Determine the equation of a line given its gradient and a point, or given two points.

Given $m = -3$, $x_1 = -1$, $y_1 = 2$.

$$y - 2 = -3(x - (-1))$$ $$y - 2 = -3(x + 1)$$

$$y - 2 = -3x - 3$$ $$y = -3x - 3 + 2$$ $$y = -3x - 1$$

The equation of the line is $\boxed{y = -3x - 1}$.

1.c) Determine the equation of the line with gradient $-1$ and passing through $\left(\frac{1}{2}, -5\right)$.

Given $m = -1$, $x_1 = \frac{1}{2}$, $y_1 = -5$.

$$y - (-5) = -1\left(x - \frac{1}{2}\right)$$ $$y + 5 = -x + \frac{1}{2}$$

$$y = -x + \frac{1}{2} - 5$$ $$y = -x + \frac{1}{2} - \frac{10}{2}$$ $$y = -x - \frac{9}{2}$$

The equation of the line is $\boxed{y = -x - \frac{9}{2}}$.

2.a) Determine the equation of the line through the points $(-1, 7)$ and $(-2, -5)$.

Let $(x_1, y_1) = (-1, 7)$ and $(x_2, y_2) = (-2, -5)$.

$$m = \frac{-5 - 7}{-2 - (-1)}$$ $$m = \frac{-12}{-2 + 1}$$ $$m = \frac{-12}{-1}$$ $$m = 12$$

$$y - 7 = 12(x - (-1))$$ $$y - 7 = 12(x + 1)$$

$$y - 7 = 12x + 12$$ $$y = 12x + 12 + 7$$ $$y = 12x + 19$$

The equation of the line is $\boxed{y = 12x + 19}$.

2.c) Determine the equation of the line through the points $(5, 0)$ and $(-3, 2)$.

Let $(x_1, y_1) = (5, 0)$ and $(x_2, y_2) = (-3, 2)$.

$$m = \frac{2 - 0}{-3 - 5}$$ $$m = \frac{2}{-8}$$ $$m = -\frac{1}{4}$$

$$y - 0 = -\frac{1}{4}(x - 5)$$ $$y = -\frac{1}{4}x + \frac{5}{4}$$

The equation of the line is $\boxed{y = -\frac{1}{4}x + \frac{5}{4}}$.