The correct option is D (2,12)
Given vertices of the triangle are (k,–3k),(5,k) and (–k,2)
Now, the area of triangle is
12∣∣∣x1x2x3x1y1y2y3y1∣∣∣=28⇒12∣∣∣k5−kk−3kk2−3k∣∣∣=28⇒|(k2+15k)+(10+k2)+(3k2−2k)|=56⇒|5k2+13k+10|=56⇒5k2+13k+10=56 (∵5k2+13k+10>0)⇒5k2+13k−46=0⇒(5k+23)(k−2)=0⇒k=2 (∵k is integer)
Now vertices areA(2,–6),B(5,2),C(–2,2)
Equation of the altitude dropped from vertex A is x=2
Equation of the altitude dropped from vertex C is 3x+8y–10=0
On solving the equations of altitude we get, Orthocentre =(2,12)