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
Example:
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.
No comments:
Post a Comment