If the area of triangle ABC formed by $A( x,\ y) ,\ B( 1,\ 2)$ and $ C( 2,\ 1)$ is 6 square units, then prove that $x+y=15$

Given: A triangle ABC with the vertices $A( x,\ y) ,\ B( 1,\ 2)$ and $C( 2,\ 1)$ and area formed by the triangle is 6 square unit.

To do: To prove $x+y=15$

The given vertices are $A( x,\ y) ,\ B( 1,2)$  and $C( 2,1)$.

And we know that area of a triangle with vertices $( x_{1} ,\ y_{1}) ,\ ( x_{2} ,\ y)$ and $( x_{3} ,\ y_{3})$

$\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$ 

On subtituting the given valuues in the formula,

We have,

$\frac{1}{2}[ x( 2-1) +1( 1-y) +2( y-2)]=6$

$\Rightarrow \frac{1}{2}( x+1-y+2y-4)=6$

$\Rightarrow \frac{x+y-3}{2} =6$

$\Rightarrow x+y-3=12$

$\Rightarrow x+y=12+3=15$

Hence, proved that $x+y=15$.