Question

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

• A. $(2,\frac{1}{2})$
• B. $(2,-\frac{1}{2})$
• C. $(1,\frac{3}{4})$
• D. $(1,-\frac{3}{4})$

Answer 1

The correct option is A $(2,\frac{1}{2})$

we have,

$\frac{1}{2}\left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right|$

$\Rightarrow 5k^2+13k-46=0$

OR

$5k^2+13k+66=0$

$\therefore k=-\frac{23}{5}$

OR

$k=2$

as $k$ is an integer,

so, $k=2$

$\therefore (2,\frac{1}{2})$