
If we solve the first two equations, we get the solution:
x = (b1c2-b2c1)/(a1b2-a2b1) and y = -(a1c2-a2c1)/(a1b2-a2b1).
Substituting the above solutions in the last equation, a3x+b3x+c3 = 0, we get:
a3(b1c2-b2c1)/(a1b2-a2b1) - b3(a1c2 - a2c1)/(a1b2-a2b1) +c3 = 0. Multiplying this equation by (a1b2-a2b1), we get:
a3(b1c2-b2c1) - b3 (a1c2 - a2c1) + c3(a1b2 - a2b1) = 0....(1). This is the required condition for the concurrency of the three lines.
If we observe carefully the left of the eq (1), it is an expansion of the determinant (given below) through the 3rd row:
[(a1, b1,c1) , (a2,b2,c2) , (a3,b3,c3)].
Therefore the condition for the concurrency the 3 lines is given by:
[(a1, b1,c1) , (a2,b2,c2) , (a3,b3,c3)] = 0.
First, we'll write the 3 lines in Cartesian coordinates:
a1*x + b1*y + c1 = 0
a2*x + b2*y + c2 = 0
a3*x + b3*y + c3 = 0
The 3 lines have an intercepting point (x,y) if the coordinates of this point, substituted in each of 3 equations, verify the equations.
To determine the intercepting point, we'll have to solve the system of 3 equations. The solution of the system represents the intercepting point of 3 lines.
The condition for 3 lines to have an intercepting point is that the determinant fomed from the coefficients of the variables x and y to cancel.
a1 b1 c1
determinant = a2 b2 c2 = 0
a3 b3 c3
We want to know if the 3 lines have an intercepting point.
x + 2y = 6
2x - 2y = -6
First, we'll put them in the general form:
x + 2y - 6 = 0
3x + 4y - 12 = 0
2x - 2y + 6 = 0
Now, we'll calculate the determinant formed from the coefficients ai, bi, where i = 1,2,3.
1 2 -6
determinant = 3 4 -12
2 -2 6
We'll verify if the determinant has the zero value.
det. = 1*4*6 + 3*(-2)*(-6) + 2*2*(-12) - 2*4*(-6) - 1*(-2)*(-12) - 2*3*6
det. = 24 + 36 - 48 + 48 - 24 - 36
We'll eliminate like terms and we'll get:
determinant = 0
Since the constraint is verified, the 3 lines have an intercepting point.
It's coordinates represent the solution of the system.
For x = 0 and y = 3, all 3 equations are verified.
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