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CCEA GCE Mathematics (2210) · Coordinate geometry

Mini-Lesson

Coordinate geometry

3 missions. Guess, learn, then check. Your place is saved.

Mission 1Do $(1, 2)$ (1, 2), $(4, 11)$ (4, 11) and $(9, 27)$ (9, 27) lie on one straight line?
Mission 2 $x^{2} + y^{2} - 6 x + 4 y = 12$ x²+y²−6x+4y=12 hides a circle. Where is its centre?
Mission 3A point moves with $x = cos t$ x=cos t, $y = sin t$ y=sin t. What shape, how fast?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Do $(1, 2)$ (1, 2), $(4, 11)$ (4, 11) and $(9, 27)$ (9, 27) lie on one straight line?

Make a guess: Do $(1, 2)$ (1, 2), $(4, 11)$ (4, 11) and $(9, 27)$ (9, 27) lie on one straight line?

Coordinate geometry · straight lines

Gradients and equations of lines

Gradient between $(x_{1}, y_{1})$ (x₁, y₁) and $(x_{2}, y_{2})$ (x₂, y₂): $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ m=(y₂−y₁)/(x₂−x₁)
Line with gradient $m$ m through $(x_{1}, y_{1})$ (x₁, y₁): $y - y_{1} = m (x - x_{1})$ y−y₁=m(x−x₁)
General form $a x + b y + c = 0$ ax+by+c=0, usually with integers $a$ a, $b$ b, $c$ c. Rearrange to $y = m x + c$ y=mx+c to read off the gradient.
Points $A$ A, $B$ B, $C$ C are collinear (on one line) exactly when the gradient of $A B$ AB equals the gradient of $BC$ BC.

Worked example Coordinate geometry · straight lines

The line through two points

Question

Find the equation of the line through

(- 2, 5)

and

(4, 1)

in the form

a x + b y + c = 0

, where

a

b

and

c

are integers.

1
Gradient: $m = \frac{1 - 5}{4 - ( - 2 )} = \frac{- 4}{6} = - \frac{2}{3}$ m=(1−5)/(4−(−2))=(−4)/6=−2/3
2
Use $y - y_{1} = m (x - x_{1})$ y−y₁=m(x−x₁) with $(- 2, 5)$ (−2, 5): $y - 5 = - \frac{2}{3} (x + 2)$ y−5=−2/3(x+2)
3
Multiply by 3: $3 y - 15 = - 2 x - 4$ 3y−15=−2x−4
4
Collect on one side: $2 x + 3 y - 11 = 0$ 2x+3y−11=0

Answer

2 x + 3 y - 11 = 0

(check

(4, 1)

8 + 3 - 11 = 0

)

⚠️ Watch out

x - (- 2)

x + 2

; a sign slip here moves the whole line.

Quick check

?What is the gradient of the line

3 x - 4 y + 8 = 0

Coordinate geometry · straight lines

Parallel, perpendicular and models

Parallel lines have equal gradients: $m_{1} = m_{2}$ m₁=m₂.
Perpendicular lines: $m_{1} m_{2} = - 1$ m₁m₂=−1, so $m_{2} = - \frac{1}{m _{1}}$ m₂=−1/m₁.
Midpoint: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ ((x₁+x₂)/2, (y₁+y₂)/2). Distance: $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ √((x₂−x₁)²+(y₂−y₁)²). Where two lines meet: solve their equations simultaneously.
In a model $y = m x + c$ y=mx+c, $m$ m is the rate of change and $c$ c the value when $x = 0$ x=0.

Worked example Coordinate geometry · straight lines

A perpendicular bisector

Question

A

(1, 3)

and

B

(7, - 1)

. Find the perpendicular bisector of

A B

in the form

a x + b y + c = 0

1
Midpoint of $A B$ AB: $(\frac{1 + 7}{2}, \frac{3 + ( - 1 )}{2}) = (4, 1)$ ((1+7)/2, (3+(−1))/2)=(4, 1)
2
Gradient of $A B$ AB: $\frac{- 1 - 3}{7 - 1} = - \frac{2}{3}$ (−1−3)/(7−1)=−2/3, so the perpendicular gradient is $\frac{3}{2}$ 3/2
3
Line through $(4, 1)$ (4, 1): $y - 1 = \frac{3}{2} (x - 4)$ y−1=3/2(x−4), so $2 y - 2 = 3 x - 12$ 2y−2=3x−12

Answer

3 x - 2 y - 10 = 0

⚠️ Watch out

the perpendicular gradient is the negative reciprocal:

- \frac{2}{3}

becomes

+ \frac{3}{2}

, not

- \frac{3}{2}

Sort it

?Compare each line with

y = 2 x + 3

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?The line

l_{1}

has equation

2 x + 5 y - 3 = 0

. The line

l_{2}

is perpendicular to

l_{1}

and passes through

(4, - 1)

. Find the

y

-coordinate of the point where

l_{2}

crosses the

y

-axis.

Calculate

?The lines

y = 2 x - 1

and

x + 3 y = 11

meet at

P

. Find the distance from

P

Q (8, 11)

Calculate

?A phone plan charges a fixed monthly fee plus a charge per GB of data. Using 5 GB costs £14.00 and using 12 GB costs £24.50. Model the cost as a straight line

C = m d + c

, where

d

is the data in GB. Find the fixed monthly fee.

Give your answer in pounds.

Mission check

?Do

(1, 2)

(4, 11)

and

(9, 27)

lie on one straight line?

Mission 2

$x^{2} + y^{2} - 6 x + 4 y = 12$ x²+y²−6x+4y=12 hides a circle. Where is its centre?

Make a guess: $x^{2} + y^{2} - 6 x + 4 y = 12$ x²+y²−6x+4y=12 hides a circle. Where is its centre?

Coordinate geometry · circles

The equation of a circle

A point $(x, y)$ (x, y) is on the circle with centre $(a, b)$ (a, b) and radius $r$ r exactly when its distance from the centre is $r$ r.
Pythagoras on the horizontal and vertical gaps gives

(x - a)^{2} + (y - b)^{2} = r^{2}

Read the centre with the signs reversed: $(x - 2)^{2} + (y + 1)^{2} = 9$ (x−2)²+(y+1)²=9 has centre $(2, - 1)$ (2, −1) and radius $9 = 3$ √9=3.
The right-hand side is $r^{2}$ r², so take its square root for the radius.

Coordinate geometry · circles

Completing the square to find the centre

Expanded, a circle looks like $x^{2} + y^{2} + D x + E y + C = 0$ x²+y²+Dx+Ey+C=0, with equal coefficients of $x^{2}$ x² and $y^{2}$ y² and no $x y$ xy term.
Complete the square in $x$ x and in $y$ y separately: $x^{2} + D x = (x + \frac{D}{2})^{2} - \frac{D ^{2}}{4}$ x²+Dx=(x+D/2)²−D²/4.
Move the constants to the right; the right-hand side is $r^{2}$ r². It must be positive for a real circle.
You should also know the general form $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ x²+y²+2gx+2fy+c=0: centre $(- g, - f)$ (−g, −f), radius $g^{2} + f^{2} - c$ √(g²+f²−c).

Worked example Coordinate geometry · circles

Centre and radius from the expanded form

Question

Find the centre and radius of the circle

x^{2} + y^{2} + 8 x - 10 y + 5 = 0

1
Group the $x$ x terms and the $y$ y terms: $(x^{2} + 8 x) + (y^{2} - 10 y) + 5 = 0$ (x²+8x)+(y²−10y)+5=0
2
Complete each square: $(x + 4)^{2} - 16 + (y - 5)^{2} - 25 + 5 = 0$ (x+4)²−16+(y−5)²−25+5=0
3
Move the constants across: $(x + 4)^{2} + (y - 5)^{2} = 36$ (x+4)²+(y−5)²=36
4
Read off, reversing the signs: centre $(- 4, 5)$ (−4, 5), $r = 36 = 6$ r=√36=6

Answer

centre

(- 4, 5)

, radius 6

⚠️ Watch out

subtract the square of half the coefficient (16 and 25) inside the bracket step, then move every constant to the right before reading

r^{2}

Quick check

?Find the centre and radius of

x^{2} + y^{2} + 4 x - 6 y = 3

Coordinate geometry · circles

Three circle facts in coordinates

Tangent ⟂ radius: if the radius to a point has gradient $m$ m, the tangent there has gradient $- \frac{1}{m}$ −1/m.
The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of any chord passes through the centre.
The angle in a semicircle is a right angle: if $A B$ AB is a diameter and $C$ C is on the circle, $A C ⊥ BC$ AC ⊥ BC. Conversely, a right angle at $C$ C makes $A B$ AB a diameter of the circumcircle of triangle $A BC$ ABC.

Worked example Coordinate geometry · circles

Equation of a tangent

Question

The point

P (4, 2)

lies on the circle

(x - 1)^{2} + (y + 2)^{2} = 25

. Find the equation of the tangent at

P

in the form

a x + b y + c = 0

1
Check $P$ P is on the circle: $3^{2} + 4^{2} = 25$ 3²+4²=25
2
Gradient of the radius from the centre $(1, - 2)$ (1, −2) to $P$ P: $\frac{2 - ( - 2 )}{4 - 1} = \frac{4}{3}$ (2−(−2))/(4−1)=4/3
3
Tangent is perpendicular to the radius: gradient $- \frac{3}{4}$ −3/4
4
Use $y - y_{1} = m (x - x_{1})$ y−y₁=m(x−x₁): $y - 2 = - \frac{3}{4} (x - 4)$ y−2=−3/4(x−4), so $4 y - 8 = - 3 x + 12$ 4y−8=−3x+12

Answer

3 x + 4 y - 20 = 0

⚠️ Watch out

the tangent's gradient is the negative reciprocal of the radius's gradient, not of the circle "equation".

Calculate

A (- 2, 1)

B (6, 7)

and

C (5, 0)

. Show that angle

A CB

9 0^{\circ}

. Hence the circle through

A

B

and

C

meets the

y

-axis at

P

and

Q

. Find the length

PQ

to 3 significant figures.

Coordinate geometry · circles

Does a line meet a circle?

Substitute the line into the circle equation to get a quadratic in one variable.
Use the discriminant $b^{2} - 4 a c$ b²−4ac of that quadratic: $> 0$ >0 two intersection points, $= 0$ =0 the line is a tangent, $< 0$ <0 no intersection.
Solve the quadratic to find the actual points when they exist.

Calculate

?The line

y = x + k

is a tangent to the circle

x^{2} + y^{2} = 8

. Find the positive value of

k

Mission check

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

hides a circle. Where is its centre?

Mission 3

A point moves with $x = cos t$ x=cos t, $y = sin t$ y=sin t. What shape, how fast?

Make a guess: A point moves with $x = cos t$ x=cos t, $y = sin t$ y=sin t. What shape, how fast?

Coordinate geometry · parametric equations

Curves drawn by a parameter (A2 only)

In parametric equations both $x$ x and $y$ y are given in terms of a third variable, the parameter (often $t$ t or $θ$ θ).
Each value of $t$ t gives one point. For $x = t + 1$ x=t+1, $y = t^{2}$ y=t²:

$t$ t	$- 2$ −2	$- 1$ −1	$0$ 0	$1$ 1	$2$ 2
$(x, y)$ (x, y)	$(- 1, 4)$ (−1, 4)	$(0, 1)$ (0, 1)	$(1, 0)$ (1, 0)	$(2, 1)$ (2, 1)	$(3, 4)$ (3, 4)

Plot the points in order of $t$ t and join them; an arrow shows the direction of increasing $t$ t.
A restricted domain of $t$ t draws only part of a curve.

Coordinate geometry · parametric equations

Converting to a Cartesian equation

Eliminate the parameter to get one equation in $x$ x and $y$ y.
Algebraic parameters: make $t$ t the subject of the simpler equation and substitute. From $x = t + 1$ x=t+1, $t = x - 1$ t=x−1, so $y = (x - 1)^{2}$ y=(x−1)².
Trigonometric parameters: use an identity such as $sin^{2} t + cos^{2} t = 1$ sin² t+cos² t=1 or the double-angle formula $cos 2 t = 1 - 2 sin^{2} t$ cos 2t=1−2sin² t (see the Trigonometry lesson).
Carry the domain across: if $t \geq 0$ t ≥ 0 and $x = t + 1$ x=t+1, the Cartesian curve only applies for $x \geq 1$ x ≥ 1.

Worked example Coordinate geometry · parametric equations

Eliminating an algebraic parameter

Question

A curve has parametric equations

x = 2 t + 1

y = t^{2} - 3

t \geq 0

. Find a Cartesian equation and state the domain.

1
Make $t$ t the subject of the $x$ x equation: $t = \frac{x - 1}{2}$ t=(x−1)/2
2
Substitute into $y$ y: $y = (\frac{x - 1}{2})^{2} - 3 = \frac{( x - 1 ) ^{2}}{4} - 3$ y=((x−1)/2)²−3=((x−1)²)/4−3
3
Tidy if wanted: $4 y = x^{2} - 2 x - 11$ 4y=x²−2x−11
4
Translate the domain: $t \geq 0$ t ≥ 0 gives $x = 2 t + 1 \geq 1$ x=2t+1 ≥ 1

Answer

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

for

x \geq 1

(half of a parabola, starting at

(1, - 3)

)

⚠️ Watch out

square the whole fraction,

\frac{( x - 1 ) ^{2}}{4}

, not just the numerator.

Quick check

?A curve has parametric equations

x = 3 t

y = \frac{6}{t}

t \neq = 0

. Which is its Cartesian equation?

Coordinate geometry · parametric equations

Circles from trig parameters

$x = a + r cos t$ x=a+rcos t, $y = b + r sin t$ y=b+rsin t gives $(x - a)^{2} + (y - b)^{2} = r^{2}$ (x−a)²+(y−b)²=r², using $cos^{2} t + sin^{2} t = 1$ cos² t+sin² t=1: a circle, centre $(a, b)$ (a, b), radius $r$ r.
At $t = 0$ t=0 the point is at $(a + r, b)$ (a+r, b); as $t$ t increases it moves anticlockwise.
With $t$ t in radians the arc travelled is $r t$ rt: a steady $r$ r units per unit of $t$ t, one lap for $0 \leq t \leq 2 π$ 0 ≤ t ≤ 2π.
$0 \leq t \leq π$ 0 ≤ t ≤ π gives only the top half.

Worked example Coordinate geometry · parametric equations

Identifying a circle

Question

A curve has parametric equations

x = 2 + 5 cos t

y = - 4 + 5 sin t

0 \leq t \leq π

. Find its Cartesian equation and describe the curve.

1
Isolate the trig functions: $cos t = \frac{x - 2}{5}$ cos t=(x−2)/5, $sin t = \frac{y + 4}{5}$ sin t=(y+4)/5
2
Use $cos^{2} t + sin^{2} t = 1$ cos² t+sin² t=1: $\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 4 ) ^{2}}{25} = 1$ ((x−2)²)/25+((y+4)²)/25=1
3
Multiply by 25: $(x - 2)^{2} + (y + 4)^{2} = 25$ (x−2)²+(y+4)²=25
4
Use the domain: $t = 0$ t=0 gives $(7, - 4)$ (7, −4), $t = \frac{π}{2}$ t=π/2 gives $(2, 1)$ (2, 1), $t = π$ t=π gives $(- 3, - 4)$ (−3, −4)

Answer

(x - 2)^{2} + (y + 4)^{2} = 25

with

y \geq - 4

: the upper semicircle of the circle centre

(2, - 4)

, radius 5

⚠️ Watch out

without the domain you would describe the whole circle; the question's

0 \leq t \leq π

keeps only

sin t \geq 0

Calculate

?A curve has parametric equations

x = sin t

y = cos 2 t

. Use the double-angle identity

cos 2 t = 1 - 2 sin^{2} t

(from the Trigonometry lesson). The Cartesian equation can be written

y = a x^{2} + b

for

- 1 \leq x \leq 1

. Find the value of

a

Coordinate geometry · parametric equations

Modelling with a parameter

In models the parameter is often time. A ball has $x = 20 t$ x=20t, $y = 15 t - 5 t^{2}$ y=15t−5t² (metres, $t$ t in seconds).
It is in flight while $y \geq 0$ y ≥ 0: $15 t - 5 t^{2} = 0$ 15t−5t²=0 gives $t = 0$ t=0 or $3$ 3, so the model holds for $0 \leq t \leq 3$ 0 ≤ t ≤ 3 and it lands 60 m away.
Eliminating $t$ t gives the path $y = \frac{3}{4} x - \frac{x ^{2}}{80}$ y=3/4x−x²/80 but loses the timing.
Constant velocity from $(1, 8)$ (1, 8) at $t = 0$ t=0 to $(6, 20)$ (6, 20) at $t = 5$ t=5: $x = 1 + t$ x=1+t, $y = 8 + 2.4 t$ y=8+2.4t.

Calculate

?A ball is thrown from a cliff path. Its position is modelled by

x = 8 t

y = 1.5 + 10 t - 5 t^{2}

, where

x

and

y

are in metres and

t \geq 0

is the time in seconds. The ball lands when

y = 0

. Find the horizontal distance travelled to 3 significant figures.

Mission check

?A point moves with

x = cos t

y = sin t

. What shape does it trace, and how fast?

Recap

Big ideas

Gradient $= \frac{change in y}{change in x}$ =(change in y)/(change in x); a line is $y - y_{1} = m (x - x_{1})$ y−y₁=m(x−x₁) or $a x + b y + c = 0$ ax+by+c=0.
Parallel: equal gradients. Perpendicular: $m_{1} m_{2} = - 1$ m₁m₂=−1. Collinear: equal gradients through a shared point.
In a straight-line model, the gradient is the rate of change and the intercept is the starting value.
A circle with centre $(a, b)$ (a, b) and radius $r$ r is $(x - a)^{2} + (y - b)^{2} = r^{2}$ (x−a)²+(y−b)²=r²; complete the square to get there from the expanded form.
Tangent ⟂ radius, the perpendicular from the centre bisects a chord, and the angle in a semicircle is $9 0^{\circ}$ 90°: in coordinates these are gradient facts ( $m_{1} m_{2} = - 1$ m₁m₂=−1) and midpoints.
Substitute a line into a circle and use the discriminant: two points, tangent, or no meeting.
Parametric equations give $x$ x and $y$ y separately in terms of a parameter; each value of $t$ t is one point, and the domain of $t$ t fixes which part of the curve is drawn.
Eliminate the parameter by substitution or with a trig identity to get the Cartesian equation, and carry the domain across.
In models the parameter is usually time, so parametric form keeps "when" as well as "where".

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Coordinate geometry for CCEA GCE Mathematics (2210). 🎉

Mission 1Do $(1, 2)$ (1, 2), $(4, 11)$ (4, 11) and $(9, 27)$ (9, 27) lie on one straight line?
Mission 2 $x^{2} + y^{2} - 6 x + 4 y = 12$ x²+y²−6x+4y=12 hides a circle. Where is its centre?
Mission 3A point moves with $x = cos t$ x=cos t, $y = sin t$ y=sin t. What shape, how fast?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (3)

The specification points and sources behind this lesson.

1 CCEA GCE Mathematics (2018) specification, Unit AS 1, Co-ordinate geometry in the (x, y) plane
2 CCEA GCE Mathematics (2018) specification, Unit AS 1 Pure Mathematics, Co-ordinate geometry
3 CCEA GCE Mathematics (2018) specification, Unit A2 1 Pure Mathematics, Co-ordinate geometry

Coordinate geometry

Do (1,2)(1, 2), (4,11)(4, 11) and (9,27)(9, 27) lie on one straight line?

Gradients and equations of lines

The line through two points

Parallel, perpendicular and models

A perpendicular bisector

x2+y2−6x+4y=12x²+y²−6x+4y=12 hides a circle. Where is its centre?

The equation of a circle

Completing the square to find the centre

Centre and radius from the expanded form

Three circle facts in coordinates

Equation of a tangent

Does a line meet a circle?

A point moves with x=costx=cos t, y=sinty=sin t. What shape, how fast?

Curves drawn by a parameter (A2 only)

Converting to a Cartesian equation

Eliminating an algebraic parameter

Circles from trig parameters

Identifying a circle

Modelling with a parameter

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Do $(1, 2)$ (1, 2), $(4, 11)$ (4, 11) and $(9, 27)$ (9, 27) lie on one straight line?

$x^{2} + y^{2} - 6 x + 4 y = 12$ x²+y²−6x+4y=12 hides a circle. Where is its centre?

A point moves with $x = cos t$ x=cos t, $y = sin t$ y=sin t. What shape, how fast?