Let k be an integer such that the triangle with vertices k , - 3 k ,   5 ,   k and - k ,   2…

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Given, Area of the triangle $∆ABC=28$$\Rightarrow \frac{1}{2}\left|k\left(k-2\right)+5\left(2+3k\right)-k\left(-4k\right)\right|=28$

$\Rightarrow \left|k\left(k-2\right)+5\left(2+3k\right)-k\left(-4k\right)\right|=56$

$\Rightarrow 5k^2+13k-46=0$

$\Rightarrow k=2$ and $k=-\frac{23}{5}$ (Not possible)

Altitude from  $A \mathbf{:} x=2$

Altitude from $B \mathbf{:} y-2=\frac{1}{2}\left(x-5\right)$

So, their point of intersection is $H\left(2,\frac{1}{2}\right)$