The midpoint of AA' is P((m-2)/2, (n+3)/2). According to p:3x-y- 1 = 0 on the straight line l, we get 3(m-2)/2-(n+3)/2- 1=0.

3m-n = 1 1……①

The slope of the straight line AA' is (n-3)/(m+2). According to the fact that the straight line AA' is perpendicular to the straight line L, the product of slopes is-1, which is simplified as [(n-3)/(m+2)]*3=-1.

m+3n = 7……②

(1) (2) The simultaneous solution of the two formulas is m=4 and n= 1, so

A'(4， 1)

2. Let the point M(x, y) be any point on the line L2, and the symmetrical point of m about the point P( 1,-1) is N(xo, yo).

Because of symmetry, the midpoint of MN is P( 1,-1), so

(x+XO)/2 = 1……①

(y+yo)/2= - 1 …………②

① ② Taking xo and yo as unknowns, two equations are solved simultaneously to get xo=2-x and YO =-2-Y 2-y..

That is, N(2-x, -2-y)

Because the point n is on the straight line L 1, the coordinates are substituted.

2(2-x)+3(-2-y)-6=0

Simplified equation 2(2-x)+3(-2-y)-6=0, that is, 2x+3y+ 10=0.