Let k be an integer such that the triangle with vertices (k, - 3k),...

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

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

$\to 5{k}^{2} + 13k − 46 = 0$

$\to k = 2$ and $k = - \dfrac{23}{5}$ (Not possible)

Altitude from $A: x = 2$

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

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