Straight Lines and Coordinate Geometry

In short: A straight line has the equation $y = m x + c$ , where $m$ is the gradient (rise over run) and $c$ is where it crosses the y-axis. Parallel lines have the same gradient, perpendicular lines have gradients that multiply to $- 1$ , and the distance between two points comes from Pythagoras.

Starting point: Years 9 and 10

Gradient

The gradient $m$ tells you how steep a line is. It is the rise divided by the run.

$m = \frac{rise}{run} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Positive gradient: the line goes up to the right.
Negative gradient: the line goes down to the right.
Horizontal line: $m = 0$ . Vertical line: gradient is undefined.

The equation of a line

$y = m x + c$

$m$ is the gradient and $c$ is the y-intercept, where the line crosses the y-axis. In the taxi story $m = 2$ and $c = 4$ .

Start at the intercept, then go across 1 and up 2.

Distance and midpoint

For points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$

The distance $d$ is how far apart the two points are: the length of the straight line from one to the other. The midpoint $M$ is the point exactly halfway along that line.

The distance formula is just Pythagoras on the grid. Draw a right-angled triangle under the line. Its run is the change in $x$ , its rise is the change in $y$ , and the line between the points is the longest side.

The distance is the long side of the right-angled triangle: √(6² + 8²) = 10.

Parallel and perpendicular lines

To get the perpendicular gradient, flip the fraction and change the sign. Then check that the two gradients multiply to $- 1$ .

General form

Lines are also written as $a x + b y + c = 0$ . To find the gradient, rearrange into $y = m x + c$ . For $2 x + 3 y - 6 = 0$ : $3 y = - 2 x + 6$ , so $y = - \frac{2}{3} x + 2$ .

Straight lines

Gradient

Rise over run
Positive, negative, zero

Equation

y = mx + c
ax + by + c = 0

Two points

Distance: Pythagoras
Midpoint: average

Pairs of lines

Parallel: same m
Perpendicular: product is -1

Models

Start value and rate

Year 11Point-gradient form

When you know the gradient $m$ and one point $(x_{1}, y_{1})$ , use

$y - y_{1} = m (x - x_{1})$

Example. Find the line through $(2, - 1)$ perpendicular to $y = \frac{1}{2} x + 3$ .

The new gradient is $- 2$ . So $y + 1 = - 2 (x - 2)$ , which gives $y = - 2 x + 3$ .

Intersection of lines. Solve the two equations simultaneously. The answer is the point where they cross.

Perpendicular bisector. The perpendicular bisector of $A B$ is the line that cuts $A B$ in half ("bisects" it) and crosses it at a right angle ("perpendicular"). So it passes through the midpoint of $A B$ , with the perpendicular gradient $- \frac{1}{m _{A B}}$ . Every point on it is the same distance from $A$ and $B$ .

For $A (0, 1)$ and $B (4, 3)$ :

Midpoint: $(\frac{0 + 4}{2}, \frac{1 + 3}{2}) = (2, 2)$ .
Gradient of $A B$ : $\frac{3 - 1}{4 - 0} = \frac{1}{2}$ , so the bisector's gradient is $- 2$ .
Line through $(2, 2)$ with gradient $- 2$ : $y - 2 = - 2 (x - 2)$ , that is $y = - 2 x + 6$ .

Same distance from two points. To find the point $(0, k)$ on the $y$ -axis equally far from $A$ and $B$ , set the squared distances equal and solve for $k$ (or find where the perpendicular bisector crosses the $y$ -axis).

For the same $A (0, 1)$ and $B (4, 3)$ :

From $(0, k)$ to $A$ : $d^{2} = (0 - 0)^{2} + (k - 1)^{2} = k^{2} - 2 k + 1$ .
From $(0, k)$ to $B$ : $d^{2} = (0 - 4)^{2} + (k - 3)^{2} = k^{2} - 6 k + 25$ .
Set them equal: $k^{2} - 2 k + 1 = k^{2} - 6 k + 25$ , so $4 k = 24$ and $k = 6$ .

The point is $(0, 6)$ , which is 5 units from both $A$ and $B$ . It is also where $y = - 2 x + 6$ crosses the $y$ -axis, as it should be.

The bisector meets AB at its midpoint, at a right angle. The two marks show the halves are equal. The point (0, 6) on it is 5 units from both A and B.

Fourth corner of a parallelogram. The diagonals of a parallelogram bisect each other, so the midpoint of $A C$ equals the midpoint of $B D$ . Use it to find a missing corner $D$ .

For $A (1, 1)$ , $B (5, 2)$ and $C (6, 5)$ :

Midpoint of $A C$ : $(\frac{1 + 6}{2}, \frac{1 + 5}{2}) = (3.5, 3)$ .
This is also the midpoint of $B D$ . With $D (x, y)$ : $\frac{5 + x}{2} = 3.5$ and $\frac{2 + y}{2} = 3$ .
So $x = 2$ and $y = 4$ : the fourth corner is $D (2, 4)$ .

Both diagonals have the same midpoint M. Knowing M and B gives D.

Areas on the grid. If two sides of a triangle are perpendicular ( $m_{1} m_{2} = - 1$ ), use them as base and height: area $= \frac{1}{2} \times$ base $\times$ height, with lengths from the distance formula. A line with intercepts $a$ and $b$ makes a right-angled triangle with the axes, of area $\frac{1}{2} ∣ a ∣∣ b ∣$ .

For $P (1, 2)$ , $Q (3, 6)$ and $R (5, 0)$ :

Gradients: $m_{P Q} = \frac{6 - 2}{3 - 1} = 2$ and $m_{P R} = \frac{0 - 2}{5 - 1} = - \frac{1}{2}$ . Their product is $- 1$ , so the corner at $P$ is a right angle.
Lengths: $P Q = 2^{2} + 4^{2} = 20$ and $P R = 4^{2} + 2^{2} = 20$ .
Area $= \frac{1}{2} \times 20 \times 20 = \frac{1}{2} \times 20 = 10$ square units.

For the line $2 x + 3 y = 12$ : the $x$ -intercept is $6$ (set $y = 0$ ) and the $y$ -intercept is $4$ (set $x = 0$ ). The triangle it makes with the axes has area $\frac{1}{2} \times 6 \times 4 = 12$ square units.

PQ and PR meet at a right angle, so one is the base and the other is the height.

Year 12Linear models

A linear model has a fixed start and a constant rate of change. Write it as $y = m x + c$ where $m$ is the rate.

A phone plan costs 15 dollars a month plus 0.10 dollars per text. Cost: $C = 0.1 t + 15$ . Break-even questions are solved by setting two models equal. Plan B costs $C = 0.05 t + 25$ . Equal when $0.1 t + 15 = 0.05 t + 25$ , so $t = 200$ texts. Above 200 texts, plan B is cheaper.

In Year 12 calculus, the rate $m$ becomes the derivative. A straight line is the only graph whose rate never changes.

A model from two data points. A town had 50 000 people in 2010 and 44 000 in 2016. The rate is $\frac{44 000 - 50 000}{6} = - 1000$ people a year. Measuring $t$ in years after 2010, $P = 50000 - 1000 t$ , so the prediction for 2025 ( $t = 15$ ) is 35 000. Predictions far outside the data are less reliable.

Profit. Profit is income minus cost. With income $I = 40 x$ and cost $C = 15 x + 500$ , profit is $25 x - 500$ , which is positive when $x > 20$ . So at least 21 items must be sold to make a profit.

Quick summary

Gradient $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ : rise over run.
$y = m x + c$ : $m$ is the gradient, $c$ is the y-intercept.
Distance uses Pythagoras. The midpoint is the average of the coordinates.
Parallel lines: same gradient. Perpendicular lines: $m_{1} m_{2} = - 1$ .
Point-gradient form: $y - y_{1} = m (x - x_{1})$ .

Quiz

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What is the gradient of the line through $(2, 3)$ and $(6, 11)$ ?
1. A $\frac{1}{2}$
2. B 2
3. C 4
4. D 8
Show answer
B: 2
$m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$ .
What is the y-intercept of $y = - 3 x + 7$ ?
1. A $- 3$
2. B 7
3. C $\frac{7}{3}$
4. D 0
Show answer
B: 7
In $y = m x + c$ , $c$ is the y-intercept.
A line goes down from left to right. Its gradient is:
1. A positive
2. B negative
3. C zero
4. D undefined
Show answer
B: negative
Falling to the right means a negative gradient.

Straight Lines and Coordinate Geometry

Starting point: Years 9 and 10

Gradient

The equation of a line

Distance and midpoint

Parallel and perpendicular lines

General form

Quick summary

Quiz

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