Let A(5 t, 5 t) , B(-5 t, 5 t) and C(3, 4) be the vertices of a triangle, where t R . As t varies, the orthocentre of the triangle traces a closed curve S . If (h, k) is the centre of S and K is the area enclosed by S , then the value of h + k + K is

Step-by-step solution

Let O(0,0) be the origin. The distance of each vertex from the origin is: OA = (5 t)^2 + (5 t)^2 = 5 OB = (-5 t)^2 + (5 t)^2 = 5 OC = 3^2 + 4^2 = 5 Since OA = OB = OC = 5 , the circumcentre of the triangle ABC is the origin O(0,0) . For a triangle inscribed in a circle with its circumcentre at the origin, the position vector of the orthocentre H(x,y) is given by the sum of the position vectors of its vertices. Thus, H = A + B + C . Adding the respective coordinates, we get: x = 5 t - 5 t + 3 y = 5 t + 5 t + 4 Rearr