Straight Lines

Equation of a Straight Line

y-y_{1}=m(x-x_{1})

The above is the equation of a straight line through two points (x_{1},y_{1}),(x_{2},y_{2}). The first point is present clearly in the equation. The second point comes in for the calculation of the gradient, m.

m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Example: Find the equation of the straight line through (1,3) and (2,5).

\begin{aligned}m&=\dfrac{5-3}{2-1}\\[1.2em]&=\dfrac{2}{1}\\[1.2em]&=2\end{aligned}

y-3=2(x-1)

Converting Between Forms of Straight Line Equations

There are three forms of straight line equation. We have already met y-y_{1}=m(x-x_{1}). The other two are:

y=mx+c

ax+by+c=0

You need to know how to reach both of these from y-y_{1}=m(x-x_{1}).

\mathbf{y-y_{1}=m(x-x_{1})}\mathbf{\rightarrow y=mx+c}

y-y_{1}=m(x-x_{1})

y-y_{1}=mx-mx_{1}

y=mx+y_{1}-mx_{1}

\mathbf{y-y_{1}=m(x-x_{1})}\mathbf{\rightarrow ax+by+c=0}

y-y_{1}=m(x-x_{1})

y-y_{1}=mx-mx_{1}

mx-y+y_{1}-mx_{1}=0

Note: It is traditional to multiply through by a factor if necessary to make a,b,c whole numbers for a line in this form.

Parallel and Perpendicular Lines

Two lines are parallel if they have the same gradient.

Two lines are perpendicular if the gradient of the second line is the negative reciprocal of the gradient of the first line.

This means that lines l_{1} and l_{2} are:

\text{parallel if gradient of }l_{1}=\text{gradient of }l_{2}

\text{perpendicular if gradient of }l_{1}=\dfrac{-1}{\text{gradient of }l_{2}}

Tip: It is easiest to compare gradients if you put lines in y=mx+c form.

Midpoint and Length of a Line Segment

Consider a line segment connecting two points (x_{1},y_{1}),(x_{2},y_{2}). The midpoint and length of the line segment are:

\text{midpoint}=\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)

\text{length}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}

Example: Find the midpoint and length of the line segment connecting (1,2) and (7,10).

\begin{aligned}\text{midpoint}&=\left(\dfrac{1+7}{2},\dfrac{2+10}{2}\right)\\[1.2em]&=\left(\dfrac{8}{2},\dfrac{12}{2}\right)\\[1.2em]&=(4,6)\end{aligned}

\begin{aligned}\text{length}&=\sqrt{(7-1)^{2}+(10-2)^{2}}\\[1.2em]&=\sqrt{6^{2}+8^{2}}\\[1.2em]&=\sqrt{36+64}\\[1.2em]&=\sqrt{100}\\[1.2em]&=10\end{aligned}