This consists in isolating $$y-p_1$$ from the continuous equation of the straight line:$$$\displaystyle \begin{array}{rcl} \frac{x-p_1}{v_1}& = & \frac{y-p_2}{v_2} \\ y-p_2 & = & \frac{v_2}{v_1} (x-p_1)\\ y-p_2 & = & m \cdot (x-p_1)\end{array}$$$ where $$\displaystyle m =\frac{v_2}{v_1}$$ 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 $$OX$$
- The slope of a straight line is a measurement of the inclination of the straight line: $$m=0 \longrightarrow $$ horizontal straight line, $$m=1 \longrightarrow$$ straight line with inclination of $$45^\circ$$, $$m <0 \longrightarrow $$ 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 $$\overrightarrow{v}= (v_1,v_2)$$ is a vector director of a straight line $$r$$, the slope of the above mentioned straight line $$r$$ will be $$\displaystyle m =\frac{v_2}{v_1}$$
- If we know the slope $$m$$ of a straight line, a vector director of this one is $$\overrightarrow {v}=(1,m)$$
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: $$$y-p_2=m \cdot (x-p_1)$$$
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: $$$(x, y) = A + k \cdot \overrightarrow {AB} = (3, 4) + k \cdot (-5, 2)$$$ Therefore, the parametrical equations of the straight line are: $$$\left. \begin{array}{rcl} x=3-5 \cdot k \\ y=4+2 \cdot k \end{array} \right\}$$$ Isolating $$k$$ we obtain the continuous equation $$$\displaystyle \begin{array}{rcl} k&=&\frac{x-3}{-5} \\ k &=& \frac{y-4}{2}\end{array}$$$ and finally, isolating $$y - 4$$ and re-writing it we have: $$$y-4=\displaystyle \frac{2}{-5}(x-3)=\frac{-2}{5}(x-3)$$$ which is the equation slope-point of the straight line.
The slope of the straight line is $$m =-\displaystyle \frac{2}{5}$$.