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

Step-by-step solution

​​​​​​​ Given, Area of the triangle ∆ A B C = 28 ⇒ 1 2 k k - 2 + 5 2 + 3 k - k - 4 k = 28 ⇒ k k - 2 + 5 2 + 3 k - k - 4 k = 56 ⇒ 5 k 2 + 13 k − 46 = 0 ⇒ k = 2 and k = - 23 5 (Not possible) Altitude from A : x = 2 Altitude from B : y - 2 = 1 2 x - 5 So, their point of intersection is H 2 , 1 2