To find the equation of side BC of triangle ABC, given that point P(2, 3) is the orthocenter, we will follow these steps:
### Step 1: Identify the equations of the given lines
The equations of the sides AB and CA are given as:
1. AB: \( x + 2y = 0 \)
2. CA: \( x - y = 3 \)
### Step 2: Find the slopes of the given lines
To find the slopes of these lines, we can rewrite them in slope-intercept form (y = mx + b).
For line AB:
\[
x + 2y = 0 \implies 2y = -x \implies y = -\frac{1}{2}x
\]
Thus, the slope \( m_{AB} = -\frac{1}{2} \).
For line CA:
\[
x - y = 3 \implies y = x - 3
\]
Thus, the slope \( m_{CA} = 1 \).
### Step 3: Determine the slopes of the altitudes
Since P is the orthocenter, we need to find the slopes of the altitudes from points A and C to lines AB and CA, respectively.
The slope of the altitude from C to AB (perpendicular to AB) is the negative reciprocal of the slope of AB:
\[
m_{altitude \, from \, C} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{2}} = 2
\]
The slope of the altitude from B to CA (perpendicular to CA) is the negative reciprocal of the slope of CA:
\[
m_{altitude \, from \, B} = -\frac{1}{m_{CA}} = -\frac{1}{1} = -1
\]
### Step 4: Write the equations of the altitudes
Using point P(2, 3) and the slopes we found, we can write the equations of the altitudes.
For the altitude from C to AB:
\[
y - 3 = 2(x - 2) \implies y - 3 = 2x - 4 \implies y = 2x - 1
\]
For the altitude from B to CA:
\[
y - 3 = -1(x - 2) \implies y - 3 = -x + 2 \implies y = -x + 5
\]
### Step 5: Find the intersection of the altitudes
Now we need to find the intersection of the two altitude lines to find point B.
Set the equations equal to each other:
\[
2x - 1 = -x + 5
\]
Solving for \( x \):
\[
2x + x = 5 + 1 \implies 3x = 6 \implies x = 2
\]
Substituting \( x = 2 \) back into one of the altitude equations to find \( y \):
\[
y = 2(2) - 1 = 4 - 1 = 3
\]
Thus, point B is at (2, 3).
### Step 6: Find point C
To find point C, we can substitute \( x = 2 \) into the line CA equation:
\[
2 - y = 3 \implies y = -1
\]
Thus, point C is at (2, -1).
### Step 7: Find the equation of line BC
Now we have points B(2, 3) and C(2, -1). The slope of line BC is:
\[
m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{2 - 2} = \text{undefined}
\]
This means line BC is a vertical line at \( x = 2 \).
### Final Answer
The equation of side BC is:
\[
x = 2
\]
