Lesson 4: Coordinate Geometry | O Level Mathematics

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1. Coordinates and Key Formulae

Coordinate geometry studies geometric shapes using an algebraic framework. Every point in a plane is described by an ordered pair (x, y) measured from the origin along horizontal (x) and vertical (y) axes.

Gradient

m = (y₂ − y₁) / (x₂ − x₁)

Rise over run. Positive m → line rises left to right. Negative m → line falls.

Midpoint

M = ((x₁+x₂)/2 , (y₁+y₂)/2)

Average of x-coordinates and average of y-coordinates.

Length (Distance)

d = √[(x₂−x₁)² + (y₂−y₁)²]

Pythagoras' theorem applied to coordinates.

📐 Worked Example 1 — Gradient, Midpoint and Length

Points A(1, 3) and B(7, 11). Find: (a) gradient of AB (b) midpoint of AB (c) length of AB.

(a) Gradient:
m = (11 − 3)/(7 − 1) = 8/6 = 4/3

(b) Midpoint:
M = ((1+7)/2, (3+11)/2) = (4, 7)

(c) Length:
d = √[(7−1)² + (11−3)²] = √[36 + 64] = √100 = 10

2. Equation of a Straight Line

Standard Form: y = mx + c, where m is the gradient and c is the y-intercept (the value of y when x = 0).

Finding the Equation of a Line

Method — Given gradient m and a point (x₁, y₁):
Use: y − y₁ = m(x − x₁) then rearrange to y = mx + c.

📐 Worked Example 2 — Equation Given Two Points

Find the equation of the line passing through P(2, 5) and Q(6, 13).

Find the gradient:
m = (13−5)/(6−2) = 8/4 = 2

Use y − y₁ = m(x − x₁) with point P(2, 5):
y − 5 = 2(x − 2)
y − 5 = 2x − 4
y = 2x + 1

Check with Q(6,13): y = 2(6)+1 = 13 ✓

Reading the Equation from a Graph

📐 Worked Example 3 — From Graph to Equation

A line crosses the y-axis at (0, −3) and passes through (4, 5). Find its equation.

c = −3 (y-intercept read directly from graph).

Calculate gradient:
m = (5−(−3))/(4−0) = 8/4 = 2

Equation: y = 2x − 3

3. Parallel and Perpendicular Lines

Parallel Lines: Have equal gradients.
If line 1 has gradient m, then any parallel line also has gradient m.
y = 3x + 5 and y = 3x − 2 are parallel (both have m = 3).

Perpendicular Lines: Their gradients multiply to −1.
If line 1 has gradient m, the perpendicular line has gradient −1/m.
m₁ × m₂ = −1 → m₂ = −1/m₁

📐 Worked Example 4 — Perpendicular Line

Line L has equation y = 3x − 4. Find the equation of the line perpendicular to L passing through (6, 2).

Gradient of L = 3. Perpendicular gradient:
m⊥ = −1/3

Use y − y₁ = m(x − x₁) with (6, 2):
y − 2 = −⅓(x − 6)
y − 2 = −⅓x + 2
y = −⅓x + 4

📐 Worked Example 5 — Intersection of Two Lines

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

Set the equations equal (at intersection, y values are the same):
2x + 1 = −x + 7
3x = 6 → x = 2

Substitute x = 2 into either equation:
y = 2(2) + 1 = 5
Intersection point: (2, 5)

4. Coordinate Geometry Problems

Cambridge often combines gradient, midpoint, length, and line equations in multi-part problems involving triangles, quadrilaterals, and perpendicular bisectors.

Perpendicular Bisector

Perpendicular Bisector of AB: The line that is perpendicular to AB AND passes through the midpoint of AB. Every point on the perpendicular bisector is equidistant from A and B.

📐 Worked Example 6 — Perpendicular Bisector

Find the equation of the perpendicular bisector of the line segment joining A(1, 4) and B(5, 2).

Find the midpoint of AB:
M = ((1+5)/2, (4+2)/2) = (3, 3)

Find gradient of AB:
m_AB = (2−4)/(5−1) = −2/4 = −½

Perpendicular gradient = 2 (negative reciprocal of −½).
Line through M(3,3) with gradient 2:
y − 3 = 2(x − 3)
y = 2x − 3

Proving Properties of Geometric Shapes

Property to Prove	Method
Lines are parallel	Show gradients are equal.
Lines are perpendicular	Show product of gradients = −1.
Triangle is right-angled	Show two sides have gradients with product −1, OR use Pythagoras on the lengths.
Triangle is isosceles	Show two sides have equal length using the distance formula.
Quadrilateral is a parallelogram	Show both pairs of opposite sides are parallel (equal gradients).
Quadrilateral is a rectangle	Show it is a parallelogram AND one angle is 90° (adjacent sides perpendicular).
Point lies on a line	Substitute the point's coordinates into the line equation — if it satisfies the equation, the point lies on the line.

📐 Worked Example 7 — Proving a Triangle is Right-Angled

A(0, 2), B(4, 0), C(5, 5). Show that angle ABC = 90°.

Find gradient of BA:
m_BA = (2−0)/(0−4) = 2/(−4) = −½

Find gradient of BC:
m_BC = (5−0)/(5−4) = 5/1 = 5

Check product: m_BA × m_BC = −½ × 5 = −2.5 ≠ −1
Hmm — let us recalculate. Actually angle ABC: vectors from B.
BA = A−B = (−4, 2), BC = C−B = (1, 5).
Dot product = (−4)(1)+(2)(5) = −4+10 = 6 ≠ 0.
Try angle BAC: grad AB = (0−2)/(4−0) = −½; grad AC = (5−2)/(5−0) = 3/5.
Product: −½ × 3/5 = −3/10 ≠ −1. Try angle ACB:
grad CA = (2−5)/(0−5)=−3/(−5)=3/5; grad CB=(0−5)/(4−5)=−5/−1=5.
Product: 3/5 × 5 = 3 ≠ −1. Use Pythagoras:
AB² = 16+4=20, BC²=1+25=26, AC²=25+9=34.
AB²+BC²=46≠AC². Check AC²+AB²: 34+20=54≠26. Try BC²+AB²=26+20=46≠34.
This triangle is not right-angled. Using a corrected example:
A(0,0), B(4,0), C(4,3):
grad AB = 0, grad BC = (3−0)/(4−4) = undefined (vertical line).
AB is horizontal, BC is vertical → they are perpendicular → angle ABC = 90°.

Dividing a Line in a Given Ratio

Section Formula: If point P divides AB in the ratio m : n, then:
P = ((mx₂ + nx₁)/(m+n) , (my₂ + ny₁)/(m+n))

📐 Worked Example 8 — Section Formula

A is (1, 2) and B is (7, 8). Point P divides AB in the ratio 2:1. Find the coordinates of P.

m=2, n=1. Apply the section formula:
P_x = (2×7 + 1×1)/(2+1) = (14+1)/3 = 15/3 = 5
P_y = (2×8 + 1×2)/(2+1) = (16+2)/3 = 18/3 = 6

P = (5, 6)
Check: AP = √[(5−1)²+(6−2)²] = √32 = 4√2
PB = √[(7−5)²+(8−6)²] = √8 = 2√2
Ratio AP:PB = 4√2 : 2√2 = 2:1 ✓

5. Linear Graphs — Practical Applications

Many real-world relationships are linear and can be analysed using coordinate geometry. Cambridge Paper 2 frequently asks you to find equations from graphs representing costs, conversions, and rates.

Interpreting Gradient and Intercept in Context

Context	What Gradient Means	What y-intercept Means
Cost vs quantity	Cost per unit (price per item)	Fixed cost (cost even when quantity = 0)
Distance vs time	Speed	Starting distance from reference point
Temperature conversion	Rate of conversion between units	Offset between scales
Hire charge	Rate per hour/day	Fixed hire fee regardless of usage

📐 Worked Example 9 — Real-World Linear Graph

A plumber charges a fixed call-out fee plus an hourly rate. For 2 hours the total charge is $95. For 5 hours it is $185. Find the call-out fee and the hourly rate. Find the equation connecting total charge C and hours h.

Two points: (2, 95) and (5, 185). Gradient = hourly rate:
m = (185−95)/(5−2) = 90/3 = $30 per hour

Use C − 95 = 30(h − 2):
C = 30h + 35
Call-out fee = c-intercept = $35. Hourly rate = $30/hour.

Check: h=5: C = 150+35 = $185 ✓

Converting Between Linear Forms

Form	Equation	Use
Slope-intercept	y = mx + c	Reading gradient and y-intercept directly
Point-slope	y − y₁ = m(x − x₁)	Finding equation given gradient and one point
General form	ax + by + c = 0	Cambridge sometimes asks for this form — rearrange from y=mx+c

General Form: To convert y = ⅔x − 4 to general form ax+by+c=0:
Multiply through by 3: 3y = 2x − 12
Rearrange: 2x − 3y − 12 = 0 (where a=2, b=−3, c=−12)

6. Locus in Coordinate Geometry

Locus: The set of all points satisfying a given condition. In coordinate geometry, a locus is often described by an equation.

Condition	Locus	Equation
All points at distance r from point (a, b)	Circle with centre (a,b) and radius r	(x−a)² + (y−b)² = r²
All points equidistant from two fixed points A and B	Perpendicular bisector of AB	Find using midpoint and perpendicular gradient
All points at a fixed distance from a line	Two parallel lines, one on each side	Parallel lines displaced by that distance

📐 Worked Example 10 — Circle as a Locus

A point P moves so that its distance from the point C(3, −1) is always 5. Write the equation of the locus of P. Does the point (7, 2) lie on this locus?

The locus is a circle with centre (3, −1) and radius 5:
(x−3)² + (y+1)² = 25

Test (7, 2): substitute into equation:
(7−3)² + (2+1)² = 16 + 9 = 25 ✓
Yes, (7, 2) lies on the locus.

📝 Exam Practice Questions

Q1 [3 marks] — A(−2, 3) and B(4, −1). Find: (a) the gradient of AB (b) the midpoint of AB (c) the length of AB.

(a) m = (−1−3)/(4−(−2)) = −4/6 = −2/3

(b) M = ((−2+4)/2, (3+(−1))/2) = (1, 1)

(c) d = √[(4−(−2))² + (−1−3)²] = √[36+16] = √52 = 2√13

Q2 [3 marks] — Find the equation of the line passing through (−1, 4) with gradient −2. Give your answer in the form ax + by + c = 0.

Working:
y − 4 = −2(x − (−1))
y − 4 = −2x − 2
y = −2x + 2
Rearrange: 2x + y − 2 = 0 → 2x + y − 2 = 0

Q3 [4 marks] — The line L₁ passes through A(0, 5) and B(4, 3). Line L₂ is perpendicular to L₁ and passes through B. Find the equation of L₂. Find where L₂ crosses the x-axis.

Gradient of L₁: m = (3−5)/(4−0) = −2/4 = −½
Gradient of L₂: m⊥ = 2
L₂ through B(4, 3): y − 3 = 2(x − 4) → y = 2x − 5
Equation of L₂: y = 2x − 5

x-intercept: Set y = 0: 0 = 2x − 5 → x = 2.5
L₂ crosses x-axis at (2.5, 0)

Exam Tip: Always check: perpendicular gradient × original gradient = −1. Here: 2 × (−½) = −1 ✓

Q4 [4 marks] — Find the equation of the perpendicular bisector of the segment joining C(−3, 2) and D(5, 6).

Midpoint: M = (1, 4)
Gradient of CD: m = (6−2)/(5−(−3)) = 4/8 = ½
Perpendicular gradient: −2
Line through M(1, 4): y − 4 = −2(x − 1) → y = −2x + 6
y = −2x + 6

Q5 [4 marks] — Points P(1, 1), Q(5, 3), R(4, 5) form a triangle. Show that PQ is perpendicular to QR and find the area of triangle PQR.

Gradient PQ: (3−1)/(5−1) = 2/4 = ½
Gradient QR: (5−3)/(4−5) = 2/(−1) = −2
Product: ½ × (−2) = −1 ✓ → PQ ⊥ QR

Area: Right angle at Q, so legs are PQ and QR.
PQ = √[(5−1)²+(3−1)²] = √[16+4] = √20 = 2√5
QR = √[(4−5)²+(5−3)²] = √[1+4] = √5
Area = ½ × 2√5 × √5 = ½ × 2 × 5 = 5 square units

Exam Tip: When a right angle is confirmed, use ½ × base × height directly using the two perpendicular legs as base and height — much faster than the general area formula.

Q6 [3 marks] — A taxi charges a fixed booking fee plus a rate per kilometre. For 5 km the fare is $11.50. For 12 km the fare is $19.00. Find the booking fee and the rate per km. Hence find the distance travelled if the fare was $23.50.

Two points: (5, 11.50) and (12, 19.00)
Rate per km (gradient): (19.00−11.50)/(12−5) = 7.50/7 = $1.07.../km... Let me use exact: 7.5/7 — check if integer: try (12−5)=7, diff=7.5. Hmm, let us restate with cleaner numbers:
Actually: m = 7.5/7 ≈ 1.071. Let us use the given values exactly:
Fare = m×d + c. 5m+c=11.5 and 12m+c=19.
Subtract: 7m = 7.5 → m = $1.07/km (not clean). Revised: use 5m+c=11.5, m=1.5 (if 7m=7.5/... Actually 7.5÷7 is not clean. Restate: if m=1.50: 7×1.5=10.5≠7.5. Let m=7.5/7. c=11.5−5(7.5/7)=11.5−37.5/7=(80.5−37.5)/7=43/7≈6.14.
Fare = $23.50: 23.5 = (7.5/7)d + 43/7 → 7×23.5 = 7.5d+43 → 164.5−43=7.5d → 121.5=7.5d → d=16.2 km

Exam Tip: Always set up two equations from the two data points, subtract to eliminate c, find m, then back-substitute for c. Verify both original conditions before answering the final part.

Q7 [3 marks] — A point P(k, 3k) lies on the line 2x − 3y + 6 = 0. Find the value of k. Hence find the distance from P to the origin.

Substitute (k, 3k): 2k − 3(3k) + 6 = 0
2k − 9k + 6 = 0 → −7k = −6 → k = 6/7

P = (6/7, 18/7)
Distance from origin = √[(6/7)² + (18/7)²] = √[(36+324)/49] = √(360/49) = (6√10)/7 ≈ 2.71 units

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