1、设AC上的高交AC于D点,即BD为y=2x+1 (1)AC垂直BD则斜率为-1/2且过A(-2,0)AC为y=-1/2*x-1 (2)由AB、AC的角平分线是x轴AB为y=1/2*x+1 (3)AB交BD于B,(1)(3)得B(0,1)2、由三角形ABC面积是4,且高为BD由(1)(2)得D(-4/5,-3/5)BD=4/5*5^(1/2)由(2)设C(x, -1/2*x-1)由S=1/2*BD*AC=4得AC=2*5^(1/2)把AC两点坐标带入求得x=-6或x=2C(-6,2)或(2,-2)由角平分线为x轴舍去(-6,2)则C(2,-2)... |