S△A1A2A3 1yy33 v31111 1 (a1,a2)+1( (a2,a1)+1(a2,a2)+1(a2,a3)+1 4|(34)+1(a32)+1(a303)+1 41(a2,a1)(a2a2)(a 1(a3,a1)(a3,a2)(as3,a3) 0111 161 0a2 0 因此,用边长表示的三角形面积公式为 D(A1,A2,A3) 这里 0111 D(A1,A2,A3) 220 叫做点集{A1,A2,Aa3}的 Cayley- Menger行列式 由推论4(2)式可得 推论5设在平面直角坐标系下,四边形A1A2A3A4四个顶点的坐标为(xk,k) (k=1,2,3,4),则它的面积 y1 S0A142434= r2y01 3为811的绝对值.用复数表示,命x1=+i0mk= 01 1,2,3,4),则 21z11 SOAlA2A3A4 =4 3211的模 4S4A1A2A3 2 = 1 4        x1 y1 1 x2 y2 1 x3 y3 1               x1 x2 x3 y1 y2 y3 1 1 1        = 1 4        (α1, α1) + 1 (α1, α2) + 1 (α1, α3) + 1 (α2, α1) + 1 (α2, α2) + 1 (α2, α3) + 1 (α3, α1) + 1 (α3, α2) + 1 (α3, α3) + 1        = − 1 4          0 1 1 1 1 (α1, α1) (α1, α2) (α1, α3) 1 (α2, α1) (α2, α2) (α2, α3) 1 (α3, α1) (α3, α2) (α3, α3)          = − 1 16          0 1 1 1 1 0 a 2 12 a 2 13 1 a 2 21 0 a 2 23 1 a 2 31 a 2 32 0          . ❏❑✣✭➛➵✳✴✘❤❥✸♠❄❼●❧ S4A1A2A3 = − 1 16 D(A1, A2, A3), ❈➸ D(A1, A2, A3) =          0 1 1 1 1 0 a 2 12 a 2 13 1 a 2 21 0 a 2 23 1 a 2 31 a 2 32 0          ➺➻✬➼ {A1, A2, A3} ✘ Cayley-Menger ❊❋●❨ q❻① 4(2) ●✹✫ ②③ 5 ⑧ ④⑤♠⑥❥✜✢✩⑦✣➽➛✸ A1A2A3A4 ➽★❦✬✘✜✢❧ (xk, yk) ,(k = 1, 2, 3, 4), ❣➾✘♠❄ S♦A1A2A3A4= 1 2          x1 y1 1 1 x2 y2 0 1 x3 y3 1 1 x4 y4 0 1          ✘⑨⑩❶❨✭➔✱ ✳✴✣➚ zk = xk + iyk,(k = 1, 2, 3, 4), ❣ S♦A1A2A3A4 = i 4          z1 z1 1 1 z2 z2 0 1 z3 z3 1 1 z4 z4 0 1          ✘↕❨ 4