Straight Lines

A LevelAQAEdexcelOCR

Straight Lines

The equation of a straight line is y=mx+cy=mx+c, but it can also be written in other forms, such as yy1=m(xx1)y-y_{1}=m(x-x_{1}) and ax+by+c=0ax+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.

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Equation of a Straight Line

yy1=m(xx1)y-y_{1}=m(x-x_{1})

The above is the equation of a straight line through two points (x1,y1),(x2,y2)(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, mm.

m=y2y1x2x1m=dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

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

m=5321=21=2begin{aligned}m&=dfrac{5-3}{2-1}[1.2em]&=dfrac{2}{1}[1.2em]&=2end{aligned}

y3=2(x1)y-3=2(x-1)

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Converting Between Forms of Straight Line Equations

There are three forms of straight line equation. We have already met yy1=m(xx1)y-y_{1}=m(x-x_{1}). The other two are:

y=mx+cy=mx+c

ax+by+c=0ax+by+c=0

You need to know how to reach both of these from yy1=m(xx1)y-y_{1}=m(x-x_{1}).

 

yy1=m(xx1)mathbf{y-y_{1}=m(x-x_{1})}y=mx+cmathbf{rightarrow y=mx+c}

yy1=m(xx1)y-y_{1}=m(x-x_{1})

yy1=mxmx1y-y_{1}=mx-mx_{1}

y=mx+y1mx1y=mx+y_{1}-mx_{1}

 

yy1=m(xx1)mathbf{y-y_{1}=m(x-x_{1})}ax+by+c=0mathbf{rightarrow ax+by+c=0}

yy1=m(xx1)y-y_{1}=m(x-x_{1})

yy1=mxmx1y-y_{1}=mx-mx_{1}

mxy+y1mx1=0mx-y+y_{1}-mx_{1}=0

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

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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 l1l_{1} and l2l_{2} are:

parallel if gradient of l1=gradient of l2text{parallel if gradient of }l_{1}=text{gradient of }l_{2}

perpendicular if gradient of l1=1gradient of l2text{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+cy=mx+c form.

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A LevelAQAEdexcelOCR

Midpoint and Length of a Line Segment

Consider a line segment connecting two points (x1,y1),(x2,y2)(x_{1},y_{1}),(x_{2},y_{2}). The midpoint and length of the line segment are:

midpoint=(x1+x22,y1+y22)text{midpoint}=left(dfrac{x_{1}+x_{2}}{2},dfrac{y_{1}+y_{2}}{2}right)

length=(x2x1)2+(y2y1)2text{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)(1,2) and (7,10)(7,10).

midpoint=(1+72,2+102)=(82,122)=(4,6)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}

 

length=(71)2+(102)2=62+82=36+64=100=10begin{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]&=10end{aligned}

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A LevelAQAEdexcelOCR

Example 1: Converting Between Forms of Straight Line Equations

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

[3 marks]

yy1=m(xx1)y-y_{1}=m(x-x_{1})

Find mm:

m=y2y1x2x1=0131=12begin{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:

y1=12(x1)y-1= – dfrac{1}{2}(x-1)

y1=12x+12y-1= – dfrac{1}{2}x+dfrac{1}{2}

12x12+y1=0dfrac{1}{2}x-dfrac{1}{2}+y-1=0

12x+y32=0dfrac{1}{2}x+y-dfrac{3}{2}=0

Multiply by 22 to turn into whole numbers:

x+2y3=0x+2y-3=0

 

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Example 2: Parallel and Perpendicular Lines

The line l1l_{1} has a gradient of 22. Find equations for:

i) l2l_{2}, a parallel line that passes through (1,1)(1,1)

ii) l3l_{3}, a perpendicular line that passes through (2,3)(2,3)

in the form y=mx+cy=mx+c.

[4 marks]

i) l2l_{2} is parallel to l1l_{1} so has the same gradient as l1l_{1} so has a gradient of 22.

yy1=m(xx1)y-y_{1}=m(x-x_{1})

y1=2(x1)y-1=2(x-1)

y1=2x2y – 1 = 2x – 2

y=2x1y = 2x – 1

 

ii) l3l_{3} is perpendicular to l1l_{1} so has gradient 1gradient of l1=12dfrac{-1}{text{gradient of }l_{1}}=dfrac{-1}{2}

yy1=m(xx1)y-y_{1}=m(x-x_{1})

y3=12(x2)y-3=-dfrac{1}{2}(x-2)

y3=12x+1y-3 = -dfrac{1}{2}x + 1

y=12x+4y = – dfrac{1}{2}x + 4

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Straight Lines Example Questions

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

[2 marks]

A Level AQAEdexcelOCR

yy1=m(xx1)y-y_{1}=m(x-x_{1})

 

Find mm:

 

m=343(1)=14begin{aligned}m&=dfrac{3-4}{3-(-1)}[1.2em]&=dfrac{-1}{4}end{aligned}

 

Substitute into equation:

 

y4=14(x(1))=14(x+1)=14x14begin{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}

 

y=14x14+4=14x+154begin{aligned}y&= -dfrac{1}{4}x-dfrac{1}{4}+4[1.2em]&= -dfrac{1}{4}x+dfrac{15}{4}end{aligned}

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

[2 marks]

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yy1=m(xx1)y-y_{1}=m(x-x_{1})

 

Find mm:

 

m=y2y1x2x1=32153112=1719begin{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:

 

y15=1719(x12)y-15=dfrac{17}{19}(x-12)

 

y15=1719x20419y-15=dfrac{17}{19}x-dfrac{204}{19}

 

1719xy20419+15=0dfrac{17}{19}x-y-dfrac{204}{19}+15=0

 

1719xy+8119=0dfrac{17}{19}x-y+dfrac{81}{19}=0

 

Multiply by 1919 to get integers:

 

17x19y+81=017x-19y+81=0

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Question 3: The points (40,80)(-40,-80) and (10,40)(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 Level AQAEdexcelOCR

a) midpoint=(40+102,80+402)=(302,202)=(15,10)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) length=(10(40))2+(40(80))2=502+1202=2500+14400=16900=130begin{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]&=130end{aligned}

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Question 4: The line l1l_{1} has the form 3x+2y+1=03x+2y+1=0.

a) What is the gradient of l1l_{1}

b) Find in ax+by+c=0ax+by+c=0 form the equation for the line l2l_{2}, which passes through (4,1)(4,-1) and is parallel to l1l_{1}

c) Find in ax+by+c=0ax+by+c=0 form the equation for the line l3l_{3}, which passes through (5,9)(-5,-9) and is perpendicular to l1l_{1}

[8 marks]

A Level AQAEdexcelOCR

a) 3x+2y+1=03x+2y+1=0

 

32x+y+1=0dfrac{3}{2}x+y+1=0

 

y=32x1y=-dfrac{3}{2}x-1

 

gradient=32text{gradient}=-dfrac{3}{2}

 

 

b) l2l_{2} is parallel to l1l_{1} so has the same gradient as l1l_{1} so has a gradient of 32-dfrac{3}{2}

 

yy1=m(xx1)y-y_{1}=m(x-x_{1})

 

y(1)=32(x4)y-(-1)=-dfrac{3}{2}(x-4)

 

y+1=32x+6y+1=-dfrac{3}{2}x+6

 

y+1+32x6=0y+1+dfrac{3}{2}x-6=0

 

32x+y5=0dfrac{3}{2}x+y-5=0

 

3x+2y10=03x+2y-10=0

 

 

c) l3l_{3} is perpendicular to l1l_{1} so has a gradient of 1gradient of l1=1(32)=23dfrac{-1}{text{gradient of }l_{1}}=dfrac{-1}{left( dfrac{-3}{2}right)}=dfrac{2}{3}

 

yy1=m(xx1)y-y_{1}=m(x-x_{1})

 

y(9)=23(x(5))y-(-9)=dfrac{2}{3}(x-(-5))

 

y+9=23(x+5)y+9=dfrac{2}{3}(x+5)

 

y+9=23x+103y+9=dfrac{2}{3}x+dfrac{10}{3}

 

23x+103y9=0dfrac{2}{3}x+dfrac{10}{3}-y-9=0

 

23xy173=0dfrac{2}{3}x-y-dfrac{17}{3}=0

 

2x3y17=02x-3y-17=0

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

C1 – Understand and use the equation of a straight line, including the forms yy1=m(xx1)y – y_1 = m(x-x_1) and ax+by+c=0ax+by+c=0; gradient conditions for two straight lines to be parallel or perpendicular
Be able to use straight line models in a variety of contexts

Straight Lines Worksheet and Example Questions

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Straight Line Coordinate Geometry

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