Coordinate Geometry – OCR FSMQ Additional Maths

Coordinate geometry turns geometry into algebra. Once a shape has coordinates, questions about lengths, angles and shapes become calculations you can do without measuring anything.

The specification restricts this section to two dimensions only. Everything happens in the $x$–$y$ plane. (Three-dimensional problems appear instead under Pythagoras and Trigonometry.)

What is new compared with GCSE

The equation of a circle · normals as well as tangents
Sketching polynomial, trigonometric and exponential curves
Linear programming — an entirely new application

Word	Meaning
Gradient	Steepness: the change in $y$ divided by the change in $x$.
Intercept	Where a graph crosses an axis.
Line segment	The finite piece of a line between two points.
Mid-point	The point halfway along a line segment.
Tangent	A line touching a curve, with the same gradient as the curve there.
Normal	The line perpendicular to the tangent at the point of contact.
Chord	A straight line joining two points on a curve.
Feasible region	The set of points satisfying every constraint in a linear programming problem.
Objective function	The quantity being maximised or minimised.

Distance between two points (CG1)

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Mid-point of a line segment (CG2)

$\left(\dfrac{x_1+x_2}{2},\; \dfrac{y_1+y_2}{2}\right)$

Gradient

$m = \dfrac{y_2-y_1}{x_2-x_1}$

Equation of a straight line

$y = mx + c$ or $y - y_1 = m(x - x_1)$

Parallel and perpendicular

Parallel: $m_1 = m_2$ · Perpendicular: $m_1 m_2 = -1$

Equation of a circle (CG3)

$(x-a)^2 + (y-b)^2 = r^2$, centre $(a, b)$, radius $r$

Worked Example — Four formulae in one question

$A(1, 2)$ and $B(7, 10)$. Find (a) the length $AB$, (b) the mid-point, (c) the equation of $AB$, (d) the equation of the circle with $AB$ as a diameter.

①(a) $AB = \sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36+64} = \sqrt{100} = 10$

②(b) Mid-point $= \left(\tfrac{1+7}{2}, \tfrac{2+10}{2}\right) = (4, 6)$

③(c) $m = \dfrac{10-2}{7-1} = \dfrac{8}{6} = \dfrac{4}{3}$

④$y - 2 = \tfrac43(x-1)$, so $y = \tfrac43 x + \tfrac23$, or $3y = 4x + 2$.

⑤(d) The centre is the mid-point $(4,6)$ and the radius is half the diameter: $r = 5$.

⑥$(x-4)^2 + (y-6)^2 = 25$

Check a circle equation by substituting a known point. $A(1,2)$: $(1-4)^2 + (2-6)^2 = 9 + 16 = 25$ ✓

CG1
CG2Straight LinesDistance, mid-point, gradient, and the equation of a line.
CG3The Geometry of Circles$(x-a)^2+(y-b)^2=r^2$, and finding centre and radius.
CG4Sketching CurvesLinear, polynomial, trigonometric and exponential graphs.
CG5Tangents and NormalsGradient, intercept, tangent and normal in problems.
CG6
CG7
CG8Linear ProgrammingFeasible regions, objective functions, and 2-D optimisation.

Subtracting in a different order top and bottom. The gradient is $\dfrac{y_2-y_1}{x_2-x_1}$. Doing $\dfrac{y_2-y_1}{x_1-x_2}$ gives the wrong sign.

Getting the circle's centre sign wrong. $(x-3)^2 + (y+2)^2 = 16$ has centre $(3, -2)$, not $(-3, 2)$. The signs in the brackets are opposite to the centre's coordinates.

Forgetting to square-root for the radius. In $(x-a)^2+(y-b)^2 = 25$, the radius is $5$, not $25$.

Confusing tangent and normal. The tangent has the curve's gradient; the normal has gradient $-1 \div$ that. Read which one is asked for.

Reading vertices off a hand-drawn graph. In linear programming, solve the boundary lines simultaneously instead.

Distance

Pythagoras on the differences.

Mid-point

Average each coordinate.

Gradient

Rise over run, consistent order.

Line equation

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

Perpendicular

Negative reciprocal gradient.

Circle

$(x-a)^2+(y-b)^2=r^2$.

Centre signs

Opposite to those in the brackets.

Diameter

Mid-point is the centre; half the length is $r$.

Tangent

Touches; same gradient as the curve.

Optimum

In linear programming, always at a vertex.

Question 1

Find the distance between $(2, 3)$ and $(6, 6)$.

▶ Show solution

$\sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$

Question 2

Find the mid-point of the segment joining $(-3, 5)$ and $(7, -1)$.

▶ Show solution

$\left(\dfrac{-3+7}{2}, \dfrac{5+(-1)}{2}\right) = (2, 2)$

Question 3

Find the equation of the line through $(2, 7)$ with gradient $3$.

▶ Show solution

$y - 7 = 3(x - 2)$

$y = 3x + 1$

Question 4

Write down the centre and radius of the circle $(x+1)^2 + (y-4)^2 = 36$.

▶ Show solution

Centre $(-1, 4)$ — opposite signs to the brackets.

Radius $\sqrt{36} = 6$.

Question 5

Find the gradient of a line perpendicular to $y = 4x - 3$.

▶ Show solution

$m_1 m_2 = -1$, so $m_2 = -\dfrac{1}{4}$.

Question 6

Find the equation of the circle with centre $(3, -2)$ and radius $\sqrt{13}$.

▶ Show solution

$(x-3)^2 + (y+2)^2 = 13$

Question 7

Show that the triangle with vertices $A(0,0)$, $B(4,2)$ and $C(-1,2)$ is right-angled at $A$.

▸ Show solution

Gradient of $AB = \dfrac{2-0}{4-0} = \dfrac{1}{2}$

Gradient of $AC = \dfrac{2-0}{-1-0} = -2$

Product $= \dfrac{1}{2} \times (-2) = -1$, so $AB$ and $AC$ are perpendicular.

Therefore the angle at $A$ is $90°$.

Check with Pythagoras: $AB^2 = 20$, $AC^2 = 5$, $BC^2 = 5^2 + 0^2 = 25$, and $20 + 5 = 25$ ✓

Question 8

Does the point $(5, 1)$ lie on, inside or outside the circle $(x-2)^2 + (y-1)^2 = 16$?

▶ Show solution

Substitute: $(5-2)^2 + (1-1)^2 = 9 + 0 = 9$.

$9 < 16$, so the point is inside the circle.

(Its distance from the centre is $3$, less than the radius $4$.)

Question 9

$P(1,4)$ and $Q(7,2)$. Find the equation of the perpendicular bisector of $PQ$.

▶ Show solution

Mid-point of $PQ$: $\left(\tfrac{1+7}{2}, \tfrac{4+2}{2}\right) = (4, 3)$

Gradient of $PQ$: $\dfrac{2-4}{7-1} = -\dfrac13$

Perpendicular gradient: $3$

$y - 3 = 3(x - 4)$, so $y = 3x - 9$.

Question 10

A circle has centre $C(2, 3)$ and passes through $A(6, 6)$.

(a) Find the radius and the equation of the circle. (b) Find the equation of the tangent to the circle at $A$. (c) Find the coordinates of the other end $B$ of the diameter through $A$. (d) Verify that the angle $\angle AXB$ is $90°$ for the point $X(2, 8)$ on the circle.

▶ Show solution

(a) $r = CA = \sqrt{(6-2)^2 + (6-3)^2} = \sqrt{16+9} = 5$

$(x-2)^2 + (y-3)^2 = 25$

(b) The tangent at $A$ is perpendicular to the radius $CA$.

Gradient of $CA = \dfrac{6-3}{6-2} = \dfrac34$, so the tangent gradient is $-\dfrac43$.

$y - 6 = -\tfrac43(x-6)$, giving $y = -\tfrac43 x + 14$, or $3y + 4x = 42$.

(c) $C$ is the mid-point of $AB$, so $B = (2 \times 2 - 6,\; 2 \times 3 - 6) = (-2, 0)$.

Check: $CB = \sqrt{16 + 9} = 5$ ✓

(d) First confirm $X(2,8)$ is on the circle: $(2-2)^2 + (8-3)^2 = 0 + 25 = 25$ ✓

Gradient of $XA = \dfrac{6-8}{6-2} = -\dfrac12$

Gradient of $XB = \dfrac{0-8}{-2-2} = \dfrac{-8}{-4} = 2$

Product $= -\tfrac12 \times 2 = -1$, so $XA \perp XB$ and the angle at $X$ is $90°$ ✓

This is the circle theorem "the angle in a semicircle is a right angle", verified by coordinate geometry rather than by a geometric argument.