If the sides BC, CA, AB of `Delta`ABC are respectively x + 2y = 1, 3x + y + 5 = 0, x - y + 2 = 0. Then, the altitude through B is

Options

A. x - 3y + 1 = 0
B. x - 3y + 4 = 0
C. 3x -y +4= 0
D. x - y +2=0

Answer

To find the altitude through point B in triangle ABC, where the sides BC, CA, and AB are given by the equations:

1. BC: \( x + 2y = 1 \)
2. CA: \( 3x + y + 5 = 0 \)
3. AB: \( x - y + 2 = 0 \)

we will follow these steps:

### Step 1: Find the coordinates of point B
To find point B, we need to determine the intersection of lines AB and BC.

**Equations:**
- AB: \( x - y + 2 = 0 \) (Equation 1)
- BC: \( x + 2y - 1 = 0 \) (Equation 2)

**Solving the equations:**
From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = x + 2 \]

Now, substitute \( y \) into Equation 2:
\[ x + 2(x + 2) - 1 = 0 \]
\[ x + 2x + 4 - 1 = 0 \]
\[ 3x + 3 = 0 \]
\[ x = -1 \]

Now substitute \( x = -1 \) back into the equation for \( y \):
\[ y = -1 + 2 = 1 \]

Thus, the coordinates of point B are \( B(-1, 1) \).

### Step 2: Find the slope of line AC
Next, we need to find the slope of line AC.

**Equation of AC:**
\[ 3x + y + 5 = 0 \quad \Rightarrow \quad y = -3x - 5 \]

The slope \( m_{AC} \) of line AC is:
\[ m_{AC} = -3 \]

### Step 3: Find the slope of line BD (altitude from B)
Since BD is perpendicular to AC, the slope of BD, \( m_{BD} \), can be found using the negative reciprocal of \( m_{AC} \):
\[ m_{BD} = \frac{1}{3} \]

### Step 4: Write the equation of line BD
Using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (-1, 1) \) and \( m = \frac{1}{3} \):
\[ y - 1 = \frac{1}{3}(x + 1) \]

Multiplying through by 3 to eliminate the fraction:
\[ 3(y - 1) = x + 1 \]
\[ 3y - 3 = x + 1 \]
\[ x - 3y + 4 = 0 \]

### Step 5: Conclusion
The equation of the altitude through point B is:
\[ x - 3y + 4 = 0 \]