Let k be an integer such that the triangle with vertices (k, −3k),(5, k) and (−k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:

Let ABC be the triangle the coordinates of whose vertices are ;4 (k, -3k), B, (5, k)  and  C(-k, 2). It is given that Area of∆ABC = 28 sq. units =12k    −3k    15    k    1−k    2    1=±28⇒ 12k    −3k    15−k    4k    0−2k    2+3k    0=±28=(5−k)(2+3k)+8k2=±56=5k2+13k+66=0 or 5k2+13k−46=0⇒5k2+13k−46=0     ∵5k2+13k+66=0 has imaginary roots ⇒(k−2)(5k+23)=0⇒k=2 [∵k∈Z]Hence, the coordinates of vertices are A ( 2, - 6), B (5, 2) and(−2, 2)The equations of altitudes through vertices A. andC are .:x= 2 and 3x + By -10 = 0 respectively. These two altitudes intersect : (2, 1/2). Hence, the coordinates of the orthocentre are ( 2, 1/ 2).