Question

已知A(5,-1),B(1,1),C(x,3)三点,且△ABC为直角三角形,求x的值.

Answer 1

Kab=(1+1)/(1-5)=-1/2

Kac=(3+1)/(x-5)=4/(x-5)

Kbc=(3-1)/(x-1)=2/(x-1)

当 AB⊥AC

Kab*Kac=-1

-1/2*4/(x-5)=-1

x-5=2

x=7

当 AB⊥BC

-1/2*2/(x-1)=-1

x-1=1

x=2

当 AC⊥BC

4/(x-5)*2/(x-1)=-1

(x-1)(x-5)+8=0

x²-6x+13=0 无解

故 x=2 或 x=7