Question

如图,在平面直角坐标系内,A(0,2)B(-4,0)OD=3OA,点B.C关于Y轴对称,DE⊥AB于E,DM=AB

Answer 1

有两种求法,

一、连接AC,求出AM,CM,AC,cos(∠AMC)=(AM^2+CM^2-AC^2)/(2.AM.CM);

二、求出AM、CM的直线斜率K1,K2,tan(∠AMC)=(K1-K2)/(1+K1.K2);

A(0,2),B(-4,0),C(4,0),D(0,-6),M(m,m1);

DE⊥AB,KDE=-1/KAB=-1/(2/4)=-2;

(m1+6)/m=-2;(1)

DM=AB=2√5; √((m1+6)^2+m^2)=2√5;(2)

由(1),(2)解得m=-2,m1=-2;

KAM=k1=(2-m1)/(-m)=2; KCM=k2=(-m1)/(4-m)=1/3;

tan(∠AMC)=(k1-k2)/(1+k1.k2)=1; ∠AMC=45°;

或者,AM=2√5; CM=2√10; AC=2√2;

cos(∠AMC)=(AM^2+CM^2-AC^2)/(2.AM.CM)=(√2/2); ∠AMC=45°;