Question

一道数学几何题.如图,在平面直角坐标系中,A(a,0),B（0,b）,且a、b满足a^2-8a+16+√b+4=0,点C

Answer 1

(1)a^2-8a+16+√(b+4)=0变为

(a-4)^2+√(b+4)=0,

∴a=4,b=-4,A(4,0),B(0,-4).

(2)C(0,4),

S△AMN=(3/2)S△AMB,

AN=(3/2)AB,

xN-xA=(3/2)(xA-xB),即xN-4=6,xN=10,

AB；y=x-4,

∴yN=10-4=6.

设M(m,0),m>4,由MN⊥CM得CM^2+MN^2=CN^2,

即m^2+16+(m-10)^2+36=104,

整理得2m^2-20m+48=0,

∴m^2-10m+24=0,m>4,

解得m=6,M(6,0).

(3)您是哪个年级的学生?