已知a，b，c为某一直角三角形的三边长，c为斜边，若点（m，n）在直线ax+by+2c=0上，则m2+n2的最小值为多少？需具体步骤。
热心网友
ma+nb+2c=0-m=(b/a)*n + 2c/am^2 = (b^2/a^2)*n^2 + (4bc/a^2)*n + 4c^2/a^2TT = m^2+n^2 =(c^2/a^2)*n^2 + (4bc/a^2)*n + 4c^2/a^2=(c^2/a^2) * [ n^2 + (4b/c)*n + 4] TTT = n^2 + (4b/c)*n + 4= n^2 + 2*(2b/c)*n + (2b/c)^2 - (2b/c)^2 + 4= (n+2b/c)^2 + 4*a^2/c^2当 n=-2b/c 时 TTT 取最小值 4*a^2/c^2TT最小值 即为c^2/a^2 * 4 a^2/c^2 =4