Let k be an integer such that the triangle with vertices (k, -k), (5, k) and (-k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point. (2, -2), (1, 34), (2, 12), (1, -9)

Let the vertices of the triangle be (x1, y1) = (k, -k); (x2, y2) = (5, k) and (x3, y3) = (-k, 2)
Hence, 1/2 |k(k - 2) + 5(2 + k) + (-k)(-k - k)| = 28
|k^2 - 2k + 10 + 5k + k^2| = 56
|2k^2 + 3k + 10| = 56
2k^2 + 3k + 10 = ±56
2k^2 + 3k - 46 = 0 or 2k^2 + 3k + 66 = 0
The first equation above has two solutions i.e k = 2 or k = -23/2 whereas the second equation has no real solution.
Since k is an integer, k = 2
Hence, the vertices of the triangle are (2, -2); (5, 2) and (-2, 2)
Now, we know that the orthocenter lies on the altitude drawn from the vertex to the opposite side.
We have to find the equation of two altitudes and then their point of intersection.
A1(2, -2): y + 2 = ∞(x + 2) ⇒ A1(2, -2): x = 2
A2(5, 2): y - 2 = 1/2(x - 5)