Coordinate Geometry

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Notes

Coordinates

The **Cartesian plane** has a horizontal x-axis and vertical y-axis meeting at the **origin** (0,0).
Coordinates are written as **(x, y)** – x is horizontal, y is vertical.
Positive x: right of origin; negative x: left. Positive y: above; negative y: below.
Mnemonic: "Along the corridor, up the stairs" (x then y).
Check the scale on axes – 1 square may not equal 1 unit.

Midpoint of a Line

The **midpoint** is the average (mean) of the endpoints' coordinates.
Formula: midpoint = **( (x₁+x₂)/2 , (y₁+y₂)/2 )**.
Example: A(-4,3) and B(8,-12) → midpoint $= (2, -4.5)$ .

Gradient of a Line

**Gradient** measures steepness: **rise/run** (change in y over change in x).
Positive gradient: line goes uphill (bottom-left to top-right).
Negative gradient: line goes downhill (top-left to bottom-right).
Formula: gradient **m = (y₂ - y₁)/(x₂ - x₁)**.
To draw a gradient, e.g. $\frac{2}{3}$ : move 3 right, 2 up. For -5: move 1 right, 5 down.
A gradient of 3 is steeper than 2; -5 is steeper than -4.

Length of a Line

Distance between two points uses **Pythagoras' theorem**.
Formula: **d = √[(x₁ - x₂)² + (y₁ - y₂)²]**.
Be careful with negative coordinates – use brackets to avoid errors.
Example: A(3,-4) and B(-5,2) → $d = √(8^{2} + (-6)^{2}) = \sqrt{100} = 10$ units.

Cartesian Plane with Points

Midpoint Example

Gradient as Rise over Run

Length of a Line (Pythagoras)

Practice questions

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1.What is the midpoint of the line segment joining A(1, 3) and B(5, 8)?
Easy
- A(3, 5.5)
- B(2, 2.5)
- C(6, 11)
- D(4, 5)
2.A line joins A(1, 3) to B(5, 8). Find the midpoint of AB.
Easy
- A(3, 5.5)
- B(2, 2.5)
- C(4, 5)
- D(6, 11)
3.The point A has coordinates (5, −4). The point B has coordinates (13, 1). Work out the coordinates of the midpoint of AB.
Easy
- A(9, -1.5)
- B(8, -3)
- C(18, -3)
- D(9, 2.5)
4.Find the gradient of the straight line with equation $5x + 2y = 7$ .
Easy
- A $-\frac{5}{2}$
- B $\frac{5}{2}$
- C $-\frac{2}{5}$
- D $\frac{2}{5}$
5.Line L has equation $y = 2 - 3x$ . Write down the gradient of line L.
Easy
- A-3
- B3
- C2
- D-2
6.Find the mid-point of AB where $A = (w, r)$ and $B = (3w, t)$ . Give your answer in its simplest form in terms of w, r and t.
Medium
- A(2w, (r+t)/2)
- B(4w, r+t)
- C(2w, r+t)
- D(w, (r+t)/2)
7.A is the point (2, 3) and B is the point (7, −5). Find the coordinates of the midpoint of AB.
Medium
- A(4.5, -1)
- B(9, -2)
- C(4.5, 1)
- D(5, -2)
8.Point A has coordinates (5, 8) and point B has coordinates (9, −4). Work out the gradient of AB.
Medium
- A-3
- B3
- C $-\frac{1}{3}$
- D $\frac{1}{3}$

Summary

## Coordinate Geometry - Quick Revision ### Coordinates - Cartesian plane: x-axis (horizontal), y-axis (vertical), origin (0,0). - Coordinates are (x, y): "along the corridor, up the stairs". ### Midpoint of a Line - Midpoint = average of coordinates: $$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$ ### Gradient of a Line - Gradient = steepness: $\frac{\text{change in }y}{\text{change in }x} = \frac{\text{rise}}{\text{run}}$. - Between two points $(x_1,y_1)$ and $(x_2,y_2)$: $$\frac{y_2-y_1}{x_2-x_1}$$ - Positive gradient: uphill (left to right). Negative: downhill. - To draw gradient $\frac{a}{b}$: move $b$ right, $a$ up (if positive) or down (if negative). ### Length of a Line - Distance between $(x_1,y_1)$ and $(x_2,y_2)$: $$d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$$ - Uses Pythagoras' theorem. Be careful with negative coordinates. ### Key Tips - Check scale on axes. - For gradient, always check sign (uphill/downhill). - For length, square differences, add, then square root.

Key terms

Coordinates: A pair of numbers (x, y) that describe the location of a point on the Cartesian plane, where x is horizontal distance from origin and y is vertical distance.
Midpoint: The point halfway between two points, found by averaging their x and y coordinates: ((x1+x2)/2, (y1+y2)/2).
Gradient: A measure of steepness of a line, calculated as change in y divided by change in x (rise/run). Positive for uphill, negative for downhill.
Length of a line: The distance between two points, calculated using Pythagoras: d = √((x1-x2)² + (y1-y2)²).

Slides

Coordinates

The Cartesian plane has an x-axis (horizontal) and y-axis (vertical).
Coordinates are written as (x, y) – 'along the corridor, up the stairs'.
The origin is (0, 0). Positive x right, negative x left; positive y up, negative y down.

What are the coordinates of the point that is 3 units left and 4 units up from the origin?

(3, 4)
(-3, 4)
(3, -4)
(-3, -4)

Midpoint of a Line

The midpoint is the average of the coordinates of the endpoints.
Formula: midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2).
Example: For A(-4,3) and B(8,-12), midpoint $= (2, -4.5)$ .

The midpoint of (2, 5) and (6, 1) is (____, ____).

Gradient of a Line

Gradient measures steepness: rise/run.
Positive gradient: uphill (bottom left to top right).
Negative gradient: downhill (top left to bottom right).
Formula: (y₂ - y₁)/(x₂ - x₁).

Which of these are true about gradients?

A gradient of 3 means for every 1 unit right, go up 3.
A gradient of -2 is steeper than a gradient of -5.
A line with gradient 0 is horizontal.

Finding Gradient from Two Points

Use formula: gradient = (y₂ - y₁)/(x₂ - x₁).
Example: A(1,3) and B(5,8): gradient $= (8-3)/(5-1) = \frac{5}{4}$ .
Always check sign: uphill positive, downhill negative.

Find the gradient of the line through (2,3) and (7,-5).

$\frac{8}{5}$
$-\frac{8}{5}$
$\frac{5}{8}$
$-\frac{5}{8}$

Length of a Line

Use Pythagoras: distance = √[(x₁ - x₂)² + (y₁ - y₂)²].
Example: A(3,-4), B(-5,2): distance $= √[(3-(-5))^{2} + (-4-2)^{2}] = \sqrt{64+36} = 10$ .
Be careful with negative coordinates – use brackets.

The distance between (1,2) and (4,6) $is ____$ .

Drawing a Line with a Given Gradient

Write gradient as a fraction: rise/run.
Positive: move run right, rise up. Negative: move run right, rise down.
Example: gradient $-2 = -\frac{2}{1}$ : from a point, go 1 right and 2 down.

Which statements about drawing lines are true?

To draw gradient $\frac{2}{3}$ , go 3 right and 2 up.
To draw gradient -5, go 1 right and 5 down.
A line with gradient 0 is vertical.

Midpoint with Variables

Midpoint formula works with algebraic coordinates.
Example: A(w, r) and B(3w, t): midpoint $= ((w+3w)/2, (r+t)/2) = (2w, (r+t)/2)$ .

Find the midpoint of A(2,5) and B(4,p).

(3, (5+p)/2)
(6, 5+p)
(3, 5+p)
((5+p)/2, 3)

Gradient from Equation

Rearrange equation to $y = mx +$ c; m is the gradient.
Example: $5x + 2y = 7$ → $2y = 7 - 5x$ → $y = 3.5 - 2.5x$ , gradient $= -2.5$ .
Example: $y = 2 - 3x$ , gradient $= -3$ .

The gradient of the line $2y + 7x = 10$ $is ____$ .

Parallel and Perpendicular Lines

Parallel lines have equal gradients.
Perpendicular lines have gradients that multiply to -1 (m₁ × m₂ $= -1)$ .
Example: gradient 2 is perpendicular to gradient -1/2.

Which statements are true?

Lines with gradients 3 and 3 are parallel.
Lines with gradients 2 and -2 are perpendicular.
Lines with gradients 4 and -0.25 are perpendicular.

Section Formula (Extension)

If point N divides PQ in ratio m:n, coordinates are weighted averages.
Example: $P(16,9), Q(22,24), PN:NQ = 2:1$ → $N = ((1\times 16+2\times 22)/3, (1\times 9+2\times 24)/3) = (20, 19)$ .

Given A(2,5), B(4,p), C(q,17) on a line with $AC = 4AB$ , find p and q.

$p=7, q=10$
$p=8, q=10$
$p=7, q=12$
$p=8, q=12$

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