Question

Let k be an integer such that the triangle with vertices $\left( {k, - 3k} \right),\left( {5,x} \right)\,\,and\,\,\left( { - k,2} \right)$ has are 28 sq. units. Then the orthocenter of this triangle is at the point.
A. $\left( {-2,\dfrac{1}{2}} \right)$
B. $\left( {1,\dfrac{3}{4}} \right)$
C. $\left( {1, - \dfrac{3}{4}} \right)$
D. $\left( {2,\dfrac{1}{2}} \right)$

Answer 1

The orthocenter of the triangle is the intersection of the triangle’s three altitudes. It has several important properties and relations with other parts of the triangle. In simple words the orthocenter of a triangle lies on the altitudes.

Complete step by step solution:

Given: The vertices of a triangle are $\left( {k, - 3k} \right),\left( {5,k} \right)\,\,and\,\,\left( { - k,2} \right)$ and are is given to be 28 sq. units.

Let o us assume the vertices of triangle $\left( {{x_1},{y_1}} \right)\left( {{x_2},{y_2}} \right)\left( {{x_3},{y_3}} \right)$ where $\left( {{x_2},{y_2}} \right) = \left( {5k} \right)\,\,\,\,\,\left( {{x_3},{y_3}} \right) = \left( { - k,2} \right)\,\,\,and\,\,\left( {{x_1},{y_1}} \right) = \left( {k, - 3k} \right)$.

Hence, we known the area of a triangle in determinant form is

$\Rightarrow \left| {k\left( {k - 2} \right) + 3k\left( {5 + k} \right) + 1\left( {10 + {k^2}} \right)} \right| = 56$

$\Rightarrow \left| {{k^2} - 2k + 15k + 3{k^2} + 10 + {k^2}} \right| = 56$

$\Rightarrow \left| {5{k^2} + 13k + 10} \right| = 56$

since when $\left| x \right| = 3$ than the value of $x = \pm 3$

Now, solving the quadratic equation taking the values + 56 and – 56 we get

$5{k^2} + 13k + 10 = 56$ and

$5{k^2} + 13k + 10 = - 56$

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

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

Now solving the 1st equation, we get to two solution justifying the quadratic equation, the two values we get more $k = 2$k = 2 or $k = - \dfrac{{23}}{5}$

Whereas the second equation i.e. $5{k^2} + 13k + 66 = 0$ has no real solution.

Now according to question since k is an integer therefore, we will neglect the value of $k = - \dfrac{{23}}{5}$

Therefore, $k = 2$

Hence the vertices of the leangle are $\left( {2, - 6} \right),\,\,\left( {5,2} \right)\,\,and\,\,\left( { - 2,2} \right)$

Now, we know that the orthocenter is on the altitude drawn from the vertex to the opposite side.

We have to find the equation of two altitudes and then their point of intersection

${A_1}\left( {2, - 6} \right);\,y + 6 = - \dfrac{7}{0}\left( {x - 2} \right)$

$\Rightarrow {A_1}\left( {2, - 6} \right);\,\,x = 2$

Now,

${A_2}\left( {5,2} \right);\,\,y - 2 = \dfrac{1}{2}\left( {x - 5} \right)$

$\Rightarrow {A_2}\left( {5,2} \right);\,\,x - 2y = 1$

Solving these we get the coordinates of the orthocenter as $\left( {2,\dfrac{1}{2}} \right)$.

Thus option (D) is correct.

Note: In these types of questions students often get confused regarding the term orthocenter as they O misconception it with circumcenter, incenter or eccentric. But these all are totally different things. Students should make a chart of this important formula and keep them while solving these types of questions.