Question

在直角坐标系中,三角形ABC的顶点A,B的坐标分别为(-1,-2),(3,-2),定点C在直线y=x+2上移动.

Answer 1

C点坐标(x, x+2)

AB^2=[3-(-1)]^2+[-2-(-2)]^2=4^2

AB=4

1, AB的直线方程：y=-2, y+2=0

C到AB的距离：2*S(ABC)/AB=2*6/4=3

点到直线距离公式：│AXo＋BYo＋C│／√（A²＋B²)=|1*(x+2)+2|/1=3, x=x=-1

y=x+2=-1+2=1

C点坐标(-1, 1)

2, AC^2=(x+1)^2+(x+2+2)^2=2x^2+10x+17

BC^2=(x-3)^2+(x+2+2)^2=2x^2+2x+25

AC=BC, AC^2=BC^2

2x^2+10x+17=2x^2+2x+25

x=1 y=x+2=1+2=3

C点坐标(1, 3)