Q: What does IP Maths Notes (Lower Sec, Year 1-2): 04) Coordinate Geometry & Graphs cover? A: Understand gradients, intercepts, distance and midpoint formulas, and sketch linear graphs with contextual interpretations.
The core idea is simple: Graphs turn algebra into a picture of direction, position, and change.
Use it as a working check: Gradient measures change, intercepts show where a graph crosses an axis, and distance or midpoint formulas describe points without drawing perfectly.
Then go one layer deeper: Use the examples to practise finding a line from two points, comparing parallel or perpendicular lines, and explaining what a graph means in context.
Coordinate geometry links algebra to visuals. This note refreshes gradient logic, distance and midpoint formulas, equation forms, and graph interpretation.
These notes align with MOE Lower Secondary Mathematics syllabus used in IP pathways (aligned to O-Level Mathematics 4052 foundations).
Status: MOE Lower Secondary Mathematics syllabus (latest release) checked 2025-11-30 - scope unchanged; remains the reference for these lower-sec notes.
Slope-intercept form: y=mx+c, where c is the y-intercept.
Point-slope form: y−y1=m(x−x1).
Gradient meaning map
Before calculating, decide what the gradient should mean.
What the line does from left to right
Gradient sign
What it tells you
Rises
Positive
y increases as x increases.
Falls
Negative
y decreases as x increases.
Flat
Zero
y stays constant.
Vertical
Undefined
x stays constant, so the gradient formula would divide by zero.
For a context graph, read the units too:
gradient=change in horizontal quantitychange in vertical quantity.
That is why the gradient of a distance-time graph is speed, while the gradient of a cost-quantity graph is cost per item.
Worked example - Gradient & intercept
Find the gradient and y-intercept of the line passing through (2,−3) and (5,6).
m=5−26−(−3)=39=3.
Use point-slope form with (2,−3):
y+3=3(x−2)⟹y=3x−9.
Y-intercept is −9.
Worked example - Reading a line before calculating
A graph of total cost C against number of notebooks n passes through (0,2) and (5,17).
The point (0,2) means the starting cost is $2, even before buying any notebooks.
The gradient is
m=5−017−2=3.
The equation is C=3n+2.
In context, each notebook costs $3, and $2 is the fixed starting charge.
Misconception check: the y-intercept is not always "free". It is the value of the vertical quantity when the horizontal quantity is zero.
2 Distance and midpoint formulas
Distance between two points: d=(x2−x1)2+(y2−y1)2.
Midpoint: M(2x1+x2,2y1+y2)
Coordinate step diagram checkpoint
Before choosing a formula, decide whether the question wants length or location.
A(x1, y1) -------------------- B(x2, y2)
horizontal change: x2 - x1
vertical change: y2 - y1
Distance: use the two changes as sides of a right-angled triangle.
Midpoint: average the two x-values, then average the two y-values.
Question asks for
What to calculate
Why this works
Common trap
Length of AB
Differences: x2−x1 and y2−y1.
These are the horizontal and vertical sides of the right-angled triangle.
Adding coordinates instead of finding changes.
Midpoint of AB
Averages: 2x1+x2
Sign check
Keep negative values inside the brackets first.
Squaring removes signs for distance, but averaging keeps the sign information.
Turning −6 into 6 before finding the midpoint.
Worked check: for A(−2,5) and B(4,−1), the horizontal change is 4−(−2)=6, while the vertical change is −1−5=−6. Distance uses 62+(−6)2, but midpoint uses the original coordinate averages: (−2+4)/2=1 and (5+(−1))/2=2.
Example: Points A(−2,5) and B(4,−1).
d=(4−(−2))2+(−1−5)2=62+(−6)2=72=62.
Midpoint:
M(2−2+4,25+(−1))=M(1,2).
3 Parallel and perpendicular lines
Parallel lines share the same gradient.
Perpendicular gradients multiply to −1: if m1×m2=−1, the lines are perpendicular.
Example: The line through (3,−4) perpendicular to y=−21x+1 has gradient 2 and equation y+4=2(x−3)⟹y=2x−10.
Line relationship checkpoint
Before writing an equation, identify the relationship between the new line and the given line. This prevents the common error of copying a gradient when the question asks for a perpendicular line.
Question clue
First move
New gradient or equation
Common trap
Parallel to y=mx+c
Keep the same gradient
m
Changing the y-intercept but also changing the gradient.
Perpendicular to y=mx+c
Take the negative reciprocal
−m1
Using m1 without changing the sign.
Parallel to x=a
Keep it vertical
x=b
Trying to assign a finite gradient to a vertical line.
Perpendicular to x=a
Make it horizontal
y=b
Saying the perpendicular gradient is zero without naming the horizontal equation.
Worked check: a line perpendicular to y=43x−2 has gradient −34. If it passes through (6,1), start with y−1=−34(x−6), then simplify only if the question asks for a final equation in a specific form.
4 Graph interpretation
Always label axes with units.
Identify intercepts, turning points, and intersection points.
Connect slope to rate e.g.,distance−timegraphs:slope=speed.
Worked example - Interpreting a distance-time graph
A student jogs 3 km in 18 minutes, rests for 6 minutes, then jogs another 2 km in 10 minutes.
Segment 1 gradient: 0.33=10 km h−1.
Segment 2 gradient: 0 (rest).
Segment 3 gradient: 60102=12 km h−1.
Explain what each segment means in sentences - examiners award communication marks.
Distance-time unit checkpoint
Before finding a gradient on a distance-time graph, make the distance unit and time unit match the answer unit you want.
Graph units
Convert first?
Gradient unit
Common trap
Distance in km, time in h
No conversion needed.
km h−1
Adding "per hour" without checking the time axis.
Distance in km, time in min
Convert minutes to hours if speed is wanted in km h−1.
km h−1
Dividing by minutes and still calling the answer km per hour.
Distance in m, time in s
No conversion needed for SI speed.
m⋅s−1
Converting metres to kilometres when the answer asks for metres per second.
Worked check: 18min=18/60=0.30h, so a 3km change over 18min gives
speed=0.303=10 km h−1.
Misconception check: the gradient is rise over run, but the units come from vertical unit divided by horizontal unit. Convert the horizontal unit first if the requested speed unit needs it.
5 System of equations via intersection
Graphing y=2x+1 and y=−x+7 reveals intersection at (2,5). Solving algebraically verifies the result.
Intersection-solving checkpoint
The intersection point is the pair of values that makes both equations true at the same time. Read it from the graph when a sketch is required, then verify it algebraically when exact coordinates are needed.
Step
What to do
Why it matters
1
Put both equations in a comparable form, usually y=mx+c.
You can see each line's gradient and intercept before sketching.
2
Set the two right-hand sides equal.
At the intersection, both lines have the same x-value and the same y-value.
3
Solve for x.
This gives the horizontal coordinate of the intersection.
4
Substitute x into either original equation.
This gives the vertical coordinate and checks consistency.
Worked check: solve y=2x+1 and y=−x+7.
2x+1=−x+7⇒3x=6⇒x=2.
Substitute into y=2x+1: y=2(2)+1=5. The intersection is (2,5).
Misconception check: the answer is not just x=2. A graph intersection is a coordinate pair, so include both x and y.
Practice Quiz
Consolidate gradient calculations, distance and midpoint work, and graph interpretation with the interactive quiz.
Try it yourself
Find the equation of the line through (4,−5) with gradient −43.
Determine the midpoint of the segment joining (7,2) and (−1,−6).
Show that the line joining (2,3) and (6,7) is parallel to the line 2x−2y=1.
Two lines intersect at (1,−2). One has equation y=4x−6. Find the equation of the perpendicular line through the same point.
