The image you uploaded contains several questions related to finding equations of lines based on parallel and perpendicular relationships. I will solve each problem step by step and provide the equations in slope-intercept form.

1. Equation passing through (1, -10) parallel to $y = -3x + 2$:

The given line has a slope of -3 (from the equation $y = mx + b$, where $m$ is the slope). Since parallel lines have the same slope, the slope of the new line is also -3.

To find the equation:

• The slope $m = -3$
• The point is $(1, -10)$

Use the point-slope form of the equation:
$y - y_1 = m(x - x_1)$

Substitute: $y - (-10) = -3(x - 1)$

Simplify: $y + 10 = -3(x - 1)$
$y + 10 = -3x + 3$
$y = -3x - 7$

Equation: $y = -3x - 7$

2. Equation passing through (4, 1) parallel to $y = -x - 5$:

The slope of the given line is -1 (since the equation is $y = -x - 5$, which has slope -1). The new line will also have a slope of -1 because they are parallel.

To find the equation:

• The slope $m = -1$
• The point is $(4, 1)$

Use point-slope form:
$y - 1 = -1(x - 4)$

Simplify: $y - 1 = -x + 4$
$y = -x + 5$

Equation: $y = -x + 5$

3. Equation passing through (15, 1) perpendicular to $y = -3x$:

The slope of the given line is -3. For perpendicular lines, the slopes are negative reciprocals. So, the slope of the perpendicular line will be $\frac{1}{3}$.

To find the equation:

• The slope $m = \frac{1}{3}$
• The point is $(15, 1)$

Use point-slope form:
$y - 1 = \frac{1}{3}(x - 15)$

Simplify: $y - 1 = \frac{1}{3}x - 5$
$y = \frac{1}{3}x - 4$

Equation: $y = \frac{1}{3}x - 4$

4. Equation passing through (12, -7) perpendicular to $y = \frac{6}{7}x + 8$:

The slope of the given line is $\frac{6}{7}$. For perpendicular lines, the slope will be the negative reciprocal, which is $-\frac{7}{6}$.

To find the equation:

• The slope $m = -\frac{7}{6}$
• The point is $(12, -7)$

Use point-slope form:
$y + 7 = -\frac{7}{6}(x - 12)$

Simplify: $y + 7 = -\frac{7}{6}x + 14$
$y = -\frac{7}{6}x + 7$

Equation: $y = -\frac{7}{6}x + 7$

5. Equation passing through (-6, -3) parallel to $y = 6$:

The line $y = 6$ is a horizontal line with a slope of 0. A parallel line will also have a slope of 0.

To find the equation:

• The slope $m = 0$
• The point is $(-6, -3)$

The equation of a horizontal line is just $y =$ the y-coordinate of the point.
So, the equation is:
Equation: $y = -3$

These are the solutions for the given problems. Would you like any further explanation on any of these steps?