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

$open parentheses 2 comma 1 half close parentheses$
$open parentheses 2 comma negative 1 half close parentheses$
$open parentheses 1 comma negative 3 over 2 close parentheses$
$open parentheses 1 comma 3 over 2 close parentheses$

Solution

open parentheses 2 comma 1 half close parentheses

We have,
$1 half open vertical bar table row straight k cell negative 3 straight k end cell 1 row 5 straight k 1 row cell negative straight k end cell 2 1 end table close vertical bar space equals space 28$

⇒ 5k² + 13k – 46 = 0
or
5k² + 13k + 66 = 0 (no real solution exist)
∴ k =–23/5
or k = 2
k is an integer, so k =2
As
Therefore (2, 1/2)

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