24.已知ΔABC\Delta ABCΔABC的顶点A(1,2)A(1,2)A(1,2),ABABAB边上的中线CMCMCM所在的直线方程为x+2y−1=0x+2y-1=0x+2y−1=0,∠ABC\angle ABC∠ABC的平分线BHBHBH所在直线方程为y=xy=xy=x,则直线BCBCBC的方程为 ___2x−3y−1=02x-3y-1=02x−3y−1=0___.