知道三角形三点坐标A(0.0)B(1.0)C(b.c)，求重心，外心，垂心！
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（一）重心。易知重心G((1+b)/3,c/3).重心坐标公式：三角形ABC,A(x1,y1),B(x2,y2),C(x3,y3),则重心G((x1+x2+x3)/3,(y1+y2+y3)/3).(二）外心。外心就是两边中垂线的交点。易知，AB边的中垂线为x=1/2.AC边的中垂线方程为2bx+2cy=b²+c².===>两中垂线的交点为外心：(1/2,(b²+c²-b)/(2c)).（三）垂心就是两条高线的交点。易知，AB边上的高线为x=b,AC边上的高线方程为：bx+cy=b.两条高线的交点就是垂心(b,(b-b²)/c).绛旓細c鐐瑰潗鏍囷紙3,3鏍瑰彿3锛锛岃y=k/x锛屽甫鍏ュ緱k锛屾眰鍑鸿В鏋愬紡銆傚悜涓婂钩绉涓夎褰灏辨槸褰撴í鍧愭爣涓6鏃讹紝鍙嶆瘮渚嬪嚱鏁扮殑鍊硷紝甯﹀叆涓婂紡姹傚緱鐨勮В鏋愬紡锛屾眰鍑虹殑y灏辨槸n
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绛旓細杩嘽鐐逛綔鍨傜洿,鍨傝冻涓篸,鎵浠d=5鍙堝洜涓篴b涓ょ偣鐨鍧愭爣,鍙互寰楀嚭ab=6鎵浠涓夎褰闈㈢Н涓6*5*1/2=15
绛旓細A (0.0)B (4.0)C (2.-3鏍瑰彿2)
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绛旓細C1鐐瑰潗鏍=锛-4.5锛0.0锛C2鐐瑰潗鏍=锛7.51锛0.0锛夛紝濡傚浘鎵绀猴細
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绛旓細鈻矨BC鐨勯噸蹇僄鐨鍧愭爣鏄(a/3,b/3,c/3),璁锯柍ABC鐨勫瀭蹇僅鐨勫潗鏍囦负(m,n,p),鍒 鍚戦噺AH鈯C,BH鈯A,H鍦ㄥ钩闈BC涓婏紝寰(m-a,n,p)*(0,-b,c)=-bn+cp=0,(m,n-b,p)*(a,0,-c)=am-cp=0,m/a+n/b+p/c=1,瑙ｈ繖涓叧浜巑,n,p鐨勬柟绋嬬粍銆傝绠椾粠鐣ャ備綑鑰呬豢涓娿
绛旓細O(0, 0), A(2, 0), B(0, 4) 涓虹洿瑙涓夎褰锛屽垯杩囨涓夌偣鐨勫渾蹇冨湪鏂滆竟AB鐨勪腑鐐筂 M鐨鍧愭爣涓(1, 2)r²=MA²=(2-1)²+2²=5 鍥犳鍦嗙殑鏂圭▼涓(x-1)²+(y-2)²=5
