equation of locus

the required locus is a straight line L3
let L3 be y = mx+c

slope of L1 = -3, slope of L2 = -3
so m =-3 too.

consider the line L4 : y=x/3 is perpendicular to L1 and L2
L4 intersects L1 and L2 at A(9/5,3/5) and B(-27/20, -9/20) respectively
mid pt of AB = (9/40, 3/40)
note that L3 passes through this point

so put (x,y) = (9/40, 3/40) into y=-3x+c
we get c=3/4 after solving
so the required locus is y=-3x+3/4, ie, 12x+4y-3=0

2013-01-26 13:01:06 補充：
corrected version:

the required locus is a straight line L3
let L3 be y = mx+c

slope of L1 = -3, slope of L2 = -3
so m =-3 too.

consider the line x=0
set x=0 for L1 and L2, y-intercepts of L1 and L2 are 6 and -9/2 respectively
c = [6+(-9/2)]/2 = 3/4
so y=-3x+3/4, ie, 12x+4y-3=0

For L1, when y = 0, x = 2, point A(2,0) is on L1.
For L2, when y = 0, x = -3/2, point B(-3/2, 0) is on L2.
Mid - point of A and B is [ (2 - 3/2)/2, 0] = (1/2, 0) which is on the locus straight line.
Slope of the line is also - 3, so equation of the locus is
y = - 3(x - 1/2)
2y = - 6x + 3
6x + 2y - 3 = 0

2013-01-26 07:50:05 補充：
Correction : mid - point of AB should be (1/4,0), so locus should be y = - 3( x - 1/4)
or 4y = - 12x + 3, that is 12x + 4y - 3 = 0