Find the area of a triangle whose vertices are
(i) (6, 3), (-3, 5) and (4, - 2)
(ii) 
(iii) (a, c + a), (a, c) and (-a, c - a)
(i) (6, 3), (-3, 5) and (4, - 2)
Let A≡(x1, y1 ) ≡ (6, 3), B ≡ (x2, y2 ) ≡ (-3, 5) and C ≡ (x3, y3 ) ≡ (4, -2)
Area of ∆ABC = 
|x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)| sq. units
∴ Area of ∆ABC = 
| { 6 (5 – (-2)) – 3 ( -2 – 3) + 4 (3 – 5)} |
=
| { 6 × 7 + 15 – 8 } |
= 
| 57 – 8|
= 
sq. units
(ii) 
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Here, (x1, y1)=(at12,2at1), (x2,y2)=(at22,2at2), (x3,y3)=(at32,2at3)
∴, area=
|at12(2at2-2at3)+at22(2at3-2at1)+at32(2at1-2at2)|
=
|2a2t12t2-2a2t12t3+2a2t22t3-2a2t22t1+2a2t32t1-2a2t32t2|
=
× 2a2|t12t2-t12t3+t22t3-t22t1+t32t1-t32t2|
=a2|t12t2-t12t3+t22t3-t22t1+t32t1-t32t2|
=a2|t12(t2-t3)+t2t3(t2-t3)-t1(t22-t32)|
=a2|t12(t2-t3)+t2t3(t2-t3)-t1(t2+t3)(t2-t3)|
=a2|(t2-t3)(t12+t2t3-t1t2-t1t3)|
=a2|(t2-t3){t1(t1-t2)-t3(t1-t2)}|
=a2|(t2-t3)(t1-t2)(t1-t3)|
∴ Area is a2|(t2-t3)(t1-t2)(t1-t3)| sq. units
(iii) (a, c + a), (a, c) and (-a, c - a)
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area = 
| a (c – c + a) + a( c – a –c – a) – a (c + a – c)|
= 
| a (a ) + a(-2a) – a( a)|
= 
| -2a2 |
= a2
∴ Area is a2 sq. units