QJEEM17Ma74b2: JEE Main 2017 Mathematics - Class 11 JEEM17Ma74b2 Solution: Let k be an integer such that the triangle with vertices (k, –…

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

$$\frac{1}{2}\left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right|=\pm 28$$

$\Rightarrow$ 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 $\left( {2,{1 \over 2}} \right)$ .