QUESTION 1 1.1 The line PQ has equation $$ y=3 x-8 $$ and point $$ P $$ has coordination $$ (6,10) $$. find the equation of the line that passes through $$ P $$ and is perpendicular to $$ P Q $$.

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To find the equation of the line that passes through point $$ P(6, 10) $$ and is perpendicular to the line $$ PQ $$ with equation $$ y = 3x - 8 $$, follow these steps:

1. **Determine the slope of line $$ PQ $$:**
   
   The given equation $$ y = 3x - 8 $$ is in slope-intercept form $$ y = mx + b $$, where $$ m $$ is the slope. Therefore, the slope of $$ PQ $$ is:
   $$
   m_{PQ} = 3
   $$

2. **Find the slope of the perpendicular line:**
   
   The slope of a line perpendicular to another is the negative reciprocal of the original slope. Thus, the slope $$ m_{\perp} $$ of the perpendicular line is:
   $$
   m_{\perp} = -\frac{1}{3}
   $$

3. **Use the point-slope form to find the equation of the perpendicular line:**
   
   The point-slope form of a line is:
   $$
   y - y_1 = m (x - x_1)
   $$
   Plugging in $$ P(6, 10) $$ and $$ m_{\perp} = -\frac{1}{3} $$:
   $$
   y - 10 = -\frac{1}{3}(x - 6)
   $$

4. **Simplify to slope-intercept form:**
   
   $$
   y - 10 = -\frac{1}{3}x + 2 \
   y = -\frac{1}{3}x + 12
   $$

**Final Equation:**
$$
y = -\frac{1}{3}x + 12
$$
The equation of the line that passes through point $$ P(6, 10) $$ and is perpendicular to $$ PQ $$ is: $$ y = -\frac{1}{3}x + 12 $$

QUESTION 1 1.1 The line PQ has equation \( y=3 x-8 \) and point \( P \) has coordination \( (6,10) \). find the equation of the line that passes through \( P \) and is perpendicular to \( P Q \).

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