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: $x - 9y = -27$

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!