Key Concepts and Formulas

• Intersection of Lines: Solving the system of linear equations representing two lines gives their intersection point.
• Orthocenter and Altitudes: The orthocenter is the intersection of the altitudes of a triangle. An altitude is a line segment from a vertex perpendicular to the opposite side. The product of the slopes of perpendicular lines is $-1$.
• Centroid: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
• Distance Formula: The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Let the two given sides of the triangle be $L_1: x - 2y + 1 = 0$ and $L_2: 2x - y - 1 = 0$. Let the vertices of the triangle be $A, B, C$. The orthocenter is given as $H = \left(\frac{7}{3}, \frac{7}{3}\right)$.

1. Find Vertex A (Intersection of the Two Given Sides)

• Why: Vertex A is the intersection of the two lines $L_1$ and $L_2$. We find its coordinates by solving the system of equations.
• We solve the system of equations: $x - 2y + 1 = 0 \quad \text{(Equation 1)}$ $2x - y - 1 = 0 \quad \text{(Equation 2)}$
• From Equation 1, $x = 2y - 1$. Substituting this into Equation 2: $2(2y - 1) - y - 1 = 0$ $4y - 2 - y - 1 = 0$ $3y - 3 = 0$ $y = 1$
• Substituting $y = 1$ back into $x = 2y - 1$, we get $x = 2(1) - 1 = 1$.
• So, the coordinates of vertex A are $(1, 1)$.

2. Find Equations of the Altitudes

• Why: To find vertices B and C, we need to use the orthocenter. We find the equations of the altitudes from B to AC and C to AB, which pass through the orthocenter.
• Altitude from B to AC (BH):
  • Side AC is represented by $L_1: x - 2y + 1 = 0$. The slope of AC is $m_{AC} = -\frac{1}{-2} = \frac{1}{2}$.
  • Since BH is perpendicular to AC, its slope is $m_{BH} = -\frac{1}{m_{AC}} = -2$.
  • BH passes through $H = \left(\frac{7}{3}, \frac{7}{3}\right)$. Using the point-slope form: $y - \frac{7}{3} = -2\left(x - \frac{7}{3}\right)$ $3y - 7 = -6\left(x - \frac{7}{3}\right)$ $3y - 7 = -6x + 14$ $6x + 3y - 21 = 0$ $2x + y - 7 = 0 \quad \text{(Equation of Altitude BH)}$
• Altitude from C to AB (CH):
  • Side AB is represented by $L_2: 2x - y - 1 = 0$. The slope of AB is $m_{AB} = -\frac{2}{-1} = 2$.
  • Since CH is perpendicular to AB, its slope is $m_{CH} = -\frac{1}{m_{AB}} = -\frac{1}{2}$.
  • CH passes through $H = \left(\frac{7}{3}, \frac{7}{3}\right)$. Using the point-slope form: $y - \frac{7}{3} = -\frac{1}{2}\left(x - \frac{7}{3}\right)$ $6y - 14 = -3\left(x - \frac{7}{3}\right)$ $6y - 14 = -3x + 7$ $3x + 6y - 21 = 0$ $x + 2y - 7 = 0 \quad \text{(Equation of Altitude CH)}$

3. Find Vertices B and C

• Why: Vertex B is the intersection of side AB and altitude BH. Vertex C is the intersection of side AC and altitude CH.
• Find Vertex B:
  • Solve the system of equations for side AB and altitude BH: $2x - y - 1 = 0 \quad \text{(Side AB)}$ $2x + y - 7 = 0 \quad \text{(Altitude BH)}$
  • Adding the two equations: $(2x - y - 1) + (2x + y - 7) = 0$ $4x - 8 = 0$ $x = 2$
  • Substituting $x = 2$ into the equation for side AB: $2(2) - y - 1 = 0$ $3 - y = 0$ $y = 3$
  • So, the coordinates of vertex B are $(2, 3)$.
• Find Vertex C:
  • Solve the system of equations for side AC and altitude CH: $x - 2y + 1 = 0 \quad \text{(Side AC)}$ $x + 2y - 7 = 0 \quad \text{(Altitude CH)}$
  • Adding the two equations: $(x - 2y + 1) + (x + 2y - 7) = 0$ $2x - 6 = 0$ $x = 3$
  • Substituting $x = 3$ into the equation for side AC: $3 - 2y + 1 = 0$ $4 - 2y = 0$ $y = 2$
  • So, the coordinates of vertex C are $(3, 2)$.

4. Calculate the Centroid G

• Why: The centroid is the average of the coordinates of the vertices.
• The vertices are $A=(1,1)$, $B=(2,3)$, and $C=(3,2)$.
• Using the centroid formula: $G = \left(\frac{1 + 2 + 3}{3}, \frac{1 + 3 + 2}{3}\right)$ $G = \left(\frac{6}{3}, \frac{6}{3}\right)$ $G = (2, 2)$
• The centroid of the triangle is $(2, 2)$.

5. Calculate the Distance from the Origin to the Centroid G

• Why: We calculate the distance between the origin and the centroid using the distance formula.
• The origin is $O=(0,0)$ and the centroid is $G=(2,2)$.
• Using the distance formula: $Distance = \sqrt{(2 - 0)^2 + (2 - 0)^2}$ $Distance = \sqrt{2^2 + 2^2}$ $Distance = \sqrt{4 + 4}$ $Distance = \sqrt{8} = 2\sqrt{2}$

Common Mistakes & Tips

• Be careful with signs when calculating slopes and solving linear equations. A small mistake can lead to a completely wrong answer.
• Ensure you're using the correct slope for the perpendicular lines. If the slope of a line is $m$, the slope of a perpendicular line is $-1/m$.
• Double-check your arithmetic, especially when dealing with fractions and square roots.

Summary

We found the vertices of the triangle by intersecting the given sides and using the orthocenter to determine the equations of altitudes. Then, we found the centroid of the triangle using the coordinates of the vertices. Finally, we calculated the distance from the origin to the centroid. The vertices are $A=(1,1)$, $B=(2,3)$, and $C=(3,2)$, the centroid is $G=(2,2)$, and the distance of the centroid from the origin is $2\sqrt{2}$.

The final answer is \boxed{2\sqrt{2}}, which corresponds to option (C).