点A(1,1)到直线L:y=k(x-2)+2的距离为√2,求k?

问题描述:

点A(1,1)到直线L:y=k(x-2)+2的距离为√2,求k?

点A(1,1)到直线L:y=k(x-2)+2的距离为√2,求k?解:y-kx+2k-2=0d=│1-k+2k-2│/√(1+k²)=√2即 (k-1)²/(1+k²)=2k²-2k+1=2k²+2k²+2k+1=(k+1)²=0故k=-1