# Straight Lines

## Straight Lines Revision

**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.

**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}

**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]**

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

**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

## Straight Lines Example Questions

**Question 1: **A straight line passes through (-1,4) and (3,3). Find its equation in y=mx+c form.

**[2 marks]**

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}

**Question 2: **A straight line passes through the points (12,15) and (31,32). Find its equation in ax+by+c=0 form, where a,b,c are integers.

**[2 marks]**

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

**Question 3: **The points (-40,-80) and (10,40) have a line segment between them.

a) What is the midpoint of the line segment?

b) What is the length of the line segment?

**[4 marks]**

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}

**Question 4: **The line l_{1} has the form 3x+2y+1=0.

a) What is the gradient of l_{1}

b) Find in ax+by+c=0 form the equation for the line l_{2}, which passes through (4,-1) and is parallel to l_{1}

c) Find in ax+by+c=0 form the equation for the line l_{3}, which passes through (-5,-9) and is perpendicular to l_{1}

**[8 marks]**

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