MCQ
Let $k$ be an integer such that triangle with vertices $\left( {k, - 3k} \right),\left( {5,k} \right)$ and $\left( { - k,2} \right)$ has area $28$ sq. units. Then the orthocentre of his triangle is at the point :
  • $\left( {2,\frac{1}{2}\;} \right)\;$
  • B
    $\left( {2,\frac{{ - 1}}{2}\;} \right)$
  • C
    $\left( {1,\frac{3}{4}\;} \right)$
  • D
    $\left( {1,\frac{{ - 3}}{4}\;} \right)$

Answer

Correct option: A.
$\left( {2,\frac{1}{2}\;} \right)\;$
We  have $\frac{1}{2}\,\left\| \begin{array}{l} \,k\,\,\, - 3k\,\,\,1\\ \,5\,\,\,\,\,\,\,\,k\,\,\,\,\,1\\  - k\,\,\,\,\,2\,\,\,\,\,1 \end{array} \right\| = 28$
$ \Rightarrow 5{k^2} + 13k - 46 = 0$
or $5{k^2} + 13k - 66 = 0$
Now, $5{k^2} + 13k - 46 = 0$
$ \Rightarrow k = \frac{{ - 13 \pm \sqrt {1089} }}{{10}}$
$\therefore k = \frac{{ - 23}}{5};k = 2$
since $k$ is an integer,
$\therefore k = 2$
Also $5{k^2} + 13k + 66 = 0$
$ \Rightarrow k = \frac{{ - 13 \pm \sqrt { - 1151} }}{{10}}$
So no real solution exist
For cothocentre
$BH \bot AC$
$\therefore \left( {\frac{{\beta - 2}}{{\alpha - 5}}} \right)\left( {\frac{8}{{ - 4}}} \right) = - 1$
$ \Rightarrow \,\,\alpha - 2\beta = 1.......\left( 1 \right)$
Also $CH \bot AB$
$\therefore \left( {\frac{{\beta - 2}}{{\alpha + 2}}} \right)\left( {\frac{8}{3}} \right) = - 1$
$ \Rightarrow \,\,3\alpha - 8\beta = 1\,0.......\left( 2 \right)$
Solving $(1)$ and $(2)$ , we get
$\alpha = 2,\beta = \frac{1}{2}$
orthocentre is $\left( {2,\frac{1}{2}} \right)$

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