Coordinate geometry in the (x, y) plane

Section 3

Both years

4 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section 3

Checked against Edexcel 9MA0 section 3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

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3.1

Understand and use the equation of a straight line, including the forms y − y1 = m(x − x1) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; use straight line models in a variety of contexts.

Notes

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Explanation

A non-vertical straight line has constant gradient $m$ and may be written as $y-y_1=m(x-x_1)$ through a known point or rearranged into $ax+by+c=0$ .
Parallel lines have equal gradients, while non-vertical perpendicular lines have gradients whose product is $-1$ .
For two known points, first calculate $m=(y_2-y_1)/(x_2-x_1)$ , then substitute one point into a line equation and check the other point.
In a linear model, interpret the gradient and intercept in context and respect practical restrictions.
A common error is to round a limiting input upwards when this breaks the stated constraint.

Worked example

The line $L$ has equation $2x-5y+7=0$ . Find an equation of the line through $(4,-1)$ that is perpendicular to $L$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.Rearrange $L$ to $y=\dfrac{2}{5}x+\dfrac{7}{5}$ , so its gradient is $\dfrac{2}{5}$ .
2.A perpendicular line has gradient $-\dfrac{5}{2}$ .
3.Hence $y+1=-\dfrac{5}{2}(x-4)$ , which rearranges to $5x+2y-18=0$ .

Answer: $5x+2y-18=0$

Common mistakes

Don't treat the constant $c$ in $ax+by+c=0$ as the $y$ -intercept without first rearranging the equation.
Don't use the negative reciprocal rule on the coefficients without first finding the line's gradient.

Exam tip

When asked for a perpendicular line, extract the original gradient, use its negative reciprocal, then substitute the given point.

Tier 1 · Easy

ORIGINAL

Find the equation of the line with gradient

-3

that passes through

(2,4)

. Write it as

y=mx+c

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

The lines

y=2x+1

and

y=-x+7

meet at

P

. Find the coordinates of

P

and an equation of the line through

P

that is parallel to

3x+y=4

. Write your equation as

ax+by+c=0

. Use integer coefficients with greatest common divisor

1

, and choose

a>0

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

A delivery company models the charge

C

pounds for a journey of

d

kilometres by a straight line. Journeys of

8

km and

14

km cost

\pounds 19.60

and

\pounds 29.80

respectively. Find the model for

C

in terms of

d

, interpret its gradient, and find the greatest whole number of kilometres that can be travelled for at most

\pounds 45

(5)

(Total for Question 1 is 5 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

3.2

Understand and use the coordinate geometry of the circle, including the equation (x − a)² + (y − b)² = r²; complete the square for centre and radius; use circle properties (semicircle angle, chord bisector, tangent-radius).

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Explanation

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a)^2+(y-b)^2=r^2$ .
Complete the square separately in $x$ and $y$ to reveal the centre and radius of a circle given in expanded form.
A radius is perpendicular to the tangent at its endpoint; perpendicular chord bisectors meet at the circumcentre, and an angle subtended by a diameter at the circumference is $90^\circ$ .
When testing tangency, require exactly one intersection or set the perpendicular distance from the centre to the line equal to the radius.
A common error is to read the centre as $(a,b)$ from $(x+a)^2+(y+b)^2$ instead of $(-a,-b)$ .

A point

(x,y)

lies on the circle when its distance from centre

(a,b)

r

Worked example

The point $P(5,3)$ lies on the circle with centre $(2,-1)$ . Determine the tangent at $P$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.The gradient of the radius from $(2,-1)$ to $(5,3)$ is $4/3$ , so the tangent gradient is $-3/4$ .
2.Hence $y-3=-\dfrac{3}{4}(x-5)$ .
3.Multiplying by $4$ and rearranging gives $3x+4y-27=0$ .

Answer: $3x+4y-27=0$

Common mistakes

Don't complete the square but treat $r^2$ as the radius instead of taking its positive square root.
Don't find the radius gradient but forget that the tangent gradient is its negative reciprocal.

Exam tip

For a tangent equation, join the centre to the contact point first, then use perpendicular gradients.

Tier 1 · Easy

ORIGINAL

Find the centre and radius of the circle

x^2+y^2-6x+8y-11=0

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

The points

A(-2,1)

and

B(6,5)

are the endpoints of a diameter of a circle. Find the equation of the circle.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

The points

A(-1,2)

B(5,2)

and

C(1,6)

lie on a circle. Use perpendicular bisectors to find the equation of this circumcircle.

(5)

(Total for Question 1 is 5 marks)

3.3

Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms.

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Explanation

Parametric equations express both coordinates in terms of a third variable: $x=f(t)$ and $y=g(t)$ .
To obtain a Cartesian equation, eliminate the parameter by making $t$ the subject of one equation and substituting into the other, or by using a suitable identity.
To parametrise a Cartesian curve, choose expressions that satisfy it identically, such as $x=a\cos t$ and $y=b\sin t$ for $x^2/a^2+y^2/b^2=1$ .
Record any parameter interval because it determines which part of the Cartesian curve is traced.
A common error is to eliminate $t$ but lose a domain restriction.

Worked example

The curve $x=2t+1$ , $y=t^2-3$ is defined for $t\ge0$ . Find a Cartesian equation and state the corresponding restriction on $x$ .

1.Rearrange $x=2t+1$ to $t=(x-1)/2$ .
2.Therefore $y=((x-1)/2)^2-3=(x-1)^2/4-3$ .
3.Since $t\ge0$ , $x=2t+1\ge1$ .

Answer: $y=\dfrac{(x-1)^2}{4}-3$ , with $x\ge1$

Common mistakes

Don't square an equation to eliminate the parameter and retain Cartesian points not generated by any allowed parameter.
Don't eliminate the parameter correctly but lose the restriction inherited from its stated domain.

Exam tip

After converting to Cartesian form, translate the parameter interval into the corresponding restriction on the Cartesian variable.

Tier 1 · Easy

ORIGINAL

For

x=2t-1

and

y=t+4

, eliminate

t

to obtain a Cartesian equation.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

The curve

y=(x-1)^2+2

is parameterised so that

x=2t+1

. Express

y

in terms of

t

and find the parameter values where the curve meets the line

y=6

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

The curve

x=t+1

y=t^2-2t

meets the line

y=3x-6

at two points. Find the exact coordinates of both points.

(5)

(Total for Question 1 is 5 marks)

3.4

Use parametric equations in modelling in a variety of contexts.

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Explanation

In a parametric model, the parameter often represents time and the pair $(x(t),y(t))$ gives the object's position at that time.
Translate a contextual event into equations: meeting a boundary fixes one coordinate, while a collision requires both coordinates of two objects to agree at the same time.
Eliminating the parameter can reveal the path, but the parameter range still controls the physically modelled section of that path.
Check units, permitted times and whether a calculated event occurs within the model's domain.
A common error is to find where two paths cross without checking that the objects arrive there simultaneously.

Worked example

An arch is modelled by $x=6t$ , $y=12t-3t^2$ for $0\le t\le4$ , with $x$ and $y$ in metres. Find a Cartesian equation for the arch and its horizontal span.

1.Since $x=6t$ , $t=x/6$ .
2.Substitute into $y$ : $y=12(x/6)-3(x/6)^2=2x-x^2/12$ .
3.The parameter interval gives $0\le x\le24$ .
4.The arch meets ground level at its two endpoints, so its horizontal span is $24-0=24$ m.

Answer: $y=2x-\dfrac{x^2}{12}$ for $0\le x\le24$ ; Span $24$ m

Common mistakes

Don't solve for a geometric intersection but assign the two objects different parameter values at the same time.
Don't treat the parameter as a physical coordinate rather than eliminating it and interpreting the stated interval.

Exam tip

In a parametric model, connect the parameter endpoints to the physical endpoints before reporting dimensions.

Tier 1 · Easy

ORIGINAL

A particle's position after

t

seconds is modelled by

x=12t

y=20t-5t^2

, where distances are in metres. Find its position when

t=1.5

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

A robot moves with position

x=2t+1

y=t^2

for

0\leq t\leq4

, where distances are in metres and

t

is in seconds. Find the time at which it crosses the line

y=3x-5

and give the exact coordinates of the crossing point.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

Two particles move for

t\ge0

seconds. Particle

P

has position

(3t+2,t^2+2)

and particle

Q

has position

(12-2t,t+4)

. Determine whether they collide and, if they do, find the time and position of the collision.

(4)

(Total for Question 1 is 4 marks)

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