∫[1(x+a)3(x+b)5]1/4dx
A triangle has vertices A, (x1,y1) for i=1,2,3. Then the determinant
Δ=∣∣ ∣ ∣∣x2−x3y2−y3y1(y2−y3)+x1(x2−x3)x3−x1y3−y1y2(y3−y1)+x2(x3−x1)x1−x2y1−y2y3(y1−y2)+x3(x1−x2)∣∣ ∣ ∣∣=0 means