Equation slope-point of the straight line

This consists in isolating $y - p_{1}$ from the continuous equation of the straight line: $\begin{array}{rcl} \frac{x - p_{1}}{v_{1}} & = & \frac{y - p_{2}}{v_{2}} \\ y - p_{2} & = & \frac{v_{2}}{v_{1}} (x - p_{1}) \\ y - p_{2} & = & m \cdot (x - p_{1}) \end{array}$ where $m = \frac{v_{2}}{v_{1}}$ is the slope of the straight line.

Some remarkable properties of the slope are:

The slope of a straight line is the tangent of the angle that forms the straight line with the axis $O X$
The slope of a straight line is a measurement of the inclination of the straight line: $m = 0 ⟶$ horizontal straight line, $m = 1 ⟶$ straight line with inclination of $45^{\circ}$ , $m < 0 ⟶$ sloping straight line down.
Two straight lines that have the same slope are parallel (they can be the same).
We can know the angle between two straight lines from their respective slopes.
If $\vec{v} = (v_{1}, v_{2})$ is a vector director of a straight line $r$ , the slope of the above mentioned straight line $r$ will be $m = \frac{v_{2}}{v_{1}}$
If we know the slope $m$ of a straight line, a vector director of this one is $\vec{v} = (1, m)$

An important property of the equation slope-point is that it allows us to write the equation of the straight line just using the slope and a point of the straight line.

Precisely, if we want a straight line with a slope $m$ that crosses point $P = (p_{1}, p_{2})$ we will have to write: $y - p_{2} = m \cdot (x - p_{1})$

Example

Find the equation slope-point of the straight line $r$ that crosses points $(3, 4)$ and $(- 2, 6)$ .

The vector equation with $A = (3, 4)$ and $B = (- 2, 6)$ is: $(x, y) = A + k \cdot \vec{A B} = (3, 4) + k \cdot (- 5, 2)$ Therefore, the parametrical equations of the straight line are: $\begin{array}{rcl} x = 3 - 5 \cdot k \\ y = 4 + 2 \cdot k \end{array}}$ Isolating $k$ we obtain the continuous equation $\begin{array}{rcl} k & = & \frac{x - 3}{- 5} \\ k & = & \frac{y - 4}{2} \end{array}$ and finally, isolating $y - 4$ and re-writing it we have: $y - 4 = \frac{2}{- 5} (x - 3) = \frac{- 2}{5} (x - 3)$ which is the equation slope-point of the straight line.

The slope of the straight line is $m = - \frac{2}{5}$ .