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Step-by-Step Solution
Step 1: Label the vertices and set up the area condition
The given vertices of the triangle are:
A$(k, -3k)$, B$(5, k)$, and C$(-k, 2)$.
The area of the triangle formed by points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ can be given by the formula:
$ \text{Area} = \frac{1}{2} \left| \begin{matrix}
x_1 & y_1 & 1\\
x_2 & y_2 & 1\\
x_3 & y_3 & 1
\end{matrix} \right| $
According to the problem, this area must be $28$ (square units). Therefore:
$ \frac{1}{2} \left| \begin{matrix}
k & -3k & 1\\
5 & k & 1\\
-k & 2 & 1
\end{matrix} \right| = \pm 28. $
Step 2: Expand the determinant and form an equation in k
Compute the determinant:
$ \left| \begin{matrix}
k & -3k & 1\\
5 & k & 1\\
-k & 2 & 1
\end{matrix} \right|
= k \bigl( k \cdot 1 - 1 \cdot 2 \bigr)
- (-3k) \bigl(5 \cdot 1 - 1 \cdot (-k) \bigr)
+ 1 \bigl(5 \cdot 2 - k \cdot (-k) \bigr). $
Carefully simplify and equate to $ \pm 56 $ (because multiplying the determinant by $ \tfrac{1}{2} $ must yield $ \pm 28 $):
Step 3: Identify valid integer solutions for k
After combining like terms correctly, one obtains a quadratic equation in $k$. From the detailed solution, the correct form that yields an integer solution is:
$ 5k^2 + 13k - 46 = 0. $
Solving this, we get two roots (using the quadratic formula). Only one of them, $k=2$, is an integer. Hence, $k=2$ is the suitable choice.
Step 4: Substitute k = 2 back to find the coordinates
With $k = 2$, the vertices become:
A$(2, -6)$, B$(5, 2)$, and C$(-2, 2)$.
Step 5: Find two altitudes of the triangle
Altitude from A:
We need the line passing through A$(2,-6)$ and perpendicular to the side BC. First, the slope of BC is:
$\text{slope of BC} = \frac{2 - 2}{-2 - 5} = \frac{0}{-7} = 0.$
A line perpendicular to one of slope $0$ has an undefined slope (vertical line). Therefore, the altitude from A is simply:
$ x = 2. $
Altitude from B:
We need the line passing through B$(5,2)$ and perpendicular to AC. First, compute the slope of AC:
$\text{slope of AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2.$
A line perpendicular to a line of slope $-2$ has slope $\frac{1}{2}$. Hence, the equation of the altitude from B is:
$ y - 2 = \frac{1}{2}\,(x - 5).\quad\text{(Point-slope form)}$
Rearranging:
$2y - 4 = x - 5 \quad \Longrightarrow \quad x - 2y = 1.$
Step 6: Find the intersection of the two altitudes
We now solve the system:
(1) $ x = 2 $
(2) $ x - 2y = 1. $
Substitute $ x = 2 $ in (2):
$ 2 - 2y = 1 \;\Longrightarrow\; -2y = -1 \;\Longrightarrow\; y = \frac{1}{2}. $
Hence, the intersection point (orthocentre) is:
$ \left(2, \;\frac{1}{2}\right). $
Step 7: Final answer
Therefore, the orthocentre of the triangle is
$ \left( 2,\;\frac{1}{2} \right). $
