G11: Coordinate Geometry Problems
Solve geometrical problems on coordinate axes
📋 Key Concepts
Coordinate geometry uses coordinates to describe points, lines, and shapes on a grid. The x-axis runs horizontally, the y-axis runs vertically.
Coordinates are written as (x, y) where x is the horizontal position and y is the vertical position.
📝 Finding the Midpoint
Midpoint formula: The midpoint of a line segment joining (x₁, y₁) and (x₂, y₂) is found by averaging the x-coordinates and y-coordinates.
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example 1
Find the midpoint of the line segment joining A(2, 4) and B(8, 10).
Solution:
Midpoint = ((2 + 8)/2, (4 + 10)/2)
= (10/2, 14/2)
= (5, 7)
Example 2
If M(3, 5) is the midpoint of AB, and A is at (1, 2), find the coordinates of B.
Solution:
Let B = (x, y)
Midpoint: ((1 + x)/2, (2 + y)/2) = (3, 5)
(1 + x)/2 = 3 → 1 + x = 6 → x = 5
(2 + y)/2 = 5 → 2 + y = 10 → y = 8
B = (5, 8)
📝 Distance Between Two Points
Distance formula: The distance between (x₁, y₁) and (x₂, y₂) is found using Pythagoras' theorem.
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Example 3
Find the distance between A(1, 2) and B(4, 6).
Solution:
Distance = √[(4 - 1)² + (6 - 2)²]
= √[3² + 4²]
= √[9 + 16]
= √25 = 5 units
📝 Gradient of a Line
Gradient formula: The gradient (slope) of a line joining two points measures the steepness.
Gradient = (y₂ - y₁)/(x₂ - x₁)
Example 4
Find the gradient of the line joining (2, 3) and (6, 11).
Solution:
Gradient = (11 - 3)/(6 - 2)
= 8/4 = 2
📝 Equation of a Line
Equation form: y = mx + c where m is the gradient and c is the y-intercept.
Example 5
Find the equation of the line with gradient 3 passing through (2, 7).
Solution:
y = mx + c
7 = 3(2) + c
7 = 6 + c
c = 1
Equation: y = 3x + 1
Example 6
Find the equation of the line passing through (1, 4) and (5, 12).
Solution:
Gradient = (12 - 4)/(5 - 1) = 8/4 = 2
y = 2x + c
Using (1, 4): 4 = 2(1) + c → c = 2
Equation: y = 2x + 2
📝 Parallel and Perpendicular Lines
Parallel lines: Have the same gradient.
Perpendicular lines: The product of their gradients equals -1.
If gradient of line 1 = m, then gradient of perpendicular line = -1/m
Example 7
Find the equation of the line parallel to y = 2x + 5 passing through (3, 4).
Solution:
Parallel lines have same gradient: m = 2
y = 2x + c
4 = 2(3) + c → c = -2
Equation: y = 2x - 2
Example 8
Find the equation of the line perpendicular to y = 3x + 1 passing through (0, 5).
Solution:
Gradient of perpendicular line = -1/3
Passing through (0, 5): this is the y-intercept, c = 5
Equation: y = -1/3x + 5
❓ Practice Questions
Q1: Find the midpoint of (4, 2) and (10, 8).
Q2: Find the distance between (0, 0) and (3, 4).
Q3: Find the gradient of the line joining (-2, 1) and (4, 13).
Q4: Find the equation of the line with gradient -2 passing through (1, 6).
Q5: Find the gradient of a line perpendicular to y = 4x + 3.
✅ Answers
- (7, 5)
- 5 units
- 12/6 = 2
- y = -2x + 8
- -1/4
🎯 Exam Tips
- Always substitute coordinates back to check your equation
- For perpendicular gradients, flip and change the sign
- Draw a quick sketch to visualise the problem
- Watch out for negative coordinates
- Remember: parallel = same gradient, perpendicular = gradients multiply to -1
🧠 Problem-Solving Strategies
Problem-Solving
Gradient = (y2 - y1)/(x2 - x1). Midpoint = ((x1+x2)/2, (y1+y2)/2). Distance = sqrt((x2-x1) squared + (y2-y1) squared). For parallel lines: same gradient. For perpendicular lines: gradients multiply to give -1. Always sketch the diagram to check your answer makes sense.
Multi-Step Problem
A(2,3) and B(8,7) are endpoints of a diameter of a circle. Find the centre and radius of the circle, and the equation of the perpendicular bisector of AB.
Solution: Centre = midpoint = ((2+8)/2, (3+7)/2) = (5, 5). Radius = half the distance AB. AB = sqrt(36 + 16) = sqrt(52). Radius = sqrt(52)/2 = sqrt(13). Gradient of AB = 4/6 = 2/3. Perpendicular gradient = -3/2. Perpendicular bisector through (5,5): y - 5 = -3/2(x - 5), so y = -3x/2 + 25/2.
⚠️ Common Errors
Watch Out!
1. Wrong: Calculating gradient as (x2 - x1)/(y2 - y1) Correct: Gradient = (change in y)/(change in x) = (y2 - y1)/(x2 - x1). Put y on top, x on bottom.
2. Wrong: Saying perpendicular lines have the same gradient Correct: Parallel lines have the same gradient. Perpendicular lines have gradients that multiply to -1 (negative reciprocal).
3. Wrong: Forgetting to square both differences when using the distance formula Correct: Distance = sqrt((x2-x1) squared + (y2-y1) squared). Both the x-difference and y-difference must be squared before adding.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Points P(1,2), Q(7,5) and R(4,k) form a triangle. (a) Find the gradient of PQ. (b) If PR is perpendicular to PQ, find k. (c) Find the area of triangle PQR.
(a) Gradient of PQ = (5-2)/(7-1) = 3/6 = 1/2.
(b) Gradient of PR = (k-2)/(4-1) = (k-2)/3. For perpendicular: gradient of PR x gradient of PQ = -1. So (k-2)/3 x 1/2 = -1. (k-2)/6 = -1. k - 2 = -6. k = -4.
(c) R = (4, -4). Base PQ = sqrt(36 + 9) = sqrt(45) = 3sqrt(5). Height: distance from R to line PQ. Line PQ: y - 2 = 1/2(x - 1), so 2y = x + 3, or x - 2y + 3 = 0. Distance = |4 - 2(-4) + 3|/sqrt(1 + 4) = |4 + 8 + 3|/sqrt(5) = 15/sqrt(5) = 3sqrt(5). Area = 1/2 x 3sqrt(5) x 3sqrt(5) = 1/2 x 45 = 22.5 square units.
Mark scheme: M1 gradient formula, A1 m = 1/2, M1 perpendicular condition, A1 k = -4, M1 distance/area method, A1 22.5
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Three towns A(0,0), B(6,8), C(12,0) are connected by straight roads.
(a) Show that triangle ABC is isosceles.
(b) Find the midpoint M of AC and the gradient of BM. Is BM perpendicular to AC?
(c) A planner says "Building a road from B to M is the shortest route from B to any point on road AC." Is this correct? Explain mathematically.
Answers: (a) AB = sqrt(36+64) = 10. BC = sqrt(36+64) = 10. AC = 12. AB = BC so isosceles. (b) M = (6,0). Gradient of BM = (8-0)/(6-6) = undefined (vertical line). Gradient of AC = 0 (horizontal). Vertical and horizontal lines are perpendicular. Yes, BM is perpendicular to AC. (c) Yes — the perpendicular from a point to a line gives the shortest distance. Since BM is perpendicular to AC, M is the closest point on AC to B, so this is the shortest route from B to any point on road AC.
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