Question

Let$k$ be an integer such that the triangle with vertices $(k,-3k),(5,k)$and $(-k,2)$ has area $28\mathrm{sq}.\mathrm{units}$. Then the orthocenter of this triangle is at the point

• A. $\left(1,\frac{3}{4}\right)$
• B. $\left(1,\frac{-3}{4}\right)$
• C. $\left(2,\frac{1}{2}\right)$
• D. $\left(2,\frac{-1}{2}\right)$

Answer 1

The correct option is C

$\left(2,\frac{1}{2}\right)$

Explanation for correct option

Determine the Orthocenter of the triangle.

Given the three vertices of a triangle are $\left(k,-3k\right),\left(5,k\right)$ and $\left(-k,2\right)$, and area of a triangle is $28\mathrm{sq}.\mathrm{units}$

We know that the area of a triangle in determinate form is calculated by placing $x$ coordinates in the first column, $y$ coordinates in the second column and $1$ in the third column. Hence,

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right| \\ \Rightarrow 28=\frac{1}{2}\left|\begin{array}{ccc}k& -3k& 1\\ 5& k& 1\\ -k& 2& 1\end{array}\right| \\ \Rightarrow 56=\left|k\left(k-2\right)+3k\left(5+k\right)+\left(10+k^2\right)\right| \\ \Rightarrow 56=\left|5k^2+13k+10\right| \\ \Rightarrow \pm 56=5k^2+13k+10 \end{gathered}$$

Now consider the negative value $\left(-56\right)$ we get,

$$\begin{gathered} 5k^2+13k+10=-56 \\ \Rightarrow 5k^2+13k+66=0 \end{gathered}$$

Here, the value of discriminant will be negative, hence this is rejected.

Now consider the positive value $\left(+56\right)$ we get,

$$\begin{gathered} 5k^2+13k+10=56 \\ \Rightarrow 5k^2+13k-46=0 \end{gathered}$$

Finding its roots we get,

$$\begin{gathered} k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\left(a=5,b=13,c=-46\right) \\ \Rightarrow k=\frac{-13\pm \sqrt{\left({13}^2-4\times 5\times -46\right)}}{2\times 5} \\ \Rightarrow k=\frac{-13\pm \sqrt{1089}}{10} \\ \Rightarrow k=\frac{-13+33}{10},\frac{-13-33}{10} \\ \Rightarrow k=2,\frac{-46}{10}\left[\mathrm{rejected}\mathrm{as}\mathrm{k}\mathrm{is}\mathrm{an}\mathrm{integer}\right] \end{gathered}$$

Now the coordinates of vertices are $\left(2,-6\right),\left(5,2\right),\left(-2,2\right)$

The orthocenter of a triangle is the point of intersection of lines $AD$ and $BE$.Let its coordinates be $\left(x,y\right)$

Now slope of line $AD$ is infinite, hence the $x$ coordinate of its orthocenter will be $2$ because the line $AD$ passes through $\left(2,-6\right)$

Line $BE$ is perpendicular on $AC$, hence its slope will be equal to,

$$\begin{gathered} m_{BE}\times m_{AC}=-1\left[\because BE\perp AC\right] \\ \Rightarrow m_{BE}\times -2=-1 \\ \Rightarrow m_{BE}=\frac{1}{2} \end{gathered}$$

Now the equation of line $BE$ will be,

$$\begin{gathered} y-y_1=m_{BE}\left(x-x_1\right) \\ \Rightarrow y-2=\frac{1}{2}\left(x-5\right) \\ \Rightarrow y=\frac{1}{2}\left(2-5\right)+2\left[x=2\right] \\ \Rightarrow y=\frac{-3}{2}+2 \\ \Rightarrow y=\frac{1}{2} \end{gathered}$$

Therefore the coordinates of orthocenter are $\left(2,\frac{1}{2}\right)$.

Hence, the correct answer is option (C)