# Straight Lines

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Straight Lines

The equation of a straight line is $y=mx+c$, but it can also be written in other forms, such as $y-y_{1}=m(x-x_{1})$ and $ax+by+c=0$. On this page you will learn how to find the equation of a straight line and how to convert between the forms of straight line equation, as well as finding the length and midpoint of straight line segments. Finally, we will put all of this knowledge together to study parallel and perpendicular lines.

A Level

## 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}

$y-3=2(x-1)$

A Level

## 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.

A Level
A Level

## 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}

A Level

## 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.

A Level
GCSE

## Example 1: Converting Between Forms of Straight Line Equations

Find the equation of the straight line passing through $(1,1)$ and $(3,0)$, in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers.

[3 marks]

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

Find $m$:

\begin{aligned}m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\[1.2em]&=\dfrac{0-1}{3-1}\\[1.2em]&=\dfrac{-1}{2}\end{aligned}

Substitute into equation:

$y-1= - \dfrac{1}{2}(x-1)$

$y-1= - \dfrac{1}{2}x+\dfrac{1}{2}$

$\dfrac{1}{2}x-\dfrac{1}{2}+y-1=0$

$\dfrac{1}{2}x+y-\dfrac{3}{2}=0$

Multiply by $2$ to turn into whole numbers:

$x+2y-3=0$

A Level

## Example 2: Parallel and Perpendicular Lines

The line $l_{1}$ has a gradient of $2$. Find equations for:

i) $l_{2}$, a parallel line that passes through $(1,1)$

ii) $l_{3}$, a perpendicular line that passes through $(2,3)$

in the form $y=mx+c$.

[4 marks]

i) $l_{2}$ is parallel to $l_{1}$ so has the same gradient as $l_{1}$ so has a gradient of $2$.

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

$y-1=2(x-1)$

$y - 1 = 2x - 2$

$y = 2x - 1$

ii) $l_{3}$ is perpendicular to $l_{1}$ so has gradient $\dfrac{-1}{\text{gradient of }l_{1}}=\dfrac{-1}{2}$

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

$y-3=-\dfrac{1}{2}(x-2)$

$y-3 = -\dfrac{1}{2}x + 1$

$y = - \dfrac{1}{2}x + 4$

A Level

## Example Questions

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

Find $m$:

\begin{aligned}m&=\dfrac{3-4}{3-(-1)}\\[1.2em]&=\dfrac{-1}{4}\end{aligned}

Substitute into equation:

\begin{aligned}y-4&=-\dfrac{1}{4}(x-(-1))\\[1.2em]&=-\dfrac{1}{4}(x+1)\\[1.2em]&=-\dfrac{1}{4}x-\dfrac{1}{4}\end{aligned}

\begin{aligned}y&= -\dfrac{1}{4}x-\dfrac{1}{4}+4\\[1.2em]&= -\dfrac{1}{4}x+\dfrac{15}{4}\end{aligned}

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

Find $m$:

\begin{aligned}m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\[1.2em]&=\dfrac{32-15}{31-12}\\[1.2em]&=\dfrac{17}{19}\end{aligned}

Substitute into equation:

$y-15=\dfrac{17}{19}(x-12)$

$y-15=\dfrac{17}{19}x-\dfrac{204}{19}$

$\dfrac{17}{19}x-y-\dfrac{204}{19}+15=0$

$\dfrac{17}{19}x-y+\dfrac{81}{19}=0$

Multiply by $19$ to get integers:

$17x-19y+81=0$

a) \begin{aligned}\text{midpoint}&=\left(\dfrac{-40+10}{2},\dfrac{-80+40}{2}\right)\\[1.2em]&=\left(\dfrac{-30}{2},\dfrac{-20}{2}\right)\\[1.2em]&=(-15,-10)\end{aligned}

b) \begin{aligned}\text{length}&=\sqrt{(10-(-40))^{2}+(40-(-80))^{2}}\\[1.2em]&=\sqrt{50^{2}+120^{2}}\\[1.2em]&=\sqrt{2500+14400}\\[1.2em]&=\sqrt{16900}\\[1.2em]&=130\end{aligned}

a) $3x+2y+1=0$

$\dfrac{3}{2}x+y+1=0$

$y=-\dfrac{3}{2}x-1$

$\text{gradient}=-\dfrac{3}{2}$

b) $l_{2}$ is parallel to $l_{1}$ so has the same gradient as $l_{1}$ so has a gradient of $-\dfrac{3}{2}$

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

$y-(-1)=-\dfrac{3}{2}(x-4)$

$y+1=-\dfrac{3}{2}x+6$

$y+1+\dfrac{3}{2}x-6=0$

$\dfrac{3}{2}x+y-5=0$

$3x+2y-10=0$

c) $l_{3}$ is perpendicular to $l_{1}$ so has a gradient of $\dfrac{-1}{\text{gradient of }l_{1}}=\dfrac{-1}{\left( \dfrac{-3}{2}\right)}=\dfrac{2}{3}$

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

$y-(-9)=\dfrac{2}{3}(x-(-5))$

$y+9=\dfrac{2}{3}(x+5)$

$y+9=\dfrac{2}{3}x+\dfrac{10}{3}$

$\dfrac{2}{3}x+\dfrac{10}{3}-y-9=0$

$\dfrac{2}{3}x-y-\dfrac{17}{3}=0$

$2x-3y-17=0$

A Level

A Level

A Level

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