(1)∵y=x2-2x-3=(x+1)(x-3) ∴A(-1,0),B(3,0);
(2)设点P的坐标是(x,y).则由题意,得 S△ABP=AB•|y|=×4•|y|=6, 解得,|y|=3. ①当y=-3时,当y=3时,x2-2x-3=-3,即x2-2x=0, 解得x1=,x2=2.则P1(0,-3),P2(2,-3); ②当y=3时,x2-2x-3=3,即x2-2x-6=0, 解得x1=1+,x2=1-; 则P3(1+,3),P4(1-,3). 综上所述,符号条件的点P的坐标分别是:P1(0,-3),P2(2,-3),P3(1+,3),P4(1-,3). |