We find the equation of a straight line passing through two points by finding the gradient and the $y$-intercept of the line.

a)

We can find the gradient ($m$) using the points $A = (x_1,y_1)=(\var{xa},\var{ya})$ and $B = (x_2,y_2)=(\var{xb},\var{yb})$.

As the definition of gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).
The equation for gradient is,

$$\begin{align} m &= \frac{y_2-y_1}{x_2-x_1} \\[0.5em] &= \frac{\simplify[!collectNumbers]{{yb}-{ya}}}{\simplify[!collectNumbers]{{xb}-{xa}}} \\[0.5em] &= \frac{\simplify[]{{yb}-{ya}}}{\simplify{{xb}-{xa}}} \\[0.5em] &= \simplify[simplifyFractions,unitDenominator]{({yb-ya})/({xb-xa})}\text{.} \end{align}$$

b)

Rearranging the equation $y=mx+c$ and substituting either of the points gives

$$c = y_1-mx_1 \quad \mathrm{or} \quad c = y_2-mx_2 \,\text{.}$$

We can then also use this equation with the other point's coordinates to check our answer.

Let's use point $A$ first:

$$\begin{align} c &= y_1-mx_1 \\ &= \var{ya}-\var[fractionnumbers]{m}\times\var{xa} \\ & = \simplify[fractionnumbers]{{ya-m*xa}}\text{.} \end{align}$$

We then check this against point $B$:

$$\begin{align} y_2 &= mx_2 + c \\[0.5em] &= \simplify[fractionNumbers]{{m}{xb}+{c}} \\[0.5em] &= \var[fractionnumbers]{m*xb+c}\text{.} \end{align}$$

c)

We can now substitute these values for $m$ and $c$ into $y=mx+c$  to get:

$$y=\simplify[!noLeadingMinus,fractionNumbers,unitFactor]{{m} x+ {c}}\text{.}$$

The green line drawn on the graph represents the above line equation.