3 Coordinate geometry in the (x, y) plane — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in 3, read straight through: 4 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

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.

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.

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).

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.

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

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.

3.4 · Use parametric equations in modelling in a variety of contexts.

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.