he line perpendicular to $$ 3 x+y=8 $$ that asses through $$ (0,-2) $$

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To find the equation of the line perpendicular to $$ 3x + y = 8 $$ that passes through the point $$ (0, -2) $$, follow these steps:

1. **Find the slope of the given line**:
   
   Start by rewriting the equation in slope-intercept form ($$ y = mx + b $$):
   $$
   3x + y = 8 \quad \Rightarrow \quad y = -3x + 8
   $$
   The slope ($$ m_1 $$) of the given line is **-3**.

2. **Determine the slope of the perpendicular line**:
   
   The slope of a line perpendicular to another is the negative reciprocal of the original slope. Therefore:
   $$
   m_2 = -\frac{1}{m_1} = -\frac{1}{-3} = \frac{1}{3}
   $$

3. **Use the point-slope form to find the equation of the perpendicular line**:
   
   The point-slope form is given by:
   $$
   y - y_1 = m(x - x_1)
   $$
   Plugging in the slope $$ m_2 = \frac{1}{3} $$ and the point $$ (0, -2) $$:
   $$
   y - (-2) = \frac{1}{3}(x - 0) \quad \Rightarrow \quad y + 2 = \frac{1}{3}x
   $$
   
4. **Simplify to the slope-intercept form**:
   $$
   y = \frac{1}{3}x - 2
   $$

**Final Answer:**
$$
y = \frac{1}{3}\,x\; -\;2
$$
The equation of the perpendicular line is $$ y = \frac{1}{3}x - 2 $$.

he line perpendicular to \( 3 x+y=8 \) that asses through \( (0,-2) \)

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