Let ABC be a triangle with equations of its sides AB , BC . CA respectively are x-2=0, y-5=0 and 5 x+2 y-10=0 . Then the orthocentre of triangle lies on the line

Step-by-step solution

The equations of the sides are L₁: x - 2 = 0 , L₂: y - 5 = 0 , and L₃: 5x + 2y - 10 = 0 . Find the vertices by solving the equations pairwise: Vertex A is the intersection of L₁ and L₃ : x = 2 , 5(2) + 2y - 10 = 0 2y = 0 y = 0 . So, A = (2, 0) . Vertex B is the intersection of L₁ and L₂ : x = 2 , y = 5 . So, B = (2, 5) . Vertex C is the intersection of L₂ and L₃ : y = 5 , 5x + 2(5) - 10 = 0 5x = 0 x = 0 . So, C = (0, 5) . The triangle ABC is a right-angled triangle at vertex B(2, 5) because side AB is vertical ( x=