finding equation of a straight line

Every straight line can be written as $y = mx + b$, with slope $m$ and $y$-intercept $b$. To build the equation you need two things: the slope, and one point the line passes through.

When you are given two points, the slope is the rise over the run: $m = \frac{y_2 - y_1}{x_2 - x_1}$. In the figure, going from $(1, 1)$ to $(4, 3)$ the run is $3$ and the rise is $2$, so the slope is $\frac{2}{3}$.

Once you have the slope and any point, use point-slope form $y - y_1 = m(x - x_1)$ and simplify. With slope $\frac{2}{3}$ and point $(1, 1)$: $y - 1 = \frac{2}{3}(x - 1)$, which gives $y = \frac{2}{3}x + \frac{1}{3}$.

Parallel lines have equal slopes; perpendicular lines have slopes that are negative reciprocals (they multiply to $-1$). A line parallel to $y = \frac{2}{3}x + 1$ has slope $\frac{2}{3}$, and one perpendicular to it has slope $-\frac{3}{2}$.

Find the equation of the line with slope $-\frac{1}{3}$ passing through $(0, 5)$.
The point $(0, 5)$ is the $y$-intercept, so $b = 5$.
The equation is $y = -\frac{1}{3}x + 5$.

Find the equation of the line through $(0, 32)$ and $(100, 212)$.
Slope $= \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}$.
The $y$-intercept is $32$ (from $(0, 32)$), so the equation is $y = \frac{9}{5}x + 32$.

Find the line parallel to $3x - 5y = 4$ that passes through $(0, 2)$.
Rewrite $3x - 5y = 4$ as $y = \frac{3}{5}x - \frac{4}{5}$, so its slope is $\frac{3}{5}$.
A parallel line has the same slope, and with $y$-intercept $2$: $y = \frac{3}{5}x + 2$.