Key Concepts and Formulas

• Area of a Triangle: Given the vertices of a triangle as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, its area $A$ can be calculated using the determinant formula: $A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$
• Orthocentre: The orthocentre of a triangle is the point of intersection of its altitudes. An altitude is a line segment from a vertex perpendicular to the opposite side.
• Slope of Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1. That is, if a line has slope $m$, a line perpendicular to it will have a slope of $-\frac{1}{m}$ (provided $m \neq 0$).
• Equation of a Line (Point-Slope Form): The equation of a line passing through a point $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$.

Step-by-Step Solution

Step 1: Determine the value of 'k' using the given area.

We are given the vertices $A(k, -3k)$, $B(5, k)$, and $C(-k, 2)$ and the area of the triangle is 28. Using the area formula: $\frac{1}{2} \left| \begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} \right| = 28$ $\left| \begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} \right| = 56$ The determinant itself can be $\pm 56$: $\begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} = \pm 56$ Expanding the determinant along the first row: $k(k - 2) - (-3k)(5 + k) + 1(10 + k^2) = \pm 56$ $k^2 - 2k + 15k + 3k^2 + 10 + k^2 = \pm 56$ $5k^2 + 13k + 10 = \pm 56$

We have two cases:

Case 1: $5k^2 + 13k + 10 = 56$ $5k^2 + 13k - 46 = 0$ Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $k = \frac{-13 \pm \sqrt{13^2 - 4(5)(-46)}}{2(5)}$ $k = \frac{-13 \pm \sqrt{169 + 920}}{10}$ $k = \frac{-13 \pm \sqrt{1089}}{10}$ $k = \frac{-13 \pm 33}{10}$ So, $k_1 = \frac{20}{10} = 2$ $k_2 = \frac{-46}{10} = -\frac{23}{5}$

Case 2: $5k^2 + 13k + 10 = -56$ $5k^2 + 13k + 66 = 0$ Using the quadratic formula: $k = \frac{-13 \pm \sqrt{13^2 - 4(5)(66)}}{2(5)}$ $k = \frac{-13 \pm \sqrt{169 - 1320}}{10}$ $k = \frac{-13 \pm \sqrt{-1151}}{10}$ Since $k$ is an integer, we choose $k=2$ from Case 1.

Step 2: Determine the coordinates of the vertices.

Substituting $k=2$:

• $A(2, -3(2)) = A(2, -6)$
• $B(5, 2)$
• $C(-2, 2)$

The vertices are $A(2, -6)$, $B(5, 2)$, and $C(-2, 2)$.

Step 3: Find the equations of at least two altitudes.

Altitude from A to BC (AD): Slope of BC: $m_{BC} = \frac{2 - 2}{-2 - 5} = \frac{0}{-7} = 0$ Since BC is horizontal ($y=2$), the altitude AD is vertical. The equation of the altitude AD is $x = 2$.

Altitude from B to AC (BE): Slope of AC: $m_{AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2$ Slope of BE: $m_{BE} = -\frac{1}{m_{AC}} = -\frac{1}{-2} = \frac{1}{2}$ The equation of the altitude BE passing through $B(5, 2)$ with slope $\frac{1}{2}$: $y - 2 = \frac{1}{2}(x - 5)$ $2y - 4 = x - 5$ $x - 2y = 1$

Step 4: Find the orthocentre.

We have the two altitude equations: $x = 2$ $x - 2y = 1$ Substitute $x=2$ into the second equation: $2 - 2y = 1$ $-2y = -1$ $y = \frac{1}{2}$ The orthocentre is $\left(2, \frac{1}{2}\right)$.

Common Mistakes & Tips

• Remember the $\pm$ when using the area formula to find $k$.
• Pay attention to constraints like "k is an integer."
• Be careful when finding slopes of horizontal and vertical lines.

Summary

We used the area formula to find $k=2$, then found the coordinates of the vertices. We then calculated the equations of two altitudes and solved them simultaneously to find the orthocentre, which is $\left(2, \frac{1}{2}\right)$.

The final answer is $\boxed{\left( {2,{1 \over 2}} \right)}$, which corresponds to option (C). The problem in the prompt states the correct answer is (A), however, after careful calculation, the correct answer is (C). Let us re-evaluate the problem, working backwards from option (A) to see if there is an error in the given information. If the orthocentre is at $(1, \frac{3}{4})$, then $x=1$ is the equation of the altitude from A to BC. Therefore, BC must be a horizontal line. The y-coordinate of B and C must be the same. $k=2$, therefore the y-coordinate of B is 2. The y coordinate of C is given as 2, therefore $k=2$ is correct. Slope of AC is $\frac{2 - (-3k)}{-k - k} = \frac{2+3k}{-2k} = \frac{2+6}{-4} = -2$. The altitude from B to AC has slope $\frac{1}{2}$. The equation of this altitude is: $y - k = \frac{1}{2}(x - 5)$ $y - 2 = \frac{1}{2}(x - 5)$ $2y - 4 = x - 5$ $x - 2y = 1$ Since the orthocentre is at $(1, \frac{3}{4})$: $1 - 2(\frac{3}{4}) = 1 - \frac{3}{2} = -\frac{1}{2} \neq 1$ Therefore, $(1, \frac{3}{4})$ is not the orthocentre. The given correct answer is incorrect. The correct answer is $\boxed{\left( {2,{1 \over 2}} \right)}$, which corresponds to option (C). Key Concepts and Formulas

• Area of a Triangle: Given the vertices of a triangle as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, its area $A$ can be calculated using the determinant formula: $A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$
• Orthocentre: The orthocentre of a triangle is the point of intersection of its altitudes. An altitude is a line segment from a vertex perpendicular to the opposite side.
• Slope of Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1. That is, if a line has slope $m$, a line perpendicular to it will have a slope of $-\frac{1}{m}$ (provided $m \neq 0$).
• Equation of a Line (Point-Slope Form): The equation of a line passing through a point $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$.

Step-by-Step Solution

Step 1: Determine the value of 'k' using the given area.

We are given the vertices $A(k, -3k)$, $B(5, k)$, and $C(-k, 2)$ and the area of the triangle is 28. Using the area formula: $\frac{1}{2} \left| \begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} \right| = 28$ $\left| \begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} \right| = 56$ The determinant itself can be $\pm 56$: $\begin{vmatrix} k & -3k & 1 \\ 5 & k & 1 \\ -k & 2 & 1 \end{vmatrix} = \pm 56$ Expanding the determinant along the first row: $k(k - 2) - (-3k)(5 + k) + 1(10 + k^2) = \pm 56$ $k^2 - 2k + 15k + 3k^2 + 10 + k^2 = \pm 56$ $5k^2 + 13k + 10 = \pm 56$

We have two cases:

Case 1: $5k^2 + 13k + 10 = 56$ $5k^2 + 13k - 46 = 0$ Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $k = \frac{-13 \pm \sqrt{13^2 - 4(5)(-46)}}{2(5)}$ $k = \frac{-13 \pm \sqrt{169 + 920}}{10}$ $k = \frac{-13 \pm \sqrt{1089}}{10}$ $k = \frac{-13 \pm 33}{10}$ So, $k_1 = \frac{20}{10} = 2$ $k_2 = \frac{-46}{10} = -\frac{23}{5}$

Case 2: $5k^2 + 13k + 10 = -56$ $5k^2 + 13k + 66 = 0$ Using the quadratic formula: $k = \frac{-13 \pm \sqrt{13^2 - 4(5)(66)}}{2(5)}$ $k = \frac{-13 \pm \sqrt{169 - 1320}}{10}$ $k = \frac{-13 \pm \sqrt{-1151}}{10}$ Since $k$ is an integer, we choose $k=2$ from Case 1.

Step 2: Determine the coordinates of the vertices.

Substituting $k=2$:

• $A(2, -3(2)) = A(2, -6)$
• $B(5, 2)$
• $C(-2, 2)$

The vertices are $A(2, -6)$, $B(5, 2)$, and $C(-2, 2)$.

Step 3: Find the equations of at least two altitudes.

Altitude from A to BC (AD): Slope of BC: $m_{BC} = \frac{2 - 2}{-2 - 5} = \frac{0}{-7} = 0$ Since BC is horizontal ($y=2$), the altitude AD is vertical. The equation of the altitude AD is $x = 2$.

Altitude from B to AC (BE): Slope of AC: $m_{AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2$ Slope of BE: $m_{BE} = -\frac{1}{m_{AC}} = -\frac{1}{-2} = \frac{1}{2}$ The equation of the altitude BE passing through $B(5, 2)$ with slope $\frac{1}{2}$: $y - 2 = \frac{1}{2}(x - 5)$ $2y - 4 = x - 5$ $x - 2y = 1$

Step 4: Find the orthocentre.

We have the two altitude equations: $x = 2$ $x - 2y = 1$ Substitute $x=2$ into the second equation: $2 - 2y = 1$ $-2y = -1$ $y = \frac{1}{2}$ The orthocentre is $\left(2, \frac{1}{2}\right)$.

Common Mistakes & Tips

• Remember the $\pm$ when using the area formula to find $k$.
• Pay attention to constraints like "k is an integer."
• Be careful when finding slopes of horizontal and vertical lines.

Summary

We used the area formula to find $k=2$, then found the coordinates of the vertices. We then calculated the equations of two altitudes and solved them simultaneously to find the orthocentre, which is $\left(2, \frac{1}{2}\right)$.

The final answer is $\boxed{\left( {2,{1 \over 2}} \right)}$, which corresponds to option (C).