Year 8AC9M8A02

Straight Line Graphs

Graph linear relations on the Cartesian plane; solve linear equations and one-variable inequalities using graphical and algebraic techniques; verify solutions by substitution.

You should know.

You should know

Standard form

y = mx + c

$m$ is the GRADIENT. $c$ is the $y$ INTERCEPT.

Gradient from two points

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Rise over run.

Sign of the gradient

m > 0 \ \text{rises} \qquad m < 0 \ \text{falls}

Read the line left to right.

Parallel lines

m_{1} = m_{2}

Same gradient.

Perpendicular lines

m_{1} \times m_{2} = -1

Gradients multiply to $-1$.

Worked example

e.g.

y = 3x - 4

Read off the gradient. It is the number in front of $x$ $m = 3$
Read off the $y$ intercept. It is the number on its own $c = -4$
The gradient is positive, so the line rises $m > 0$
The line crosses the $y$ axis at $-4$ $(0; -4)$

Practice

Write down the gradient and the $y$ intercept:

y = 2x + 5

Answer

m = 2,\ c = 5

y = -4x + 1

Answer

m = -4,\ c = 1

y = \tfrac{1}{2}x - 3

Answer

m = \tfrac{1}{2},\ c = -3

y = 7 - 2x

Answer

m = -2,\ c = 7

2y = 6x + 4

Answer

m = 3,\ c = 2

y = x

Answer

m = 1,\ c = 0

3y = 9x - 12

Answer

m = 3,\ c = -4

y + 2x = 5

Answer

m = -2,\ c = 5

\text{Is } y = 2x+1 \text{ parallel to } y = 2x-7?

Answer

\text{Yes, both have } m = 2

\text{Is } y = 3x \text{ perpendicular to } y = -\tfrac{1}{3}x?

Answer

\text{Yes, } 3 \times -\tfrac{1}{3} = -1