设A,B两点的坐标分别是(x1,y1),(x2,y2),直线AB的斜率为k(k≠0),求证:(1)|AB|=√(1+k^2) |x1-x2| (2)|AB|=

|AB|＝√[(x1-x2)^2+(y1-y2)^2]
=√[(x1-x2)^2+k^2(x1-x2)^2]
=√(1+k^2) *√（x1-x2）^2
=√(1+k^2) |x1-x2|
|AB|＝√[(x1-x2)^2+(y1-y2)^2]
=√【[(y1-y2)^2]/k²+(x1-x2)^2】
=√(1+1/k^2) *√（y1-y2）^2
=√(1+1/k^2) |y1-y2|