A(x1,0),B(x2,0)x1+x2=k-1 x1*x2=-(k+1)
判别式=(k-1)^2+4(k+1)=k^2+2k+5>0
/AB/=根号下(x1-x2)^2
=根号下[(x1+x2)^2-4x1x2]
=根号下[(k-1)^2+4(k+1)]
=根号下[k^2+2k+5]
/c/=/k+1/
S△ABC/AB/*/c//2={根号下[k^2+2k+5]}*/k+1//2=
A(x1,0),B(x2,0)x1+x2=k-1 x1*x2=-(k+1)
判别式=(k-1)^2+4(k+1)=k^2+2k+5>0
/AB/=根号下(x1-x2)^2
=根号下[(x1+x2)^2-4x1x2]
=根号下[(k-1)^2+4(k+1)]
=根号下[k^2+2k+5]
/c/=/k+1/
S△ABC/AB/*/c//2={根号下[k^2+2k+5]}*/k+1//2=