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Let's work through each of the problems one by one, converting the given line information into standard form with integer coefficients.

### Problem 11: The line through (2, 3) and (4, 5)
1. First, calculate the slope \((m)\) using the formula:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - 2} = 1
   \]
2. Now use the point-slope form: \(y - y_1 = m(x - x_1)\). Pick the point \((2, 3)\):
   \[
   y - 3 = 1(x - 2) \quad \Rightarrow \quad y = x + 1
   \]
3. Convert to standard form: \(Ax + By = C\):
   \[
   x - y = -1 \quad (\text{which is already in integer form!})
   \]

### Problem 12: The line through \((-4, 6)\) and \((2, -2)\)
1. Calculate the slope:
   \[
   m = \frac{-2 - 6}{2 - (-4)} = \frac{-8}{6} = -\frac{4}{3}
   \]
2. Use the point-slope form with point \((-4, 6)\):
   \[
   y - 6 = -\frac{4}{3}(x + 4)
   \]
   Simplify:
   \[
   y - 6 = -\frac{4}{3}x - \frac{16}{3} \quad \Rightarrow \quad y = -\frac{4}{3}x + \frac{2}{3}
   \]
3. Multiply everything by 3 to eliminate fractions:
   \[
   3y = -4x + 2 \quad \Rightarrow \quad 4x + 3y = 2
   \]

### Problem 13: The line through \((-4, 2)\) with slope 3
1. Start with the point-slope form using point \((-4, 2)\):
   \[
   y - 2 = 3(x + 4)
   \]
   Simplify:
   \[
   y - 2 = 3x + 12 \quad \Rightarrow \quad y = 3x + 14
   \]
2. Convert to standard form:
   \[
   3x - y = -14
   \]

### Problem 14: The line through \((1, 2)\) with slope \(\frac{4}{5}\)
1. Start with the point-slope form using point \((1, 2)\):
   \[
   y - 2 = \frac{4}{5}(x - 1)
   \]
   Simplify:
   \[
   y - 2 = \frac{4}{5}x - \frac{4}{5} \quad \Rightarrow \quad y = \frac{4}{5}x + \frac{6}{5}
   \]
2. Multiply everything by 5 to eliminate fractions:
   \[
   5y = 4x + 6 \quad \Rightarrow \quad 4x - 5y = -6
   \]

### Problem 15: A line through \((3, 1)\) with slope 0
1. Since the slope is 0, the line is horizontal, and the equation is simply \(y = 1\).
2. In standard form, this is:
   \[
   0x + y = 1
   \]

### Problem 16: A line with slope \(\frac{2}{3}\) and y-intercept \((0, 5)\)
1. The slope-intercept form is:
   \[
   y = \frac{2}{3}x + 5
   \]
2. Multiply everything by 3:
   \[
   3y = 2x + 15 \quad \Rightarrow \quad 2x - 3y = -15
   \]

### Problem 17: \(2y = -4x - 12\)
1. Divide by 2:
   \[
   y = -2x - 6
   \]
2. Rearrange into standard form:
   \[
   2x + y = -6
   \]

### Problem 18: \(\frac{2}{3}x + 3 = 6y - 15\)
1. First, eliminate the fraction by multiplying through by 3:
   \[
   2x + 9 = 18y - 45
   \]
2. Simplify and rearrange:
   \[
   2x - 18y = -54
   \]
   Divide through by 2 to get the simplest integer coefficients:
   \[
   x - 9y = -27
   \]

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Would you like further details on any of these problems? Here are five related questions to expand your understanding:

1. How do you find the equation of a line parallel to a given line?
2. What is the relationship between slopes of perpendicular lines?
3. How can you verify that two points lie on a given line equation?
4. How can you convert a line equation from slope-intercept form to point-slope form?
5. What is the significance of the standard form of a line equation?

**Tip:** Always check your final equation by plugging in one of the original points to ensure the equation is correct!

Write an equation of each line in standard form with integer coefficients for the following problems:
11. the line through (2, 3) and (4, 5)
12. the line through (-4, 6) and (2, -2)
13. the line through (-4, 2) with slope 3
14. the line through (1, 2) with slope 4/5
15. a line through (3, 1) with slope 0
16. a line with slope 2/3 and y-intercept (0, 5)
17. 2y = -4x - 12
18. 2/3x + 3 = 6y - 15

Learn how to find the standard form of a line given two points or a slope. This problem set explores converting lines from point-slope form into standard form with integer coefficients. Suitable for Grades 8-10.

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