If \((3,-2)\) and \((-2,3)\) are two vertices and \((6,-1)\) is the orthocenter of a triangle, then the third vertex would be \(\ldots \ldots\) (a) \((1,6)\) (b) \((-1,6)\) (c) \((1,-6)\) (d) none of these

The orthocenter is the point where the three altitudes of a triangle meet. An altitude of a triangle is a line segment from any vertex of the triangle, perpendicular to the line containing the opposite side. Note that the orthocenter can lie inside or outside the triangle.

We know two vertices of the triangle and the orthocenter: Vertex A: (3, -2) Vertex B: (-2, 3) Orthocenter H: (6, -1) Let the third vertex be C, and let P and Q be the feet of the altitudes from A and B, respectively. Then, the midpoints of CH and BP are the midpoints of the sides AB and BC, respectively. We find the midpoint between H and A, as an average of the coordinates: Midpoint of CH : \(\left(\frac{3+6}{2},\frac{-2+(-1)}{2}\right) = \left(\frac{9}{2},\frac{-3}{2}\right) = (4.5, -1.5)\) Similarly, we find the midpoint between B and H: Midpoint of BH : \(\left(\frac{(-2)+6}{2},\frac{3+(-1)}{2}\right) = \left(\frac{4}{2},\frac{2}{2}\right) = (2, 1)\)

We will now use the midpoint formula to find the equations of the lines containing sides AB and BC. The slope of line AB, m_AB, can be found using the slope formula: m_AB = \(\frac{(-2) - 3}{3 - (-2)} = \frac{-5}{5} = -1\) So, the equation of line AB is of the form: \(y+1 = -1(x-3) \Rightarrow y = -x+2\). Now we find the slope of line BC, m_BC. As H is the orthocenter, the altitude from vertex C must be perpendicular to line AB. So, the line containing side BC has a slope perpendicular to the line containing side AB. Recalling that two lines are perpendicular if their slopes are negative reciprocals, we have: m_BC = \(\frac{1}{1} = 1\) Thus, the equation of line BC is of the form: \(y-1 = 1(x-2) \Rightarrow y = x+1\).

Now we have the equations of the lines containing sides AB and BC. We also have the midpoints of CH and BP. We can now find the equations of the altitudes from vertices A and B, using the fact that these altitudes are perpendicular to the sides AB and BC, respectively. Then, we will solve these equations simultaneously to find the coordinates of C. Since altitude AP is perpendicular to line AB, its slope m_AP is the negative reciprocal of m_AB: m_AP = \(\frac{1}{-1} = -1\) Using the midpoint of CH (4.5, -1.5) and slope -1, the equation of altitude AP is: \(y - (-1.5) = -1(x - 4.5) \Rightarrow y = -x + 6\) Similarly, since altitude BQ is perpendicular to line BC, its slope m_BQ is: m_BQ = -\frac{1}{1} = -1 Using the midpoint of BH (2, 1) and slope -1, the equation of altitude BQ is: \(y - 1 = -1(x - 2) \Rightarrow y = -x + 3\) Now we can solve these equations simultaneously to find vertex C: \begin{cases} y = -x + 6 \\ y = -x + 3 \end{cases} Subtracting the first equation from the second, we get: 0 = 3 This implies that there is no solution for vertex C. So, the third vertex is not one of the given options.

Based on our result in Step 4, the correct answer is: (d) none of these