The area of a triangle is 6, two of its vertices are (1, 1) and (4, -1), the third vertex lies on y = x + 5. Find the third vertex.

Let us assume that X co-ordinate of the 3rd vertex is a. Then y = a + 5. So the co-ordinates of the vertex = (a, a+5) 

Ttherefore, 6 =   $$\dfrac{1}{2}\times \begin{vmatrix}1 & 1 & 1\\4 & -1 & 1\\a & a+5 & 1 \end{vmatrix}$$

$$\Rightarrow$$ $$\pm12$$ = 1(-1-a-5) -1(4 - a) + 1(4a+20+a)
$$\Rightarrow$$ $$\pm12$$ = 5a + 10
$$\Rightarrow$$ a = $$\dfrac{2}{5}$$, $$\dfrac{-22}{5}$$

Hence, the third vertex can be out of the two values ($$\dfrac{2}{5}$$, $$\dfrac{27}{5}$$) or ($$\dfrac{-22}{5}$$, $$\dfrac{3}{5}$$).

Therefore, option A is the correct answer.