For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: parallel and perpendicular conditions, midpoints, rectilinear area, and straight-line problem solving form the main route.
School-sensitive extension: distance-to-line, foot-of-perpendicular, loci, and wider line forms may be school-sensitive depth.
2027 national comparison: K341 Topic G2 includes parallel or perpendicular lines, midpoint, and area of a rectilinear figure.
Check your school: confirm accepted line forms and whether perpendicular-distance formulas are supplied or required.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 06) Coordinate Geometry of Lines cover? A: Gradients, perpendicularity, and distance formula recap for line questions in IP Additional Mathematics.
Linear coordinate geometry blends algebra with spatial reasoning. Track gradients and midpoints carefully and always label key coordinates on a sketch.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Line geometry is gradients, distances, midpoints, and perpendicular relationships.
Use it as a working check: Sketch and label the coordinates before calculating. Most errors come from using the wrong pair of points or forgetting a negative reciprocal.
Then go one layer deeper: Example: for a perpendicular bisector, find the midpoint first, then use the negative reciprocal of the original gradient for the new line.
Choosing the right line tool
Start each coordinate-geometry question by naming the unknown: a gradient, a point, a length, or an equation. That choice determines the first calculation.
Question cue
First tool to use
Follow-up check
Find the equation of a line through one point
Need a gradient first, then use point-slope form y−y1=m(x−x1).
Substitute the given point back into the final equation.
Find a perpendicular bisector
Find the midpoint, then the negative reciprocal gradient.
The final line must pass through the midpoint, not either endpoint unless the segment is special.
Find the foot of a perpendicular
Write the original line and perpendicular line, then solve them simultaneously.
The foot must lie on the original line.
Find distance from a point to a line
Use the point-to-line distance formula only after the line is in Ax+By+C=0 form.
Distance is never negative, so keep the absolute value.
Prove two lines are parallel or perpendicular
Compare gradients before forming equations.
Parallel lines have equal gradients; perpendicular non-vertical lines satisfy m1m2=−1.
Common trap: A perpendicular gradient is the negative reciprocal only for non-vertical, non-horizontal lines. Vertical and horizontal lines swap directly: x=a is perpendicular to y=b.
1 Essentials
Gradient between (x1,y1) and (x2,y2): m=x2−x1y2−y1.
Equation forms: point-slope y−y1=m(x−x1), two-point, intercept, and normal forms are interchangeable.
Rectilinear-area checkpoint
When vertices are given as coordinates, order them around the boundary before finding the area. You can split the figure into triangles and rectangles, or use the shoelace method if your school teaches it. A quick check is that reversing the vertex order changes only the sign of the shoelace total, not the geometric area.
For a triangle with vertices A(x1,y1), B(x2,y2), and C(x3,y3),
Area=21∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.
Misconception check: distance formulae give side lengths, not area. Use an area method unless the shape is already known to be a rectangle or a right triangle.
2 Worked example - Perpendicular bisector
Find the equation of the perpendicular bisector of segment joining A(2,−1) and B(8,5).
Midpoint:M=(22+8,2−1+5)=(5,2).
Gradient of AB: mAB=8−25−(−1)=66=1
Perpendicular gradient:m⊥=−1.
Equation through M: y−2=−1(x−5) so y=−x+7
3 Worked example - Distance from point to line
Compute the distance from P(4,−3) to the line 3x−4y−20=0.
Find the foot of the perpendicular from P(4,1) onto the line 2x−y+5=0, and state the equation of the perpendicular.
Rewrite the line: y=2x+5. Its gradient is 2; the perpendicular gradient is −21.
Equation of perpendicular through P: y−1=−21(x−4), so y=−21x+3
Solve the simultaneous system with y=2x+5: −21x+3=2x+5
Substitute into y=2x+5: y=2(−54)+5=517
Therefore the foot is F(−54,517) and the perpendicular line is y=−21x+3
Answer: F(−54,517); perpendicular line y=−21x+3.
Unknown-coordinate checkpoint
When a coordinate is written as a letter, turn each geometry condition into an equation before trying to solve for the letter.
line condition -> equation of line -> intersection condition -> unknown coordinate
Prompt clue
First equation to build
Next condition to apply
Common trap
A line has a given gradient and passes through one point
Use point-slope form.
Substitute another point or an axis condition.
Treating the given gradient as an x-coordinate change.
A line is perpendicular to another line
Find the negative reciprocal gradient first.
Use the given point to form the new line.
Using the original line's gradient again.
Intersection lies on the x-axis
Set y=0.
Solve the line equation for the x-coordinate of the intersection.
Setting x=0, which is the y-axis condition.
A point (k,a) lies on a line
Substitute x=k and y=a.
Solve for k
Worked check: if L1 has gradient −23, a perpendicular line has gradient 32. If L2 passes through (k,−2) and meets the x-axis at (1,0), then
1−k0−(−2)=32.
So 2/(1−k)=2/3, hence 1−k=3 and k=−2. The key step is using the x-axis condition to create the second point on L2.
Misconception check: "perpendicular" tells you the gradient of L2, not the value of k directly. The value of k comes only after you combine that gradient with a second point or axis condition.
5 Practice Quiz
Test your gradient, midpoint, and distance formulas together with perpendicular/bisector reasoning before tackling harder locus problems.
6 Try this
Line L1 passes through (1,7) and has gradient −23. Another line L2 passes through (k,−2) and is perpendicular to L1. Determine k if the intersection lies on the x-axis.
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gives
−25x=2
and
x=−54
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only after the line equation is known.
Solving for k before finding which line the point lies on.
