Answer:
(7,2)
Explanation:
The orthocenter of a triangle is the point of intersection of its altitudes. The altitude of a triangle is a line segment from a vertex perpendicular to the line containing the opposite side.
To find the orthocenter of triangle ABC with vertices A(2,7), B(8,7), and C(7,2), we can follow these steps:
1. Find the slopes of the lines containing each side of the triangle.
2. Find the slopes of the perpendicular lines (altitudes) to each side passing through the opposite vertex.
3. Use the slope-point form to find the equations of the altitudes.
4. Find the point of intersection of the altitudes (the orthocenter).
Let's go through the calculations:
**Slopes of the Sides:**
- Slope of AB: \((7-7)/(8-2) = 0\)
- Slope of BC: \((2-7)/(7-8) = 5\)
- Slope of AC: \((2-7)/(7-2) = -5/5 = -1\)
**Slopes of Perpendicular Lines (Altitudes):**
- Perpendicular to AB (through C): Slope is \(m_{\text{perp}} = -1/0\) (undefined, as AB is a horizontal line). The altitude passes through point C(7,2).
- Perpendicular to BC (through A): Slope is \(m_{\text{perp}} = -1/5\). The altitude passes through point A(2,7).
- Perpendicular to AC (through B): Slope is \(m_{\text{perp}} = 5\). The altitude passes through point B(8,7).
**Equations of Altitudes:**
- Altitude through C: \(x = 7\)
- Altitude through A: \(y = -\frac{1}{5}x + \frac{39}{5}\)
- Altitude through B: \(y = 5x - 33\)
**Orthocenter:**
The orthocenter is the point of intersection of these altitudes. Since the altitude through C is a vertical line at \(x = 7\), it intersects with the altitude through A at (7, 2). Therefore, the orthocenter of triangle ABC is (7, 2).