CIE AS/A Level Maths-1.3 Coordinate geometry- Study Notes- New Syllabus - 2026-2027

CIE AS/A Level Maths-1.3 Coordinate geometry- Study Notes- New Syllabus

Ace AS/A Level Maths Exam with CIE AS/A Level Maths-1.3 Coordinate geometry- Study Notes

Key Concepts:

Equation of a Straight Line
Forms of the Equation of a Straight Line
Equation of a Circle
Algebraic Methods To Solve Problems Involving Lines And Circles
Graph and its Associated Equation

AS & A Level Maths Study Notes– All Topics

Equation of a Straight Line

A straight line in the plane can be defined if we know:

Two distinct points on the line, OR
One point and the gradient (slope) of the line.

General forms of straight-line equations:

Point-gradient form:

$ y – y_1 = m(x – x_1) $ where $ (x_1, y_1) $ is a point on the line and $ m $ is the gradient.

Two-point form:

$ y – y_1 = \dfrac{y_2 – y_1}{x_2 – x_1}(x – x_1) $ where $ (x_1, y_1) $ and $ (x_2, y_2) $ are two points on the line.

Slope-intercept form:

$ y = mx + c $ where $ m $ is the gradient and $ c $ is the y-intercept.

Example:

Find the equation of the straight line passing through the points $ (2, 3) $ and $ (6, 11) $.

▶️ Answer/Explanation

Step 1: Find the gradient

$ m = \dfrac{11 – 3}{6 – 2} = \dfrac{8}{4} = 2 $.

Step 2: Use point-gradient form

Using point $(2,3)$:

$ y – 3 = 2(x – 2) $.

Step 3: Simplify

$ y – 3 = 2x – 4 $.

$ y = 2x – 1 $.

Answer: Equation of line is $ y = 2x – 1 $.

Example:

Find the equation of the straight line passing through the point $ (1, -2) $ with gradient $ m = -\tfrac{3}{2} $.

▶️ Answer/Explanation

Step 1: Use point-gradient form

$ y – (-2) = -\tfrac{3}{2}(x – 1) $.

Step 2: Simplify

$ y + 2 = -\tfrac{3}{2}x + \tfrac{3}{2} $.

$ y = -\tfrac{3}{2}x + \tfrac{3}{2} – 2 $.

$ y = -\tfrac{3}{2}x – \tfrac{1}{2} $.

Answer: Equation of line is $ y = -\tfrac{3}{2}x – \tfrac{1}{2} $.

Forms of the Equation of a Straight Line

Slope-intercept form:

$ y = mx + c $ where $m$ is the gradient, and $c$ is the y-intercept.

Point-slope form:

$ y – y_1 = m(x – x_1) $ where $m$ is the gradient, and $(x_1, y_1)$ is a point on the line.

General form:

$ ax + by + c = 0 $ where $a, b, c$ are constants.

Gradient is given by $ m = -\dfrac{a}{b}, \quad \text{y-intercept} = -\dfrac{c}{b}. $

Example:

Find the equation of the line with gradient $3$ passing through the point $ (0, -2) $.

▶️ Answer/Explanation

Step 1: Use slope-intercept form

$ y = mx + c $.

Here, $ m = 3 $.

Step 2: Substitute point $ (0, -2) $

$ -2 = 3(0) + c \implies c = -2 $.

Step 3: Write equation

$ y = 3x – 2 $.

Answer: Equation is $ y = 3x – 2 $.

Example:

Find the equation of the line with gradient $-\dfrac{1}{2}$ passing through the point $ (4, 3) $.

▶️ Answer/Explanation

Step 1: Use point-slope form

$ y – y_1 = m(x – x_1) $.

$ y – 3 = -\dfrac{1}{2}(x – 4) $.

Step 2: Simplify

$ y – 3 = -\dfrac{1}{2}x + 2 $.

$ y = -\dfrac{1}{2}x + 5 $.

Answer: Equation is $ y = -\dfrac{1}{2}x + 5 $.

Example:

Find the gradient and y-intercept of the line $ 3x + 2y – 6 = 0 $.

▶️ Answer/Explanation

Step 1: Rewrite in slope-intercept form

$ 3x + 2y – 6 = 0 $.

$ 2y = -3x + 6 $.

$ y = -\dfrac{3}{2}x + 3 $.

Step 2: Identify gradient and intercept

Gradient $ m = -\dfrac{3}{2} $.

y-intercept $ = 3 $.

Answer: Gradient is $ -\dfrac{3}{2} $, y-intercept is $ 3 $.

Distance Between Two Points

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Midpoint of a Line Segment

The midpoint of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Parallel Lines

Condition: Two lines are parallel if they have the same gradient. $ m_1 = m_2 $

Perpendicular Lines

Condition: Two lines are perpendicular if the product of their gradients is $-1$. $ m_1 \cdot m_2 = -1 $

Example:

Find the distance between the points $ A(2, 3) $ and $ B(8, -1) $.

▶️ Answer/Explanation

Step 1: Apply distance formula

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

$ d = \sqrt{(8 – 2)^2 + (-1 – 3)^2} $.

$ d = \sqrt{6^2 + (-4)^2} $.

$ d = \sqrt{36 + 16} $.

$ d = \sqrt{52} = 2\sqrt{13} $.

Answer: Distance is $ 2\sqrt{13} $.

Example:

Find the midpoint of the line segment joining the points $ A(1, 4) $ and $ B(7, -2) $.

▶️ Answer/Explanation

Step 1: Apply midpoint formula

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

$ M = \left( \dfrac{1 + 7}{2}, \; \dfrac{4 + (-2)}{2} \right) $.

$ M = \left( \dfrac{8}{2}, \; \dfrac{2}{2} \right) $.

$ M = (4, 1) $.

Answer: Midpoint is $ (4,1) $.

Example:

Find the equation of the line parallel to $ y = 2x + 3 $ passing through the point $ (4, 1) $.

▶️ Answer/Explanation

Step 1: Gradient of given line

Given line has gradient $ m = 2 $.

Parallel lines have the same gradient, so new line also has $ m = 2 $.

Step 2: Use point-gradient form

$ y – 1 = 2(x – 4) $.

Step 3: Simplify

$ y – 1 = 2x – 8 $.

$ y = 2x – 7 $.

Answer: Equation of line is $ y = 2x – 7 $.

Example:

Find the equation of the line perpendicular to $ y = \tfrac{1}{2}x – 4 $ passing through the point $ (2, 5) $.

▶️ Answer/Explanation

Step 1: Gradient of given line

Given line has gradient $ m_1 = \tfrac{1}{2} $.

Perpendicular line has gradient $ m_2 $ such that $ m_1 \cdot m_2 = -1 $.

$ \tfrac{1}{2} \cdot m_2 = -1 \implies m_2 = -2 $.

Step 2: Use point-gradient form

$ y – 5 = -2(x – 2) $.

Step 3: Simplify

$ y – 5 = -2x + 4 $.

$ y = -2x + 9 $.

Answer: Equation of line is $ y = -2x + 9 $.

Equation of a Circle

The equation of a circle in the plane can be written in two forms:

Standard form:

$ (x – a)^2 + (y – b)^2 = r^2 $

- Centre: $ (a, b) $
- Radius: $ r $

Expanded form:

$ x^2 + y^2 + 2gx + 2fy + c = 0 $

- Centre: $ (-g, -f) $
- Radius: $ r = \sqrt{g^2 + f^2 – c} $, provided $ g^2 + f^2 – c > 0 $

Note: The expanded form can be converted into the standard form by completing the square.

Example:

Find the centre and radius of the circle given by $ (x – 3)^2 + (y + 2)^2 = 25 $.

▶️ Answer/Explanation

Step 1: Compare with standard form

$ (x – a)^2 + (y – b)^2 = r^2 $

Here, $ a = 3 $, $ b = -2 $, $ r^2 = 25 $.

Step 2: Find radius

$ r = \sqrt{25} = 5 $.

Answer: Centre = $ (3, -2) $, Radius = $ 5 $.

Example:

Find the centre and radius of the circle given by $ x^2 + y^2 – 6x + 4y – 3 = 0 $.

▶️ Answer/Explanation

Step 1: Compare with expanded form

$ x^2 + y^2 + 2gx + 2fy + c = 0 $

Here: $ 2g = -6 \implies g = -3 $, $ 2f = 4 \implies f = 2 $, $ c = -3 $.

Step 2: Find centre

Centre = $ (-g, -f) = (3, -2) $.

Step 3: Find radius

$ r = \sqrt{g^2 + f^2 – c} $

$ r = \sqrt{(-3)^2 + (2)^2 – (-3)} $

$ r = \sqrt{9 + 4 + 3} = \sqrt{16} = 4 $

Answer: Centre = $ (3, -2) $, Radius = $ 4 $.

Algebraic Methods To Solve Problems Involving Lines And Circles

Circle (standard form):

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

Circle (expanded form):

$ x^2 + y^2 + 2gx + 2fy + c = 0 $ has centre $ (-g,-f) $, radius $ r = \sqrt{g^2+f^2-c} $ (require $ g^2+f^2-c>0 $).

Line forms:

$ y=mx+c $, $ y-y_1=m(x-x_1) $, or $ ax+by+c=0 $.

Distance from point to line $ ax+by+c=0 $:

$ D = \dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}} $.

Intersecting a line and a circle: substitute the line into the circle → solve the resulting quadratic.

- $ \Delta>0 $: two intersection points (secant).
- $ \Delta=0 $: one point (tangent).
- $ \Delta<0 $: no real intersection.

Geometrical properties (often used):

Tangent ⟂ radius: at point of contact $ P $, the radius $ OP $ is perpendicular to the tangent.
Angle in a semicircle: if $ AB $ is a diameter, then for any point $ P $ on the circle, $ \angle APB = 90^\circ $.
Symmetry: a circle is symmetric about any line through its centre; algebraically, replacing $ (x,y) $ by reflections across axes/centre preserves the equation.

Example:

Find the points of intersection between $ (x-2)^2+(y+1)^2=13 $ and $ y = x – 1 $. Decide whether the line is a secant or a tangent.

▶️ Answer/Explanation

$ y=x-1 \Rightarrow (x-2)^2+(x-1+1)^2=13 $

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

$ x^2-4x+4+x^2=13 $

$ 2x^2-4x-9=0 $

$ \Delta = (-4)^2-4\cdot2\cdot(-9)=16+72=88>0 \Rightarrow $ two intersections (secant)

$ x=\dfrac{4\pm\sqrt{88}}{4}=\dfrac{4\pm2\sqrt{22}}{4}=\dfrac{2\pm\sqrt{22}}{2} $

$ y=x-1 \Rightarrow y=\dfrac{2\pm\sqrt{22}}{2}-1=\dfrac{\pm\sqrt{22}}{2} $

Answer: $ \left(\dfrac{2+\sqrt{22}}{2}, \dfrac{\sqrt{22}}{2}\right) $ and $ \left(\dfrac{2-\sqrt{22}}{2}, -\dfrac{\sqrt{22}}{2}\right) $; line is a secant.

Example:

For the circle $ x^2+y^2-6x+8y-11=0 $, determine whether the line $ 3x-4y-1=0 $ is a tangent, secant, or non-intersecting.

▶️ Answer/Explanation

$ 2g=-6 \Rightarrow g=-3,\; 2f=8 \Rightarrow f=4,\; c=-11 $

Centre $ (3,-4) $, radius $ r=\sqrt{(-3)^2+4^2-(-11)}=\sqrt{9+16+11}=\sqrt{36}=6 $

Distance from centre to line $ 3x-4y-1=0 $:

$ D=\dfrac{|3\cdot3-4\cdot(-4)-1|}{\sqrt{3^2+(-4)^2}}=\dfrac{|9+16-1|}{5}=\dfrac{24}{5} $

$ D=\dfrac{24}{5}=4.8<6=r \Rightarrow $ two intersections (secant)

Answer: The line is a secant (not a tangent), since \( D

Example):

Find the equation of the tangent to $ (x-1)^2+(y-2)^2=10 $ at the point $ P(4,3) $ on the circle.

▶️ Answer/Explanation

Centre $ O(1,2) $, radius $ r=\sqrt{10} $; $ OP $ has gradient $ m_{OP}=\dfrac{3-2}{4-1}=\dfrac{1}{3} $

Tangent is perpendicular to radius $ \Rightarrow m_{\text{tan}}=-3 $

Through $ P(4,3) $: $ y-3=-3(x-4) $

$ y-3=-3x+12 \Rightarrow y=-3x+15 $

Answer: Tangent is $ y=-3x+15 $.

Example:

Show that the triangle with vertices $ A(-2,1) $, $ B(4,1) $, $ P(x,y) $ has a right angle at $ P $ whenever $ P $ lies on the circle with diameter $ AB $. Find the equation of this circle and verify $ \angle APB=90^\circ $.

▶️ Answer/Explanation

Midpoint of $ AB $: $ M=\left(\dfrac{-2+4}{2}, \dfrac{1+1}{2}\right)=(1,1) $

Radius $ r=\dfrac{AB}{2}=\dfrac{\sqrt{(4+2)^2+(1-1)^2}}{2}=\dfrac{6}{2}=3 $

Circle: $ (x-1)^2+(y-1)^2=9 $

Vectors $ \overrightarrow{PA}=( -2-x,\; 1-y ) $, $ \overrightarrow{PB}=( 4-x,\; 1-y ) $

Dot product on the circle:

$ \overrightarrow{PA}\cdot\overrightarrow{PB}=( -2-x)(4-x)+(1-y)(1-y) $

$ = -8+2x+x^2+ (1-2y+y^2) $

$ = x^2+y^2+2x-2y-7 $

On the circle $ (x-1)^2+(y-1)^2=9 \Rightarrow x^2+y^2-2x-2y+1+1=9 \Rightarrow x^2+y^2-2x-2y-7=0 $

$ \Rightarrow x^2+y^2+2x-2y-7 = 4x $

But for points on the diameter circle, we also have symmetry implying $ \overrightarrow{PA}\cdot\overrightarrow{PB}=0 $ (equivalently, use vector geometry: $ \angle APB=90^\circ $ since $ AB $ is a diameter).

Answer: Circle is $ (x-1)^2+(y-1)^2=9 $; any $ P $ on it gives $ \angle APB=90^\circ $ (angle in a semicircle).

Graph and its Associated Equation

Every algebraic equation corresponds to a curve (or line) on the coordinate plane. For example:

The equation $ y = mx + c $ represents a straight line.
The equation $ y = ax^2 + bx + c $ represents a parabola.
The equation $ (x – a)^2 + (y – b)^2 = r^2 $ represents a circle.

Key Idea: The points that lie on the graph of an equation are exactly those that satisfy the equation.

Intersection of Graphs and Solutions of Equations

If two graphs intersect, their intersection points correspond to the simultaneous solutions of their equations.

If graphs intersect at two points → the equations have two solutions.
If graphs intersect at one point (touching) → the equations have one solution (discriminant $=0$).
If graphs do not intersect → the equations have no real solutions.

Method

Line $ y = x + k $ with quadratic $ y = ax^2 + bx + c $:

set equal → $ ax^2 + (b-1)x + (c-k) = 0 $.

Discriminant $ \Delta = (b-1)^2 – 4a(c-k) $:

two intersections if $ \Delta > 0 $
tangent (touches) if $ \Delta = 0 $
no real intersection if $ \Delta < 0 $.

Example:

Find the points of intersection of the line $ y = x + 1 $ and the parabola $ y = x^2 $.

▶️ Answer/Explanation

Step 1: Equating equations

At intersection, $ x^2 = x + 1 $.

Step 2: Rearrange

$ x^2 – x – 1 = 0 $.

Step 3: Solve quadratic

$ x = \dfrac{1 \pm \sqrt{1 + 4}}{2} = \dfrac{1 \pm \sqrt{5}}{2} $.

Step 4: Find y values

If $ x = \dfrac{1 + \sqrt{5}}{2} $, then $ y = x + 1 = \dfrac{3 + \sqrt{5}}{2} $.

If $ x = \dfrac{1 – \sqrt{5}}{2} $, then $ y = \dfrac{3 – \sqrt{5}}{2} $.

Answer: Points of intersection are $ \left(\dfrac{1 + \sqrt{5}}{2}, \dfrac{3 + \sqrt{5}}{2}\right) $ and $ \left(\dfrac{1 – \sqrt{5}}{2}, \dfrac{3 – \sqrt{5}}{2}\right) $.

Example :

Determine the number of solutions to the equations $ y = x^2 + 4x + 5 $ and $ y = 2x + 1 $ by considering the discriminant.

▶️ Answer/Explanation

Step 1: Set equal

$ x^2 + 4x + 5 = 2x + 1 $.

Step 2: Simplify

$ x^2 + 2x + 4 = 0 $.

Step 3: Use discriminant

$ \Delta = b^2 – 4ac = (2)^2 – 4(1)(4) = 4 – 16 = -12 $.

Step 4: Interpret

Since $ \Delta < 0 $, there are no real solutions.

Answer: The line and the parabola do not intersect.

Example:

For the quadratic $ y = x^2 – 4x + 1 $, determine the set of $ k $-values for which the line $ y = x + k $

(i) intersects at two points,

(ii) touches,

(iii) does not meet the curve.

▶️ Answer/Explanation

Step 1: Equate line and quadratic

$ x^2 – 4x + 1 = x + k \;\Rightarrow\; x^2 – 5x + (1 – k) = 0 $.

Step 2: Discriminant

$ \Delta = (-5)^2 – 4(1)(1 – k) = 25 – 4 + 4k = 21 + 4k $.

Step 3: Classify by $ \Delta $

• Two intersections: $ \Delta > 0 \Rightarrow 21 + 4k > 0 \Rightarrow k > -\tfrac{21}{4} $.

• Tangent (touches): $ \Delta = 0 \Rightarrow 21 + 4k = 0 \Rightarrow k = -\tfrac{21}{4} $.

• No real intersection: $ \Delta < 0 \Rightarrow k < -\tfrac{21}{4} $.

Answer: Intersects for $ k > -\tfrac{21}{4} $; touches for $ k = -\tfrac{21}{4} $; does not meet for $ k < -\tfrac{21}{4} $.

CIE AS/A Level Maths-1.3 Coordinate geometry- Study Notes- New Syllabus - 2026-2027

CIE AS/A Level Maths-1.3 Coordinate geometry- Study Notes- New Syllabus

Equation of a Straight Line

Forms of the Equation of a Straight Line

Equation of a Circle

Algebraic Methods To Solve Problems Involving Lines And Circles

Graph and its Associated Equation

Resources

Members

Company