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一、由一道高考题引发的思考 (全国高考题)已知直线l1和l2夹角的平分线为y=x,如果l1的方程是ax+by+c=0(ab>0),那么l2的方程是( ).(A)bx+ay+C=0 (B)ax-by+C=0(C)bx+ay-c=0 (D)bx-ay+c=0
First, the thinking caused by a college entrance examination question (national college entrance examination question) is known that the bisector of the angle between lines l1 and l2 is y=x, if the equation of l1 is ax+by+c=0 (ab>0), then l2 The equation is ().(A)bx+ay+C=0 (B)ax-by+C=0(C)bx+ay-c=0 (D)bx-ay+c=0