Given, Area of the triangle ΔABC=28⇒21∣k(k−2)+5(2+3k)−k(−4k)∣=28
⇒∣k(k−2)+5(2+3k)−k(−4k)∣=56
⇒5k2+13k−46=0
⇒k=2 and k=−523 (Not possible)
Altitude from A:x=2
Altitude from B:y−2=21(x−5)
So, their point of intersection is H(2,21)
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 orthocenter of this triangle is at the point:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
(2,−21)
(1,43)
(1,−43)
(2,21)
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