The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Let ABC be a triangle with A(a, b),B(2,1) and C(3,-2).
A lies on the line y=x+3 means,
b=a+3 ...(1).
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area of ∆ABC = 5
Substituting the values of A, B and C in formula, we, get,
5 = 
| 3a + b – 7 |
Taking positive value for | 3a + b – 7 |,
3a+b=17 ….(2)
Solving 1 and 2 simultaneously,
a =
and b =
Hence coordinates of the vertex A are (
,
).
Taking negative value for | 3a + b – 7 |,
(3a+b−7) = − 5
3a+b=−3 …(3)
Solving 1 and 2 simultaneously,
A =
and b = 
and the vertex A is (
,
)
Hence the coordinates of third vertex are (
,
) or (
,
).