To find the orthocenter of a triangle, we need to determine the intersection of the altitudes of the triangle. The steps involve:

1. Finding the equations of two altitudes.
2. Solving the system of equations to find their point of intersection, which is the orthocenter.

Problem (a)

Given the vertices:

• $A = (0,0)$
• $B = (1010, 44)$
• $C = (44, 4848)$

Step 1: Finding the slopes of sides AB and AC.

• Slope of $AB$: $m_{AB} = \frac{44 - 0}{1010 - 0} = \frac{44}{1010} = \frac{22}{505}$
• Slope of $AC$: $m_{AC} = \frac{4848 - 0}{44 - 0} = \frac{4848}{44} = 110$

Step 2: Slopes of the altitudes (perpendicular to AB and AC)

• Perpendicular slope to $AB$: $m_{\perp AB} = -\frac{1}{m_{AB}} = -\frac{505}{22}$
• Perpendicular slope to $AC$: $m_{\perp AC} = -\frac{1}{m_{AC}} = -\frac{1}{110}$

Step 3: Equations of the altitudes

• The altitude from $C = (44, 4848)$ is perpendicular to $AB$. Using the point-slope form: $y - 4848 = -\frac{505}{22}(x - 44)$
• The altitude from $B = (1010, 44)$ is perpendicular to $AC$: $y - 44 = -\frac{1}{110}(x - 1010)$

Step 4: Solving these two equations gives the orthocenter of the triangle.

Problem (b)

Given the vertices:

• $A = (55, 33)$
• $B = (99, 77)$
• $C = (33, 1313)$

Step 1: Finding the slopes of sides AB and AC.

• Slope of $AB$: $m_{AB} = \frac{77 - 33}{99 - 55} = \frac{44}{44} = 1$
• Slope of $AC$: $m_{AC} = \frac{1313 - 33}{33 - 55} = \frac{1280}{-22} = -\frac{640}{11}$

Step 2: Slopes of the altitudes

• Perpendicular slope to $AB$: $m_{\perp AB} = -\frac{1}{1} = -1$
• Perpendicular slope to $AC$: $m_{\perp AC} = -\frac{1}{-\frac{640}{11}} = \frac{11}{640}$

Step 3: Equations of the altitudes

• The altitude from $C = (33, 1313)$ is perpendicular to $AB$. Using point-slope form: $y - 1313 = -1(x - 33)$
• The altitude from $B = (99, 77)$ is perpendicular to $AC$: $y - 77 = \frac{11}{640}(x - 99)$

Step 4: Solving these two equations gives the orthocenter of the triangle.

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Would you like me to go further and solve these equations explicitly for the coordinates of the orthocenters?

Related Questions:

1. What is the centroid of a triangle, and how is it calculated?
2. How do the orthocenter, centroid, and circumcenter of a triangle relate?
3. Can the orthocenter of a triangle be outside the triangle? In what case?
4. How can we derive the circumcenter of a triangle using perpendicular bisectors?
5. How does the type of triangle (right, acute, or obtuse) affect the location of the orthocenter?

Tip: For any triangle, the orthocenter, centroid, and circumcenter are always collinear on a line called the Euler line.