Equation slope-point of the straight line

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 $m =-\displaystyle \frac{2}{5}$.