b) The line perpendicular to $$ y=-x+3 $$ c) The line parallel to $$ y=-\frac{1}{2} x-1 $$ d) The line perpendicular to $$ y=\frac{2}{3} x-4 $$ et The line perpendicular to $$ y=-\frac{1}{4} x+1 $$

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To find the equations of lines that are either perpendicular or parallel to the given lines, we need to follow these steps:

1. **Identify the slope of the given line.**
2. **For parallel lines:** The slope remains the same.
3. **For perpendicular lines:** The slope is the negative reciprocal of the given line's slope.

Let's solve each part step by step.

### b) The line perpendicular to $$ y = -x + 3 $$

1. The slope of the given line $$ y = -x + 3 $$ is $$ m = -1 $$.
2. The slope of the perpendicular line is the negative reciprocal: 
   $$
   m_{\text{perpendicular}} = -\frac{1}{-1} = 1
   $$
3. The equation of the line can be written in point-slope form as $$ y - y_1 = m(x - x_1) $$. We can choose any point, but let's use the y-intercept of the original line, which is $$ (0, 3) $$:
   $$
   y - 3 = 1(x - 0) \implies y = x + 3
   $$

### c) The line parallel to $$ y = -\frac{1}{2} x - 1 $$

1. The slope of the given line $$ y = -\frac{1}{2} x - 1 $$ is $$ m = -\frac{1}{2} $$.
2. The slope of the parallel line remains the same: 
   $$
   m_{\text{parallel}} = -\frac{1}{2}
   $$
3. Using the same point as before $$ (0, -1) $$:
   $$
   y + 1 = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x - 1
   $$

### d) The line perpendicular to $$ y = \frac{2}{3} x - 4 $$

1. The slope of the given line $$ y = \frac{2}{3} x - 4 $$ is $$ m = \frac{2}{3} $$.
2. The slope of the perpendicular line is:
   $$
   m_{\text{perpendicular}} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}
   $$
3. Using the y-intercept $$ (0, -4) $$:
   $$
   y + 4 = -\frac{3}{2}(x - 0) \implies y = -\frac{3}{2}x - 4
   $$

### e) The line perpendicular to $$ y = -\frac{1}{4} x + 1 $$

1. The slope of the given line $$ y = -\frac{1}{4} x + 1 $$ is $$ m = -\frac{1}{4} $$.
2. The slope of the perpendicular line is:
   $$
   m_{\text{perpendicular}} = -\frac{1}{-\frac{1}{4}} = 4
   $$
3. Using the y-intercept $$ (0, 1) $$:
   $$
   y - 1 = 4(x - 0) \implies y = 4x + 1
   $$

### Summary of Results
- b) The line perpendicular to $$ y = -x + 3 $$ is $$ y = x + 3 $$.
- c) The line parallel to $$ y = -\frac{1}{2} x - 1 $$ is $$ y = -\frac{1}{2} x - 1 $$.
- d) The line perpendicular to $$ y = \frac{2}{3} x - 4 $$ is $$ y = -\frac{3}{2} x - 4 $$.
- e) The line perpendicular to $$ y = -\frac{1}{4} x + 1 $$ is $$ y = 4x + 1 $$.
- b) $$ y = x + 3 $$ - c) $$ y = -\frac{1}{2}x - 1 $$ - d) $$ y = -\frac{3}{2}x - 4 $$ - e) $$ y = 4x + 1 $$

b) The line perpendicular to \( y=-x+3 \) c) The line parallel to \( y=-\frac{1}{2} x-1 \) d) The line perpendicular to \( y=\frac{2}{3} x-4 \) et The line perpendicular to \( y=-\frac{1}{4} x+1 \)

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