Coordinate geometry in the (x, y) plane — Edexcel A-Level Mathematics

Coordinate geometry in the (x, y) plane involves using algebra to describe and analyse geometric shapes, focusing on straight lines and circles. Key skills include finding gradients, equations of lines, midpoints, distances, circle equations, and solving problems involving tangents and intersections.

• Substitutes a given point and gradient into y − y₁ = m(x − x₁) and rearranges correctly to a required form.
• Finds a gradient from two points using the change in y divided by the change in x, keeping signs consistent.
• Applies m₁ = m₂ for parallel lines and m₁m₂ = −1 for perpendicular lines, using the negative reciprocal where needed.
• Rearranges between y = mx + c, y − y₁ = m(x − x₁) and ax + by + c = 0 without losing terms or signs.
• Interprets gradient and intercepts in a modelling context, for example fixed charge as the intercept and rate as the gradient.
• Reads the centre and radius directly from (x − a)² + (y − b)² = r², taking care with the signs of a and b.
• Completes the square in both x and y to convert a general circle equation into centre-radius form.
• Uses the angle in a semicircle being a right angle to set up a perpendicular or right-angled triangle condition.
• Applies the perpendicular from the centre to a chord bisecting the chord to find a midpoint or a length.
• Uses the radius being perpendicular to the tangent to find a tangent gradient as the negative reciprocal of the radius gradient.
• Substitutes correctly from one parametric equation into the other to eliminate the parameter.
• Uses a trigonometric identity such as cos²θ + sin²θ = 1 to convert between Cartesian and parametric forms.
• Chooses a suitable parameter when converting a Cartesian equation to parametric form and states it clearly.
• States any restriction on the parameter or on x and y so the parametric form represents the correct portion of the curve.
• Finds coordinates of points on the curve at a given parameter value by substituting into the parametric equations.
• Correctly substitutes a given parameter value into both x(t) and y(t) to obtain a coordinate pair on the modelled curve.
• Eliminates the parameter correctly, for example rearranging x = 3t to t = x/3 and substituting into y = 2t² to obtain y = 2x²/9.
• Differentiates parametric equations using dy/dx = (dy/dt) ÷ (dx/dt), with dx/dt not equal to zero, to find a gradient at a stated point.
• Interprets the parameter in context, such as identifying t as time and restricting t to t ≥ 0 where the model requires it.
• States a suitable domain or range for the parameter and explains how the physical context limits it.

• 💡Substitute the given point back into your final equation to check it satisfies the line.
• 💡State the gradient clearly before using a parallel or perpendicular condition.
• 💡In context questions, label what the gradient and intercept represent with their units.
• 💡Sketch the circle with its centre marked to avoid sign errors.
• 💡When a chord or tangent is involved, draw the radius to the relevant point and mark the right angle.
• 💡Check the radius by substituting a known point on the circle into the equation.
• 💡Write the parameter clearly and show each substitution step.
• 💡After converting, test one parameter value in both forms to check they agree.
• 💡Write the parameter value and both coordinates clearly before substituting, so arithmetic slips are easy to spot.
• 💡When eliminating the parameter, show the rearrangement step explicitly; this earns method credit even if later algebra slips.
• 💡Check the final Cartesian equation against one known point from the parametric form to confirm the two forms agree.

• Using the reciprocal instead of the negative reciprocal for a perpendicular gradient: the correction is to flip the fraction and change the sign, so a gradient of 2 gives −1/2.
• Dropping the negative sign when rearranging into ax + by + c = 0: the correction is to move every term carefully and check by substituting the original point.
• Assuming all lines can be written as y = mx + c: the correction is to use ax + by + c = 0 for vertical lines such as x = 4.
• Reading the centre as (a, b) without reversing signs: the error is taking (x − 3)² as centre (−3, b); the correction is that (x − 3)² gives x-coordinate 3.
• Forgetting to subtract the constant when completing the square: the correction is to balance both sides so the equation becomes (x − a)² + (y − b)² = r².
• Using the radius gradient itself as the tangent gradient: the correction is to take the negative reciprocal, since radius and tangent are perpendicular.
• Eliminating the parameter incorrectly by adding equations that should be substituted: the correction is to express t from one equation and substitute into the other.
• Losing the domain restriction: the error is presenting a full curve when only an arc is defined; the correction is to state the range of the parameter.
• Forgetting the identity when converting a circle: the correction is to use x = r cos θ and y = r sin θ so that cos²θ + sin²θ = 1 holds.
• Eliminating the parameter by dividing one equation by the other without checking that the divisor is non-zero; instead rearrange one equation for t and substitute into the other.
• Forgetting to square the whole expression when substituting, for example writing y = 2x/3 instead of y = 2(x/3)²; instead apply the exponent to the entire bracket.
• Ignoring the context's restriction on the parameter, such as allowing negative time; instead state the valid interval for the parameter before interpreting the model.