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Let's solve each of the problems one by one by finding the equations of the lines in standard form, \(Ax + By = C\), with the given properties.

### 1. Slope = 3, y-intercept = 1
The slope-intercept form is:
\[
y = mx + b
\]
Here, the slope \(m = 3\) and the y-intercept \(b = 1\), so:
\[
y = 3x + 1
\]
Convert to standard form:
\[
y - 3x = 1 \quad \Rightarrow \quad 3x - y = -1
\]
Thus, the equation is:
\[
3x - y = -1
\]

### 2. Passing through (0,2), slope = -4
The slope-intercept form is:
\[
y - y_1 = m(x - x_1)
\]
Using \( (x_1, y_1) = (0, 2) \) and slope \( m = -4 \):
\[
y - 2 = -4(x - 0)
\]
\[
y - 2 = -4x \quad \Rightarrow \quad y = -4x + 2
\]
Convert to standard form:
\[
4x + y = 2
\]

### 3. Passing through (-1,3) and (1,1)
First, find the slope:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{1 - (-1)} = \frac{-2}{2} = -1
\]
Using the point-slope form with point \((-1, 3)\) and slope \(m = -1\):
\[
y - 3 = -1(x + 1)
\]
\[
y - 3 = -x - 1 \quad \Rightarrow \quad y = -x + 2
\]
Convert to standard form:
\[
x + y = 2
\]

### 4. Passing through (1,3), slope = 1/2
Using point-slope form with point \((1, 3)\) and slope \(m = \frac{1}{2}\):
\[
y - 3 = \frac{1}{2}(x - 1)
\]
\[
y - 3 = \frac{1}{2}x - \frac{1}{2} \quad \Rightarrow \quad y = \frac{1}{2}x + \frac{5}{2}
\]
Convert to standard form:
\[
2y = x + 5 \quad \Rightarrow \quad x - 2y = -5
\]

### 5. Passing through (1/2, 1) and (4, 2)
First, find the slope:
\[
m = \frac{2 - 1}{4 - \frac{1}{2}} = \frac{1}{\frac{7}{2}} = \frac{2}{7}
\]
Using point-slope form with point \((\frac{1}{2}, 1)\) and slope \(m = \frac{2}{7}\):
\[
y - 1 = \frac{2}{7}\left(x - \frac{1}{2}\right)
\]
\[
y - 1 = \frac{2}{7}x - \frac{1}{7} \quad \Rightarrow \quad y = \frac{2}{7}x + \frac{6}{7}
\]
Convert to standard form:
\[
7y = 2x + 6 \quad \Rightarrow \quad 2x - 7y = -6
\]

### Summary of the Standard Forms:
1. \( 3x - y = -1 \)
2. \( 4x + y = 2 \)
3. \( x + y = 2 \)
4. \( x - 2y = -5 \)
5. \( 2x - 7y = -6 \)

Would you like further details or explanations on any of these steps?

Here are 5 related questions:
1. How do you convert slope-intercept form to standard form?
2. What is the formula for finding the slope between two points?
3. Can a line have a slope of 0? What would that look like?
4. How would you write the equation of a vertical line?
5. What happens if two points have the same x-value? How does that affect the slope?

**Tip:** In standard form, it’s common practice to make sure the coefficient of \(x\) is positive and all coefficients are integers.

Find the equation of each line in standard form with the given properties.

Learn how to find the equation of a line in standard form given the slope, intercept, or specific points. This solution covers linear equations, slope-intercept form, and converting to standard form, suitable for students in Grades 8-10.

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