Let's solve each problem step by step.

Problem 9

We are given:

• Point: $(3, 3)$
• Slope: $m = -\frac{1}{6}$

We want to find the equation of the line in standard form. The point-slope form of a line is:

$y - y_1 = m(x - x_1)$

Substitute the given point and slope:

$y - 3 = -\frac{1}{6}(x - 3)$

Simplify:

$y - 3 = -\frac{1}{6}x + \frac{1}{2}$

Now add 3 to both sides:

$y = -\frac{1}{6}x + \frac{1}{2} + 3$

Convert 3 to a fraction:

$y = -\frac{1}{6}x + \frac{1}{2} + \frac{6}{2} = -\frac{1}{6}x + \frac{7}{2}$

To write this in standard form, we eliminate fractions by multiplying the entire equation by 6:

$6y = -x + 21$

Now rearrange to standard form $Ax + By = C$:

$x + 6y = 21$

Thus, the equation in standard form is:

$x + 6y = 21$

Problem 10

We are given:

• Point: $(2, -5)$
• Slope: $m = -\frac{5}{2}$

Using the point-slope form again:

$y - y_1 = m(x - x_1)$

Substitute the given point and slope:

$y + 5 = -\frac{5}{2}(x - 2)$

Simplify:

$y + 5 = -\frac{5}{2}x + 5$

Subtract 5 from both sides:

$y = -\frac{5}{2}x$

To eliminate the fraction, multiply the entire equation by 2:

$2y = -5x$

Rearranging to standard form:

$5x + 2y = 0$

Thus, the equation in standard form is:

$5x + 2y = 0$

Let me know if you would like further details on the solutions.

Follow-up questions:

1. How is point-slope form derived from the slope formula?
2. What steps are needed to convert slope-intercept form into standard form?
3. What is the advantage of using the standard form of a line equation?
4. How would you graph a line using its standard form equation?
5. What does it mean for coefficients in the standard form to be integral?

Tip:

When converting to standard form, make sure the coefficient of $x$ is positive and all coefficients are integers!