解ï¼1. 设M(x0,y0)æ¯ABçä¸ç¹ï¼
ç±ä¸ç¹åæ å ¬å¼å¾ï¼
x0=(2+0)/2=1,y0=(4+(-2))/2=1
⇒M(1,1) ã
设ABè¾¹ä¸çä¸çº¿CMæå¨ç´çº¿çæ¹ç¨ä¸ºï¼
y=kx+bï¼
å°Cï¼-2,3ï¼ï¼M(1,1)ä»£å ¥ï¼
解æ¹ç¨ç»ï¼
3=-2k+b ï¼1ï¼
1=k+b ï¼2ï¼
å¾k=-2/3
b=5/3
â´ABè¾¹ä¸çä¸çº¿CMæå¨ç´çº¿çæ¹ç¨ä¸ºï¼
y = - 2x/3 + 5/3ã
2. åçå¯æ±
ABè¾¹æå¨ç´çº¿çæ¹ç¨ä¸ºï¼
y-4=3(x-2)
⇒3x-y-2=0.
ç±ç¹cï¼-2,3ï¼å°ç´çº¿ABï¼3x-y-2=0çè·ç¦»
d=|3*(-2)-3-2|/â 10
=11â 10/10.
å¾Sâ³ABC=|AB|*d/2
=2â10*(11â 10/10)/2
=11.