1、将点E坐标代入直线方程,-8k + 6 = 0,解得k= 3/4;2、OA = 0 - (-6) = 6;S = ( OA/2 ) *y= (6/2)[ (3/4)x + 6,函数式 S = 9x/4 + 6;点P在二象限,所以 x取值范围是 ( -8,0 ) ;3、S是x的函数,所以将S = 27/8代入函数式,可求得x的值;再由直线方程求得y,即得到 点P位置。9x/4 + 6 = 27/8,x = -7/6;y =(3/4)x + 6 = (-7/6)( 3/4) + 6 = 41/8;点P位置为 ( -7/6,41/8 )时,S = 27/8 。