Given : The vertices of a right angled triangle $$A\left( {1,k} \right),$$

$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit



We know that, area of night angled triangle

$$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$

$$ \Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$

Clearly for point $$P,$$



$${a^2} - 3a < 0$$ and $${a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3$$

As is the mid point of $$PQ,$$ therefore

$${{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8$$

$$\therefore$$ Equation of line is $${x \over 6} + {y \over 8} = 1$$

or $$4x + 3y = 24$$

Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -

this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$



$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$

$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$