If m1 and m2 are the roots of the equation x2-ax-a-1=0 , then the area of the triangle  formed by the three straight lines y=m1x,y=m2x and y=a(a≠−1)is

Since  m1 and m2 are the roots of the equation x2-ax-a-1=0so m1+m2=a;m1m2=−(a+1)⇒(m1−m2)2=(m1+m2)2+4m1m2=a2+4(a+1)=(a+2)2∴ The required area is Δ=±a2(a+2)2(a+1)since area is positive quantity, so we get two answers