Coordinate Geometry

How to use this chapter

Coordinate geometry becomes much easier when you stop treating formulas as isolated facts. Almost every formula here comes from a small number of geometric ideas: horizontal change, vertical change, Pythagoras, symmetry, ratios and perpendicularity.

Learn the derivation first. Then practise the standard form. Finally, solve the parameter and proof problems where you have to decide which tool is appropriate.

01. Coordinate Foundations 02. Distance & Midpoint 03. Ratio & Division 04. Gradient 05. Straight Lines 06. Parallel & Perpendicular 07. Intersections 08. Coordinate Proofs 09. Circles 10. Tangents & Chords 11. Advanced Problems 12. Exam Strategy 13. Formula Atlas 14. 100Q Exam

01 — FOUNDATIONS

The Coordinate Plane

A coordinate system gives every point a numerical address. A point is written as (x,y), where x describes horizontal position and y describes vertical position.

Region	x	y
Quadrant I	+	+
Quadrant II	−	+
Quadrant III	−	−
Quadrant IV	+	−

Axes and special points

Any point on the x-axis has y = 0. Any point on the y-axis has x = 0. The origin is (0,0).

x-axis: y = 0
y-axis: x = 0
origin: (0,0)

Symmetry

Reflection across an axis changes the sign of the coordinate perpendicular to that axis.

Reflection in x-axis: (x,y) → (x,−y)
Reflection in y-axis: (x,y) → (−x,y)
Reflection in origin: (x,y) → (−x,−y)

Worked Example 1 — Reflection

Reflect P(−4,7) in the x-axis.

Reflection in the x-axis leaves x unchanged and reverses y.

P′ = (−4,−7).

Worked Example 2 — Quadrant reasoning

Point A has coordinates (−6,−2). Point B is its reflection in the y-axis. Where is B?

The y-axis changes x → −x while keeping y unchanged.

B = (6,−2), so B lies in Quadrant IV.

Common Trap

Do not say “negative x means left and negative y means down” without considering the coordinate axes. A point with x = 0 is on the y-axis, not in a quadrant.

02 — DISTANCE & MIDPOINT

Distance and Midpoint

Deriving the distance formula

Take A(x₁,y₁) and B(x₂,y₂). Draw horizontal and vertical components between them. The horizontal change is Δx and the vertical change is Δy. These form a right triangle.

Δx = x₂ − x₁
Δy = y₂ − y₁
AB² = (x₂−x₁)² + (y₂−y₁)²

Therefore:
AB = √[(x₂−x₁)² + (y₂−y₁)²]

The formula is therefore just Pythagoras applied to the coordinate plane.

Worked Example 3 — Distance

Find the distance between A(−2,5) and B(4,−3).

Δx = 4 − (−2) = 6
Δy = −3 − 5 = −8

AB = √(6² + (−8)²) = √100 = 10.

Distance squared

When comparing lengths, you often do not need the square root. Since all lengths are non-negative, comparing AB² and CD² gives the same ordering as comparing AB and CD.

AB² = (x₂−x₁)² + (y₂−y₁)²

This is especially useful in coordinate proofs because it keeps the algebra exact and avoids unnecessary radicals.

Midpoint formula

The midpoint is obtained by averaging the x-coordinates and averaging the y-coordinates.

M = ((x₁+x₂)/2 , (y₁+y₂)/2)

Worked Example 4 — Midpoint

Find the midpoint of A(−7,4) and B(3,−8).

x-coordinate = (−7+3)/2 = −2
y-coordinate = (4−8)/2 = −2

M = (−2,−2).

Challenge — Recovering an endpoint

The midpoint of A(5,−3) and B is (−2,4). Find B.

Let B=(x,y).

(5+x)/2 = −2 → 5+x = −4 → x = −9
(−3+y)/2 = 4 → −3+y = 8 → y = 11

B = (−9,11).

03 — SECTION FORMULA

Ratio and Division of a Line

Suppose P divides AB internally in the ratio AP:PB = m:n. P is m parts from A and n parts from B, so its coordinates are a weighted average of A and B.

A(x₁,y₁), B(x₂,y₂)

P = ((n x₁ + m x₂)/(m+n), (n y₁ + m y₂)/(m+n))

Notice that the coordinate closer to A receives the larger weight from A's opposite segment. This is why the coefficients may look reversed at first.

Worked Example 5 — Internal division

A(2,5), B(11,−4). P divides AB in AP:PB = 2:1.

m=2, n=1.

x = [1(2)+2(11)]/3 = 24/3 = 8
y = [1(5)+2(−4)]/3 = −3/3 = −1

P = (8,−1).

Useful sanity check

For internal division, P must lie between A and B. If your calculated point does not lie on the segment, recheck the ratio formula and signs.

04 — GRADIENT

Gradient: Change in y per change in x

Gradient measures the steepness and direction of a straight line. It is the vertical change divided by the horizontal change.

m = (y₂−y₁)/(x₂−x₁)

A positive gradient means y increases as x increases. A negative gradient means y decreases as x increases.

Why the order does not matter

You can calculate the gradient using either direction, provided you use the same order in numerator and denominator.

(y₂−y₁)/(x₂−x₁) = (y₁−y₂)/(x₁−x₂)

Both numerator and denominator change sign, so the ratio stays unchanged.

Worked Example 6

Find the gradient through (−3,8) and (5,−4).

m = (−4−8)/(5−(−3)) = −12/8 = −3/2.

Vertical and horizontal lines

Horizontal line: y = constant → gradient 0
Vertical line: x = constant → gradient undefined

Important

Never write the gradient of a vertical line as 0. A vertical line has no defined gradient because its horizontal change is zero.

05 — STRAIGHT LINES

Equations of Straight Lines

Gradient-intercept form

y = mx + c

m is the gradient and c is the y-intercept.

Point-gradient form

y − y₁ = m(x − x₁)

This is often the fastest form when a point and gradient are known.

Worked Example 7

Find the equation of the line with gradient 4 passing through (−2,7).

y−7 = 4(x+2)
y−7 = 4x+8
y = 4x+15.

Equation through two points

Worked Example 8

Find the equation through A(2,−1) and B(8,11).

First calculate gradient:
m = (11−(−1))/(8−2) = 12/6 = 2.

Using A:
y+1 = 2(x−2)
y = 2x−5.

Equation: y = 2x−5.

General form

Ax + By + C = 0

If B ≠ 0, rearrange:

y = −(A/B)x − C/B

Therefore gradient = −A/B.

Worked Example 9

Find the gradient of 6x+3y−12=0.

3y = −6x+12
y = −2x+4
Therefore m = −2.

Intercepts

To find the x-intercept, set y=0. To find the y-intercept, set x=0.

Worked Example 10

Find the intercepts of 2x+5y=20.

x-intercept:
2x=20 → x=10 → (10,0)

y-intercept:
5y=20 → y=4 → (0,4).

06 — PARALLEL & PERPENDICULAR

Line Relationships

Parallel lines: m₁ = m₂

Perpendicular lines: m₁m₂ = −1

The perpendicular relationship follows because a 90° rotation changes the direction of a line so that the new gradient is the negative reciprocal.

Worked Example 11

Find the equation of the line through (3,−5) parallel to 2x−4y+7=0.

First find the original gradient:
−4y = −2x−7
y = 1/2 x + 7/4

Parallel line has gradient 1/2.
y+5 = 1/2(x−3)
y = 1/2x − 13/2.

Worked Example 12 — Perpendicular

Find the line through (4,1) perpendicular to y=−3x+8.

Original gradient = −3.
Perpendicular gradient = 1/3.

y−1 = (1/3)(x−4)
y = x/3 − 1/3.

Parameter problems

Worked Example 13 — Find k

The line through A(2,5) and B(k,9) has gradient 2. Find k.

(9−5)/(k−2)=2
4/(k−2)=2
4=2k−4
2k=6
k=3.

Worked Example 14 — Perpendicular condition

The line through (1,4) and (k,10) is perpendicular to a line of gradient 2. Find k.

Required gradient = −1/2.
(10−4)/(k−1)=−1/2
6/(k−1)=−1/2
12 = −(k−1)
k = −11.

07 — INTERSECTIONS

Where Geometric Objects Meet

Two straight lines

At an intersection, both equations are true simultaneously. Therefore, solve them as simultaneous equations.

Worked Example 15

Find the intersection of y=2x+1 and y=−x+7.

2x+1 = −x+7
3x=6
x=2
y=5

Intersection = (2,5).

Line and quadratic

If a line intersects a quadratic curve, substitute the line equation into the quadratic. This produces a quadratic equation in x.

Two distinct intersections → discriminant > 0
One repeated intersection → discriminant = 0
No real intersections → discriminant < 0

Worked Example 16

Determine whether y=x+2 intersects y=x²−2x+2.

Set equal:
x+2=x²−2x+2
x²−3x=0
x(x−3)=0
x=0 or x=3.

Therefore there are two intersections.

Parameter and discriminant

Suppose a line depends on k. After substitution, you obtain

ax²+bx+c=0.
Δ=b²−4ac

You can then impose a geometric condition by controlling Δ. For tangency, Δ=0. For two distinct intersections, Δ>0. For no real intersection, Δ<0.

08 — COORDINATE PROOFS

Proving Geometry with Coordinates

Coordinate proofs are powerful because different geometric properties have different algebraic fingerprints.

Geometric property	Useful test
Equal sides	Compare squared distances
Parallel sides	Compare gradients
Perpendicular sides	Gradient product −1
Same midpoint	Compare midpoint coordinates
Collinear points	Equal gradients

Proving an isosceles triangle

Worked Example 17

Show that A(1,2), B(5,5), C(9,2) form an isosceles triangle.

AB²=(5−1)²+(5−2)²=16+9=25.
BC²=(9−5)²+(2−5)²=16+9=25.

Therefore AB=BC, so triangle ABC is isosceles.

Proving a parallelogram

Worked Example 18

A quadrilateral has vertices A, B, C, D. One efficient method is to show that the diagonals bisect each other.

Calculate midpoint of AC. Calculate midpoint of BD. If the two midpoint coordinates are identical, the diagonals bisect each other, which proves the quadrilateral is a parallelogram.

Proving a rectangle

A parallelogram becomes a rectangle if one interior angle is 90°. Thus a strong strategy is:

Prove opposite sides are parallel.
Prove two adjacent sides are perpendicular.
Conclude that the parallelogram is a rectangle.

Collinearity

Worked Example 19

Determine whether A(1,3), B(4,9), C(7,15) are collinear.

Gradient AB:
(9−3)/(4−1)=6/3=2.

Gradient BC:
(15−9)/(7−4)=6/3=2.

The gradients are equal, so the three points are collinear.

09 — CIRCLES

Equation of a Circle

A circle consists of all points that are a fixed distance r from its centre (a,b).

(x−a)² + (y−b)² = r²

This is simply the distance formula with the distance fixed at r.

Reading a circle

Worked Example 20

Find the centre and radius of

(x−3)²+(y+5)²=49

Compare with:
(x−a)²+(y−b)²=r².

a=3, b=−5, r=7.

Centre = (3,−5)
Radius = 7.

Completing the square

For an equation such as

x²+y²−6x+8y−11=0

Group x and y terms:

(x²−6x)+(y²+8y)=11

Complete the square:

(x−3)²−9+(y+4)²−16=11
(x−3)²+(y+4)²=36

Therefore centre = (3,−4) and radius = 6.

Finding a circle from centre and a point

Worked Example 21

A circle has centre (2,−1) and passes through (5,3). Find its equation.

r²=(5−2)²+(3−(−1))² =9+16 =25.

Therefore:
(x−2)²+(y+1)²=25.

10 — TANGENTS & CHORDS

Circle Geometry in Coordinates

The radius drawn to the point of tangency is perpendicular to the tangent. This creates a direct connection between circle geometry and gradients.

m_radius × m_tangent = −1

Worked Example 22 — Tangent at a point

Find the tangent to

(x−2)²+(y−3)²=25

at P(5,7).

Centre C=(2,3).
Gradient CP:
m=(7−3)/(5−2)=4/3.

Therefore tangent gradient = −3/4.

Using P:
y−7=−3/4(x−5).

This is the tangent equation.

Chord midpoint theorem

The perpendicular from the centre of a circle to a chord bisects the chord. Therefore, if you know the centre and chord midpoint, the line joining them is perpendicular to the chord.

Worked Example 23

A chord has midpoint M(4,1) in a circle whose centre is C(1,5). Find the gradient of the chord.

Gradient CM:
(1−5)/(4−1)=−4/3.

Chord is perpendicular to CM.
m_chord=3/4.

Gradient = 3/4.

Line tangent to a circle using discriminant

Another method is to substitute the line into the circle. A tangent produces exactly one point of intersection, so the resulting quadratic has a repeated root.

Tangency condition: Δ = b²−4ac = 0

Worked Example 24

Find the values of k for which y=2x+k is tangent to x²+y²=25.

Substitute y=2x+k:
x²+(2x+k)²=25
5x²+4kx+k²−25=0.

For tangency:
(4k)²−4(5)(k²−25)=0
16k²−20k²+500=0
−4k²+500=0
k²=125
k=±5√5.

11 — ADVANCED REASONING

Multi-Step Coordinate Problems

Parameter workflow

Translate the geometric condition into algebra.
Choose the simplest formula.
Solve for the parameter.
Substitute back into the original geometry.
Check whether the resulting object satisfies the original condition.

Worked Example 25 — Unknown coordinate

A point P(k,4) is equidistant from A(2,1) and B(8,1). Find k.

PA²=(k−2)²+(4−1)²=(k−2)²+9.
PB²=(k−8)²+9.

Equidistant:
(k−2)²+9=(k−8)²+9
(k−2)²=(k−8)²
k²−4k+4=k²−16k+64
12k=60
k=5.

Worked Example 26 — Perpendicular bisector

Find the perpendicular bisector of A(2,3) and B(8,7).

Midpoint:
M=(5,5).

Gradient AB:
(7−3)/(8−2)=4/6=2/3.

Perpendicular gradient = −3/2.

Equation:
y−5=−3/2(x−5).

3x+2y−25=0.

Worked Example 27 — Equal distance and geometry

Explain why every point on the perpendicular bisector of AB is equidistant from A and B.

The perpendicular bisector passes through the midpoint M of AB and is perpendicular to AB. For any point P on this line, triangles PMA and PMB are right triangles. They share PM and have MA=MB. Therefore the triangles are congruent, so PA=PB.

Worked Example 28 — Circle through three points

Suppose a circle is

x²+y²+Dx+Ey+F=0.

If three points are known to lie on it, substitute each point to obtain three simultaneous linear equations in D,E,F.

This is often much more efficient than guessing the centre first when three points are supplied.

Worked Example 29 — Optimising algebra

When comparing distances, avoid square roots whenever possible.

Instead of proving AB=CD by evaluating both square roots, prove

AB²=CD².

Because both distances are non-negative, this proves AB=CD.

Worked Example 30 — Choosing the method

A problem says: “Find the point on the x-axis that is equidistant from A(2,5) and B(8,1).”

Because the unknown point is on the x-axis, write P=(x,0). Then use PA²=PB²:

(x−2)²+25=(x−8)²+1.

Expand and solve for x. The key skill is not the algebra—it is translating the words into the correct geometric condition.

Mixed challenge set

Challenge 1

A line through (k,2) and (4,8) is parallel to 3x−2y+7=0. Find k.

Answer: k=0.

Challenge 2

Find the circle with centre (−3,4) passing through (1,1).

Answer: (x+3)²+(y−4)²=25.

Challenge 3

Find the equation of the perpendicular bisector of the points (−2,5) and (6,−1).

Answer: 4x−3y+1=0.

Challenge 4

Find the value of k such that y=x+k is tangent to x²+y²=8.

Answer: k=±4.

12 — EXAM STRATEGY

How to Attack Coordinate Problems

1. Identify the object

Ask whether the problem is about a point, segment, line, circle, intersection, ratio, distance or perpendicularity.

2. Translate words into equations

Words	Translate to
same gradient	m₁=m₂
parallel	m₁=m₂
perpendicular	m₁m₂=−1
equidistant	PA²=PB²
midpoint	average coordinates
tangent	one intersection / Δ=0
collinear	equal gradients

3. Keep exact values

Use fractions and surds instead of decimals whenever possible. Exact values make later algebra cleaner and reduce rounding errors.

4. Check the answer geometrically

If a point is supposed to lie on a line, substitute it back. If a point is supposed to lie on a circle, check the circle equation. If two lines are supposed to be perpendicular, check the gradients.

Major exam traps

Forgetting brackets around negative coordinates.
Changing the sign of only the numerator or denominator.
Calling a vertical line's gradient zero.
Using the wrong perpendicular gradient.
Reading the centre of a circle with the wrong sign.
Forgetting that (y+4)² means centre y-coordinate −4.
Using Δ=0 when the question asks for two distinct intersections.
Rounding too early.
Proving equal distances using unnecessary square roots.
Finding x correctly but failing to substitute back for y.

13 — FORMULA ATLAS