AA' is perpendicular to L, because A' is a symmetrical point about the straight line L.

The slope of the straight line L is 2/3, and the slope of the straight line AA' is -3/2 because the product of the slopes of vertical lines is-1.

Substitute it into the formula of point inclination: y+2=-3/2(x+ 1), and it becomes 3x+2y+7=0.

Its intersection with l is (-23/ 13,-1113).

Point A' (that is, the symmetrical point of point (-1, -2) about (-23/ 13,-113) is.

〔2×(-23/ 13)-(- 1)],[2×(- 1 1/ 13)-(-2)]

Namely (-33/13,4/13)

(2) Select two points P1(2,0) and P2 (4 4,3) on the straight line M.

If the slope of the straight line L is 2/3, then the slope of the connecting line P 1 p 1' between P1and the symmetrical point P1about L is -3/2, and the substitution point is oblique: y-0=-3/2(x-2).

Its intersection with l is (1613, 15/ 13).

The point P 1' is [2×16/13-(-1)], [2× 15/ 13-(-2)], that is.

In the same way, find the symmetry point P2' of P2 about L as (4,3).

Substitute P 1' and P2' into the two-point formula and get M' 17x+7y+89=0.

(3) Select two points Q 1( 1, 1) and Q2 (4 4,3) on L.

The symmetry point of Q 1 about a is (2×(- 1)- 1, 2×(-2)- 1), that is, (-3, -5).

The symmetry point of Q2 about a is (2×(- 1)-4, 2×(-2)-3), that is, (-6, -7).

Substitute them into the two-point formula and it becomes the general formula 2x-3y-9=0.

The straight line Q 1Q2 is l'