question_answer

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:     JEE Main Solved Paper-2017

A) \[\left( 2,\frac{1}{2} \right)\]                     

B)  \[\left( 2,-\frac{1}{2} \right)\]

C)  \[\left( 1,\frac{3}{4} \right)\]                    

D)  \[\left( 1,-\frac{3}{4} \right)\]

Correct Answer: A

Solution :

 We have \[\frac{1}{2}\left\| \begin{matrix}    k & -3k & 1  \\    5 & k & 1  \\    -k & 2 & 1  \\ \end{matrix} \right\|=28\] \[\Rightarrow \]\[5{{k}^{2}}+13k-46=0\] or \[5{{k}^{2}}+13k+66=0\] (no real solution exist) \[\therefore \]  \[k=\frac{-23}{5}\]or \[k=2\] As k is an integer, so k = 2 \[\Rightarrow \] orthocentre is \[\left( 2,\frac{1}{2} \right)\]