Let k be an integer such that the 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 :
Solutions · 1
Official solution
Given, vertices of triangle are (k, -- 3k), (5, k) and (--k, 2).
\[{1 \over 2}\left| {k5−k−3kk2111 } \right| = \pm 28\]
\[ \Rightarrow \] 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 \[\left( {2,{1 \over 2}} \right)\].