Mathematics 9709 · Pure Mathematics 1

Equations of straight lines

A gradient and one point fix a line. One formula does every case: $y-y_1=m(x-x_1)$ .
The hard part is seeing which point and which gradient the question gives you.

Point–gradient form: $y-y_1=m(x-x_1)$ . Put in a gradient and any one point on the line, then tidy up. If you only have two points, find $m$ from them first.

Perpendicular bisector = midpoint + perpendicular gradient. Find the midpoint of $AB$ . Turn the gradient of $AB$ upside down and change its sign. Then use point–gradient form. This is a common 3 to 4 mark task.

Point–gradient form: the one formula to remember

You know $y=mx+c$ , but here you usually have a point, not the intercept $c$ , so this form is faster:

$y-y_1=m(x-x_1)$

Get the gradient $m$ (given, or from two points, or from a parallel/perpendicular line).
Pick any point $(x_1,y_1)$ the line passes through.
Substitute into $y-y_1=m(x-x_1)$ and tidy to the form the question asks for.

The line through A(−4, 3) and B(6, −2): run 10 across, rise −5 down, so gradient −½.

Worked examples

Worked example

Find the line through $(-4,\ 3)$ and $(6,\ -2)$

Cambridge Pure Mathematics 1, Worked example 3.7. With two points, find the gradient first, then use either point in the formula.

Gradient first: $m=\dfrac{-2-3}{6-(-4)}=-\dfrac{1}{2}$ .
Point–gradient form through $(-4,3)$ : $y-3=-\tfrac12(x+4)$ .
Clear the fraction and rearrange: $2y-6=-x-4\;\Rightarrow\;x+2y=2$ .

Answer

x+2y=2

Worked example

Real question: $A(5,\ 2)$ , $B(10,\ -1)$ . Find the perpendicular bisector of $AB$

9709/12 Oct/Nov 2022 Q1(a). This has two parts: the midpoint gives the point, and the negative reciprocal gives the gradient.

Midpoint of $AB$ : $M=\left(\tfrac{5+10}{2},\tfrac{2+(-1)}{2}\right)=\left(\tfrac{15}{2},\tfrac12\right)$ .
Gradient of $AB$ : $m_{AB}=\dfrac{-1-2}{10-5}=-\dfrac35$ , so the bisector has gradient $\dfrac53$ .
Point–gradient form through $M$ : $y-\tfrac12=\tfrac53\!\left(x-\tfrac{15}{2}\right)$ .
Multiply by 6 and tidy to integers: $6y-3=10x-75\;\Rightarrow\;5x-3y=36$ .

Answer

5x-3y=36

Worked example

Show-that: $A(-7,\ 3)$ , $B(5,\ 11)$ . Show the perp bisector of $AB$ is $3x + 2y = 11$

9709/13 May/June 2020 Q10(a). A “show that” gives you the answer, so finish by rearranging to that exact line, with no decimals.

Midpoint: $M=\left(\tfrac{-7+5}{2},\tfrac{3+11}{2}\right)=(-1,7)$ .
Gradient of $AB$ : $m_{AB}=\dfrac{11-3}{5-(-7)}=\dfrac{8}{12}=\dfrac23$ , so the bisector has gradient $-\dfrac32$ .
Through $(-1,7)$ : $y-7=-\tfrac32(x+1)$ .
Multiply by 2 and collect: $2y-14=-3x-3\;\Rightarrow\;3x+2y=11$ .

Answer

3x+2y=11\ \checkmark

One line, many correct forms

Why your answer can look different from the mark scheme

The same straight line can be written in many correct ways. In the Q1 example above, $y=\tfrac53 x-12$ , $5x-3y=36$ and $5x-3y-36=0$ all give the same line. This is why mark schemes write “OE” (or equivalent) and “ISW” (ignore subsequent working).

Two rules to follow. First, if the question asks for the form $ax+by+c=0$ with whole numbers, clear the fractions and match that shape. Second, for a “show that,” rearrange to the printed line exactly. A decimal version gets no marks, because it does not match the line you were asked to prove.

Where the marks go

Common mistake

Check the signs inside the bracket. A line through

(-4,3)

gives

y-3=m(x+4)

, because

x-(-4)=x+4

Common mistake

A perpendicular bisector needs both the midpoint and the perpendicular gradient. If you use the gradient of

AB

itself, or forget the midpoint, you lose all the marks. (9709/13 Jun 2020 Q10)

Now you try

Points $A(5,2)$ and $B(10,-1)$ . Find the equation of the circle with centre $A$ that passes through $B$ . (9709/12 Nov 2022 Q1(b))

Your turn— tap to reveal the worked answer (9709/12 Nov 2022 Q1)

The centre is $A(5,2)$ . The radius squared is the distance squared from $A$ to $B$ , so there is no need to take a root:

$r^2=(10-5)^2+(-1-2)^2=25+9=34$ .

Answer

(x-5)^2+(y-2)^2=34

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Equations of straight lines

A gradient and one point fix a line. One formula does every case: $y-y_1=m(x-x_1)$ .
The hard part is seeing which point and which gradient the question gives you.

Point–gradient form: $y-y_1=m(x-x_1)$ . Put in a gradient and any one point on the line, then tidy up. If you only have two points, find $m$ from them first.

Point–gradient form: the one formula to remember

You know $y=mx+c$ , but here you usually have a point, not the intercept $c$ , so this form is faster:

$y-y_1=m(x-x_1)$

Get the gradient $m$ (given, or from two points, or from a parallel/perpendicular line).
Pick any point $(x_1,y_1)$ the line passes through.
Substitute into $y-y_1=m(x-x_1)$ and tidy to the form the question asks for.

The line through A(−4, 3) and B(6, −2): run 10 across, rise −5 down, so gradient −½.

Worked examples

Worked example

Find the line through $(-4,\ 3)$ and $(6,\ -2)$

Cambridge Pure Mathematics 1, Worked example 3.7. With two points, find the gradient first, then use either point in the formula.

Gradient first: $m=\dfrac{-2-3}{6-(-4)}=-\dfrac{1}{2}$ .
Point–gradient form through $(-4,3)$ : $y-3=-\tfrac12(x+4)$ .
Clear the fraction and rearrange: $2y-6=-x-4\;\Rightarrow\;x+2y=2$ .

Answer

x+2y=2

Worked example

Real question: $A(5,\ 2)$ , $B(10,\ -1)$ . Find the perpendicular bisector of $AB$

9709/12 Oct/Nov 2022 Q1(a). This has two parts: the midpoint gives the point, and the negative reciprocal gives the gradient.

Midpoint of $AB$ : $M=\left(\tfrac{5+10}{2},\tfrac{2+(-1)}{2}\right)=\left(\tfrac{15}{2},\tfrac12\right)$ .
Gradient of $AB$ : $m_{AB}=\dfrac{-1-2}{10-5}=-\dfrac35$ , so the bisector has gradient $\dfrac53$ .
Point–gradient form through $M$ : $y-\tfrac12=\tfrac53\!\left(x-\tfrac{15}{2}\right)$ .
Multiply by 6 and tidy to integers: $6y-3=10x-75\;\Rightarrow\;5x-3y=36$ .

Answer

5x-3y=36

Worked example

Show-that: $A(-7,\ 3)$ , $B(5,\ 11)$ . Show the perp bisector of $AB$ is $3x + 2y = 11$

9709/13 May/June 2020 Q10(a). A “show that” gives you the answer, so finish by rearranging to that exact line, with no decimals.

Midpoint: $M=\left(\tfrac{-7+5}{2},\tfrac{3+11}{2}\right)=(-1,7)$ .
Gradient of $AB$ : $m_{AB}=\dfrac{11-3}{5-(-7)}=\dfrac{8}{12}=\dfrac23$ , so the bisector has gradient $-\dfrac32$ .
Through $(-1,7)$ : $y-7=-\tfrac32(x+1)$ .
Multiply by 2 and collect: $2y-14=-3x-3\;\Rightarrow\;3x+2y=11$ .

Answer

3x+2y=11\ \checkmark

One line, many correct forms

Why your answer can look different from the mark scheme

Where the marks go

Common mistake

Check the signs inside the bracket. A line through

(-4,3)

gives

y-3=m(x+4)

, because

x-(-4)=x+4

Common mistake

A perpendicular bisector needs both the midpoint and the perpendicular gradient. If you use the gradient of

AB

itself, or forget the midpoint, you lose all the marks. (9709/13 Jun 2020 Q10)

Now you try

Points $A(5,2)$ and $B(10,-1)$ . Find the equation of the circle with centre $A$ that passes through $B$ . (9709/12 Nov 2022 Q1(b))

Your turn— tap to reveal the worked answer (9709/12 Nov 2022 Q1)

The centre is $A(5,2)$ . The radius squared is the distance squared from $A$ to $B$ , so there is no need to take a root:

$r^2=(10-5)^2+(-1-2)^2=25+9=34$ .

Answer

(x-5)^2+(y-2)^2=34