两直线的交点坐标教学设计人教A版高中数学选择性必修第一册（定制版）

解法一:若直线斜率不存在,则方程为x=3.由{■(x=3"," @2x"-" y"-" 2=0"," )┤得A(3,4).， 由{■(x=3"," @x+y+3=0"," )┤得B(3,-6).由于(4+"(-" 6")" )/2=-1≠0,∴P不为线段AB的中点.若直线斜率存在,设为k,则方程为y=k(x-3).由{■(y=k"(" x"-" 3")," @2x"-" y"-" 2=0"," )┤得A((3k"-" 2)/(k"-" 2),4k/(k"-" 2)).由{■(y=k"(" x"-" 3")," @x+y+3=0"," )┤得B((3k"-" 3)/(k+1),-6k/(k+1)).∵P(3,0)为线段AB的中点,∴{■((3k"-" 2)/(k"-" 2)+(3k"-" 3)/(k+1)=6"," @4k/(k"-" 2) "-" 6k/(k+1)=0"." )┤∴{■(2k"-" 16=0"," @k^2 "-" 8k=0"." )┤∴k=8.∴所求直线方程为y=8(x-3),即8x-y-24=0.分析二:设出A(x1,y1),由P(3,0)为AB的中点,易求出B的坐标,而点B在另一直线上,从而求出x1、y1的值,再由两点式求直线的方程.解法二:设A点坐标为(x1,y1),则由P(3,0)为线段AB的中点,得B点坐标为(6-x1,-y1).∵点A,B分别在已知两直线上,∴{■(2x_1 "-" y_1 "-" 2=0"," @"(" 6"-" x_1 ")" +"(-" y_1 ")" +3=0"." )┤解得{■(x_1=11/3 "," @y_1=16/3 "." )┤∴A 11/3,16/3 .∵点A,P都在直线AB上,∴直线AB的方程为(y"-" 0)/(16/3 "-" 0)=(x"-" 3)/(11/3 "-" 3),即8x-y-24=0.分析三:由于P(3,0)为线段AB的中点,可对称地将A,B坐标设为(3+a,b),(3-a,-b),代入已知方程.