This is the most useful line equation form as it only requires one point and a slope
(of course the slope acquisition might require a second point) to write the line equation.
$$y-y_{1}=m\left(x-x_{1}\right)$$
The point is $(x_1,y_1)$ and the slope is $m$.
Example: Given the parabola equation, $f(x)=x^2$, a point on the curve, $P=(1,1)$, and a slope, $m=2$, what is the line equation that goes through the point?
Answer: There is extraneous information in the problem! We don't need the curve because we have a point and a slope. Our equation is just
$$y-y_P=m(x-x_P)$$
$$y-1=2(x-1)$$
$$y=2x-1$$
If we graph this and also show the curve, we see that the line is tangent to the curve at point P, but so what? We were only asked to get the line equation.
Example: Find the line with an angle of $30^{\circ}$ up from the $x$-axis that goes through the point $(1,2)$.
Answer: Really the only thing to do here is translate a $30^{\circ}$ angle from the origin
into a slope. We know that $30^{\circ}$ is represented on the unit circle as $\frac{\pi}{6}$ and
thus the $\frac{rise}{run}$ must be $$m= \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)}\approx 0.5774$$
Given $m$, we can just plug in to get the equation: $$y-2=0.5774(x-1)$$