Given, vertices of triangle are (k, – 3k), (5, k) and (–k, 2).

$${1 \over 2}\left| {\matrix{ k & { - 3k} & 1 \cr 5 & k & 1 \cr { - k} & 2 & 1 \cr } } \right| = \pm 28$$

$$ \Rightarrow $$ k(k - 2) + 3k(5 + k) + 1(10 + k²) = $$ \pm $$ 56

$$ \Rightarrow $$ 5k² + 13k + 10 = $$ \pm $$ 56

$$ \Rightarrow $$ 5k² + 13k - 66 = 0

$$ \Rightarrow $$ k = $${{ - 13 \pm \sqrt { - 1151} } \over {10}}$$

So no real solution exist.

or 5k² + 13k - 46 = 0

$$ \therefore $$ k = $${{ - 23} \over 5}$$ or k = 2

since k is an integer $$ \therefore $$ k = 2

Thus, the coordinate of vertices of triangle are

A(2, -6), B(5, 2) and C(-2, 2).



Now, equation of altitude from vertex A is

y - (-6) = $${{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)$$

$$ \Rightarrow $$ x = 2 .......(1)

Equation of altitude from vertex B is

y - 2 = $${{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)$$

$$ \Rightarrow $$ 2y - 4 = x - 5

$$ \Rightarrow $$ x - 2y = 1 .......(2)

Point H($$\alpha $$, $$\beta $$) lies on both (1) and (2),

$$ \therefore $$ $$\alpha $$ = 2 .........(3)

$$\alpha $$ - 2$$\beta $$ = 1 ......(4)

Solving (3) and (4), we get

$$\alpha $$ = 2 , $$\beta $$ = $${1 \over 2}$$

$$ \therefore $$ Orthocentre is $$\left( {2,{1 \over 2}} \right)$$.