Coordinate geometry: A-level Maths OCR revision notes

1.03

Straight lines

Distance, midpoint and gradient 1.03

Key results

Distance between two points: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
Midpoint of the segment joining them: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .
Gradient between two points: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .

Notes

The distance formula is Pythagoras' theorem applied to the right-angled triangle with horizontal side $x_{2} - x_{1}$ and vertical side $y_{2} - y_{1}$ ; the midpoint formula is simply the average of the coordinates.
The order of the two points does not matter for distance or gradient, since both differences change sign together — but it must be consistent between numerator and denominator when finding a gradient.
A distance is never negative, so a squared term coming out negative inside the square root signals an arithmetic slip in the coordinate subtraction, not a genuine result.

The equation of a straight line 1.03

Key results

Through a known point with known gradient: $y - y_{1} = m (x - x_{1})$ .
Gradient-intercept form: $y = m x + c$ , where $c$ is the $y$ -intercept.
General form: $a x + b y + c = 0$ , which can represent every line including vertical ones.

Method

Find the gradient, either from two given points or from a parallel or perpendicular condition.
Substitute that gradient and any one known point into $y - y_{1} = m (x - x_{1})$ .
Rearrange into whatever form the question asks for, and check by substituting a second known point.

Notes

A horizontal line has gradient $0$ and equation $y = k$ ; a vertical line has undefined gradient and equation $x = k$ , which is why $y = m x + c$ cannot describe every line but $a x + b y + c = 0$ can.
Always verify a derived line equation against a second known point, not just the one used to build it — a sign error partway through rearranging is easy to miss otherwise.

Parallel and perpendicular lines 1.03

Key results

Parallel lines have equal gradients: $m_{1} = m_{2}$ .
Perpendicular lines have gradients whose product is $- 1$ : $m_{1} m_{2} = - 1$ , so the perpendicular to a line of gradient $m$ has gradient $- \frac{1}{m}$ .

Notes

The perpendicular gradient is the negative reciprocal — invert the fraction and change the sign, so gradient $- \frac{3}{4}$ gives perpendicular gradient $\frac{4}{3}$ .
A horizontal and a vertical line are perpendicular even though the product rule fails there, because the gradient of the vertical line is undefined; treat that pair as a special case.
Two lines with the same gradient but different $y$ -intercepts are parallel and never meet; the same gradient AND the same intercept means they are the same line, not merely parallel.

Intersections, perpendicular bisectors and feet of perpendiculars 1.03

Definitions

Perpendicular bisector of $A B$ : the line perpendicular to $A B$ through its midpoint — equivalently, the set of all points equidistant from $A$ and $B$ .

Method

For an intersection, solve the two equations simultaneously; substitution is usually quickest when one equation is linear.
For a perpendicular bisector, find the midpoint of $A B$ , take the negative reciprocal of the gradient of $A B$ , then form the line through the midpoint with that gradient.
For the foot of the perpendicular from a point $P$ to a line $L$ , write the line through $P$ perpendicular to $L$ , then solve it simultaneously with $L$ ; the shortest distance from $P$ to $L$ is the distance from $P$ to that foot.

Notes

Finding where two lines meet, where a line meets a curve, or the foot of a perpendicular are all the same task: the geometry sets up the equations, and the algebra does the rest.
The centre of a circle through three given points is found by intersecting the perpendicular bisectors of two of the chords, since the centre is equidistant from all three.
The shortest distance from a point to a line is always measured perpendicular to the line — any other path between the point and the line is longer, which is exactly why the foot-of-perpendicular method works.

Worked examples

Worked example

The points $A (- 2, 5)$ and $B (6, - 1)$ are given.

Find

(a) the length $A B$ ,

(b) the midpoint of $A B$ ,

Show worked solution

(a):

∣ A B ∣ = (6 - (- 2))^{2} + (- 1 - 5)^{2} = 8^{2} + (- 6)^{2} = 64 + 36 = 100 = 10

(b) The midpoint is:

(\frac{- 2 + 6}{2}, \frac{5 + ( - 1 )}{2}) = (2, 2)

m = \frac{- 1 - 5}{6 - ( - 2 )} = \frac{- 6}{8} = - \frac{3}{4}

Using:

y - y_{1} = m (x - x_{1})

at $A (- 2, 5)$ :

y - 5 = - \frac{3}{4} (x + 2)

so:

4 y - 20 = - 3 x - 6

giving $3 x + 4 y = 14$ .

Check with $B$ :

3 (6) + 4 (- 1) = 18 - 4 = 14

Worked example

Find the equation of the perpendicular bisector of $A B$ , where $A (- 2, 5)$ and $B (6, - 1)$ .

Show worked solution

The gradient of $A B$ is $- \frac{3}{4}$ , so the perpendicular gradient is the negative reciprocal $\frac{4}{3}$ .

The bisector passes through the midpoint $(2, 2)$ :

y - 2 = \frac{4}{3} (x - 2)

so $3 y - 6 = 4 x - 8$ , giving $4 x - 3 y = 2$ .

Check: the point $(5, 6)$ satisfies:

4 (5) - 3 (6) = 20 - 18 = 2

and it is indeed equidistant from $A$ and $B$ , since:

(5 + 2)^{2} + (6 - 5)^{2} = 50

and:

(6 - 5)^{2} + (- 1 - 6)^{2} = 50

1.03

Circles

The equation of a circle 1.03

Definitions

Circle: the set of all points at a fixed distance $r$ (the radius) from a fixed point $(a, b)$ (the centre).

Key results

Standard form: $(x - a)^{2} + (y - b)^{2} = r^{2}$ , which is the distance formula applied to a general point $(x, y)$ on the circle and the centre, then squared.
A circle given by a diameter with endpoints $A$ and $B$ has centre at the midpoint of $A B$ and radius $\frac{1}{2} ∣ A B ∣$ .

Notes

Read the signs carefully: $(x + 3)^{2} + (y - 1)^{2} = 16$ has centre $(- 3, 1)$ , not $(3, - 1)$ , and radius $4$ , not $16$ .
Every point on the circle satisfies the equation and no point off it does — a quick way to check a claimed point lies on a given circle is direct substitution, faster than re-deriving the equation.

The general form of a circle equation 1.03

Key results

Expanding the standard form gives $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ , with centre $(- g, - f)$ and radius $g^{2} + f^{2} - c$ .
The equation represents a real circle only when $g^{2} + f^{2} - c > 0$ ; if it equals zero the locus is a single point, and if it is negative there are no real points at all.

Method

Group the $x$ terms and the $y$ terms, leaving the constant on its own.
Complete the square on the $x$ terms and separately on the $y$ terms.
Move the two subtracted constants and the original constant to the right-hand side.
Read off the centre from the completed squares and the radius as the square root of the right-hand side.

Notes

An equation is only a circle if the coefficients of $x^{2}$ and $y^{2}$ are equal and there is no $x y$ term; if both coefficients are equal but not $1$ , divide the whole equation through first.
The centre reads off as $(- g, - f)$ , not $(g, f)$ — a sign that's easy to drop when working quickly straight from the general form.

Tangents, radii and chords 1.03

Key results

The tangent to a circle at a point is perpendicular to the radius drawn to that point.
The angle in a semicircle is $90°$ , so if $∠ A PB = 90°$ then $P$ lies on the circle with $A B$ as diameter.
The perpendicular from the centre to a chord bisects that chord, so the centre lies on the perpendicular bisector of every chord.

Method

Check first that the given point $P$ actually lies on the circle by substituting it into the circle's equation.
Find the gradient of the radius from the centre to $P$ .
Take the negative reciprocal to get the gradient of the tangent.
Write the tangent as $y - y_{P} = m (x - x_{P})$ and rearrange.

Notes

These two facts let you find a tangent's equation, or test whether a triangle inscribed in a circle is right-angled, without solving any equations simultaneously.
A chord's perpendicular bisector always passes through the centre — this is often the quickest route to the centre when only points on the circle (not the equation) are given.

Where a line meets a circle 1.03

Key results

$Δ > 0$ : the line cuts the circle at two points (a chord).
$Δ = 0$ : the line touches the circle at exactly one point (a tangent).
$Δ < 0$ : the line misses the circle entirely.

Method

Rearrange the line to make $y$ (or $x$ ) the subject.
Substitute into the circle's equation to obtain a quadratic in the remaining variable.
Evaluate the discriminant $Δ = b^{2} - 4 a c$ of that quadratic.

Notes

Equivalently, compare the perpendicular distance from the centre to the line with the radius: greater means the line misses, equal means it is a tangent, and smaller means it cuts the circle.
Once two intersection points are found, the chord length is just the distance between them.
A repeated root ( $Δ = 0$ ) gives only one $x$ -value but the tangent point still has a definite $y$ -coordinate too — don't stop at the $x$ -value without substituting back to find the full point.

Worked examples

Worked example

A circle has a diameter with endpoints $A (- 2, 5)$ and $B (6, - 1)$ .

Find the equation of the circle.

Show worked solution

The centre is the midpoint of $A B$ :

(\frac{- 2 + 6}{2}, \frac{5 + ( - 1 )}{2}) = (2, 2)

The radius is half the length of $A B$ :

∣ A B ∣ = (6 - (- 2))^{2} + (- 1 - 5)^{2} = 8^{2} + (- 6)^{2} = 64 + 36 = 100 = 10

so the radius is $5$ .

The equation of the circle is:

(x - 2)^{2} + (y - 2)^{2} = 25

Worked example

Find the centre and radius of the circle:

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

and determine whether the point $(6, 0)$ lies inside or outside it.

Show worked solution

Complete the square in $x$ and in $y$ :

(x - 3)^{2} - 9 + (y + 2)^{2} - 4 - 12 = 0

so:

(x - 3)^{2} + (y + 2)^{2} = 25

The centre is $(3, - 2)$ and the radius is $5$ .

Check against the general form:

x^{2} + y^{2} + 2 gx + 2 f y + c = 0

here $g = - 3$ , $f = 2$ , $c = - 12$ , so the centre is:

(- g, - f) = (3, - 2)

and the radius is:

g^{2} + f^{2} - c = 9 + 4 + 12 = 25 = 5

as found.

The distance from the centre to $(6, 0)$ is:

(6 - 3)^{2} + (0 + 2)^{2} = 13 \approx 3.61

which is less than $5$ , so the point lies inside the circle.

Worked example

Show that $P (5, 6)$ lies on the circle:

(x - 2)^{2} + (y - 2)^{2} = 25

and find the equation of the tangent to the circle at $P$ .

Show worked solution

Substituting $P$ :

(5 - 2)^{2} + (6 - 2)^{2} = 9 + 16 = 25

so $P$ lies on the circle.

The radius from the centre $(2, 2)$ to $P$ has gradient:

\frac{6 - 2}{5 - 2} = \frac{4}{3}

so the tangent, being perpendicular to it, has gradient $- \frac{3}{4}$ .

Then:

y - 6 = - \frac{3}{4} (x - 5)

so:

4 y - 24 = - 3 x + 15

giving $3 x + 4 y = 39$ .

Check that this line touches the circle only once: substituting:

y = \frac{39 - 3 x}{4}

gives:

(x - 2)^{2} + (\frac{31 - 3 x}{4})^{2} = 25

and multiplying by $16$ gives:

16 (x - 2)^{2} + (31 - 3 x)^{2} = 400

i.e.

25 x^{2} - 250 x + 625 = 0

dividing by $25$ gives:

x^{2} - 10 x + 25 = 0

i.e.

$(x - 5)^{2} = 0$ — a repeated root at $x = 5$ , confirming tangency at $P$ .

1.03

Parametric equations

Parametric equations 1.03

Definitions

Parametric curve: a curve on which both $x$ and $y$ are given in terms of a third variable, the parameter (commonly $t$ or $θ$ ), rather than by a direct relationship between $x$ and $y$ .

Key results

A circle of radius $r$ centred at the origin is $x = r cos θ$ , $y = r sin θ$ — a curve that no single equation $y = f (x)$ can represent over its full domain.
Translating that circle to centre $(a, b)$ gives $x = a + r cos θ$ , $y = b + r sin θ$ .

Method

To convert to Cartesian form, rearrange one equation to make the parameter the subject and substitute it into the other.
When the parameter is an angle, instead make $cos θ$ and $sin θ$ the subjects and eliminate them using $sin^{2} θ + cos^{2} θ = 1$ .
State any restriction on $x$ or $y$ that the range of the parameter imposes on the Cartesian curve.

Notes

Parametric form is especially natural for describing motion, where $t$ is time and the pair $(x (t), y (t))$ gives the position of a moving object at that instant.
Eliminating the parameter can quietly enlarge the curve: $x = t^{2}$ , $y = t^{4}$ gives $y = x^{2}$ , but only the half with $x \geq 0$ is actually traced out.
Always state the domain/range restriction the parameter imposes on the Cartesian equation as part of the final answer — an unrestricted Cartesian equation is treated as incomplete.

Worked example

A curve has parametric equations $x = 3 + 2 cos θ$ , $y = - 1 + 2 sin θ$ .

(a) Find its Cartesian equation and describe the curve.

(b) Find the coordinates of the points where it meets the line $y = x - 2$ .

Show worked solution

(a) Rearranging,

cos θ = \frac{x - 3}{2}

and:

sin θ = \frac{y + 1}{2}

Using $sin^{2} θ + cos^{2} θ = 1$ :

(\frac{x - 3}{2})^{2} + (\frac{y + 1}{2})^{2} = 1

so:

(x - 3)^{2} + (y + 1)^{2} = 4

— a circle with centre $(3, - 1)$ and radius $2$ .

(b) Substituting $y = x - 2$ :

(x - 3)^{2} + (x - 1)^{2} = 4

so:

x^{2} - 6 x + 9 + x^{2} - 2 x + 1 = 4

giving:

2 x^{2} - 8 x + 6 = 0

i.e.

$x^{2} - 4 x + 3 = 0$ , so:

(x - 1) (x - 3) = 0

and $x = 1$ or $x = 3$ .

The points are $(1, - 1)$ and $(3, 1)$ .

Check:

(1 - 3)^{2} + (- 1 + 1)^{2} = 4

and:

(3 - 3)^{2} + (1 + 1)^{2} = 4

so both lie on the circle.