Find the area of the triangle whose vertices are (i)(2,3),(-1,0),(2,-4) (ii)(-5,-1),(3,-5),(5,2)

(i) Let A(2,3), B(-1,0) and C(2,-4) are the vertices of △ABC
Area of triangle = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
Here (x1,y1) = (2,3)
(x2,y2) = (-1,0)
(x3,y3) = (2,-4)
= 1/2[2(0+4) -1(-4-3) + 2(3-0)]
= 1/2[8+7+6] = 21/2 sq.unit
(ii) Let A(-5,-1), B(3,-5) and C(5,2) are the vertices of △ABC
Area of triangle = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
Here (x1,y1) = (-5,-1)
(x2,y2) = (3,-5)
(x3,y3) = (5,2)
= 1/2[-5(-5-2) + 3(2+1) + 5(-1+5)]
= 1/2[35+9+20] = 64/2 = 32 sq.unit