已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12

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已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12
已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值
由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0
化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5
我想问的是以上最后一步y1•y2= (12+m)/5 是怎么来的啊

已知圆想x2+Y2+X-6Y+M=0与直线x+2y-3=0相交于p,q两点,o为原点,且op垂直于oq,求实数m的值由x+2y-3=0得x=3-2y代入x2+y2+x-6y+m=0化简得:5y2-20y+12+m=0y1+y2=4,y1•y2= (12+m)/5我想问的是以上最后一步y1•y2= (12
韦达定理啊,x1,x2是二元一次方程ax^2+bx+c=0的两根,x1+x2=-b/a.x1x2=c/a
所以由5y2-20y+12+m=0这个方程,知y1•y2= (12+m)/5