Question

已知直角三角形两点坐标和一边长,求另一点坐标

Answer 1

互相垂直两直线斜率的乘积=-1

kAB*kCB=-1

(b-n)/(a-m) * (y-n)/(x-m) = -1

(a-m)(x-m)+ (b-n) (y-n) = 0

y-n = - (a-m)(x-m)/(b-n) .(1)

又：BC=L,即BC^2 = L^2：

(x-m)^2+(y-n)^2 = L^2 .(2)

将（1）代入（2）：

(x-m)^2 + {-(a-m)(x-m)/(b-n)}^2 = L^2

(x-m)^2{1+(a-m)^2/(b-n)^2} = L^2

(x-m)^2 = L^2(b-n)^2 / [(a-m)^2+(b-n)^2]

x-m = ±L(b-n) /√[(a-m)^2+(b-n)^2]

x1= m - (b-n) /√[(a-m)^2+(b-n)^2]

x2= m + (b-n) /√[(a-m)^2+(b-n)^2]

y-n = - (a-m)(x-m)/(b-n)

y1 = n + L(a-m) /√[(a-m)^2+(b-n)^2]

y1 = n - L(a-m) /√[(a-m)^2+(b-n)^2]