Linear Graphs

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Notes

Equations of Straight Lines (y = mx + c)

The general equation is **y $= mx +$ c**, where **m** is the gradient and **c** is the y-intercept.
Gradient = rise/run; positive for uphill, negative for downhill.
To find the equation from a graph: find gradient using a triangle, read off y-intercept, substitute into $y = mx + c$ .
If y-intercept is not visible, substitute a known point into $y = mx + c$ and solve for c.
Horizontal lines: **y = c**; vertical lines: **x = k**.
Rearrange equations like $ax + by = c$ into $y = mx + c$ to identify gradient and intercept.

Drawing Straight Line Graphs

Create a table of values for x and y, plot points, and join with a straight line.
Use at least 3 points to ensure accuracy.
Alternatively, start at y-intercept (c) and move up m units for every 1 unit right (down if m negative).
For equations in form $ax + by = c$ , find $x-intercept (set y=0)$ and $y-intercept (set x=0)$ , then plot.
If gradient is a fraction a/b, go a units up for every b units right.

Parallel Lines

Parallel lines have the **same gradient**.
Equation of a line parallel to $y = mx + c$ is $y = mx + d (different y-intercept)$ .
To find a parallel line through a given point: use same gradient, substitute point to find d.

Perpendicular Lines

Perpendicular lines meet at right angles (90°).
Gradients m₁ and m₂ satisfy **m₁ × m₂ = -1** (negative reciprocals).
To find a perpendicular gradient: m₂ = -1/m₁.
To find equation of perpendicular line: find negative reciprocal of given gradient, substitute point to find c.
A **perpendicular bisector** cuts a segment in half at right angles; passes through midpoint.

Gradient and y-intercept

Parallel and Perpendicular Lines

Finding Equation from Graph

Practice questions

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1.A straight line, l, has equation $y = 5x + 12$ . What is the gradient of line l?
Easy
- A5
- B12
- C-5
- D-12
2.The line $y = 3x - 2$ crosses the y-axis at G. What are the coordinates of G?
Easy
- A(0, -2)
- B(0, 2)
- C(-2, 0)
- D(2, 0)
3.Find the coordinates of the point where the line $y = 3x - 8$ crosses the y-axis.
Easy
- A(0, -8)
- B(0, 8)
- C(-8, 0)
- D(8, 0)
4.Write down the gradient of the line $y = 3x - 8$ .
Easy
- A3
- B-8
- C-3
- D8
5.Find the gradient of a line that is perpendicular to $8y + 4x = 5$ .
Medium
- A2
- B-2
- C $\frac{1}{2}$
- D $-\frac{1}{2}$
6.The equation of line L is $3x - 8y + 20 = 0$ . Find the gradient of line L.
Medium
- A $\frac{3}{8}$
- B $-\frac{3}{8}$
- C $\frac{8}{3}$
- D $-\frac{8}{3}$
7.Line L passes through the points (0, -3) and (6, 9). Find the equation of line L.
Medium
- A $y = 2x - 3$
- B $y = 2x + 3$
- C $y = -2x - 3$
- D $y = -2x + 3$
8.Find the gradient of the line that is perpendicular to the line $2y = 3 + 5x$ .
Medium
- A $-\frac{2}{5}$
- B $\frac{2}{5}$
- C $-\frac{5}{2}$
- D $\frac{5}{2}$

Summary

## Linear Graphs - Quick Revision ### Equation of a Straight Line - **General form**: $y = mx + c$ - $m$ = gradient (steepness, positive uphill, negative downhill) - $c$ = y-intercept (where line crosses y-axis) - **Finding equation from graph**: 1. Find gradient using $m = \frac{\text{rise}}{\text{run}}$ (draw a triangle) 2. Read off y-intercept $c$ 3. Substitute into $y = mx + c$ - **If y-intercept not visible**: substitute a point $(x,y)$ into $y = mx + c$ and solve for $c$ - **Horizontal line**: $y = c$ - **Vertical line**: $x = k$ ### Drawing Straight Line Graphs - **Using table of values**: choose x-values (e.g. 0, 5, 10), compute y, plot points, join with ruler - **Using gradient and intercept**: start at y-intercept, move 1 unit right and m units up (or down if negative) - **If equation not in $y = mx + c$**: rearrange first, e.g. $ax + by = c$ → $y = -\frac{a}{b}x + \frac{c}{b}$ - **Alternative for $ax + by = c$**: find x-intercept (y=0) and y-intercept (x=0), plot and join ### Parallel Lines - **Same gradient** ($m_1 = m_2$) - Equation parallel to $y = mx + c$ is $y = mx + d$ (different y-intercept) - To find parallel line through a point: use same gradient, substitute point to find $d$ ### Perpendicular Lines - **Gradients multiply to -1**: $m_1 \times m_2 = -1$ - Gradients are **negative reciprocals**: $m_2 = -\frac{1}{m_1}$ - To find perpendicular line through a point: 1. Find gradient of given line 2. Compute negative reciprocal 3. Substitute point into $y = mx + c$ to find $c$ - **Perpendicular bisector**: passes through midpoint of segment, gradient is negative reciprocal of segment's gradient ### Key Tips - Always plot at least 3 points when drawing - Check if lines are parallel/perpendicular by comparing gradients after rearranging to $y = mx + c$ - For perpendicular lines, ensure you use the negative reciprocal correctly

Key terms

Gradient (m): The steepness of a line, calculated as rise/run. Positive for uphill, negative for downhill.
y-intercept (c): The y-coordinate where the line crosses the y-axis (x=0).
Equation of a straight line: y = mx + c, where m is the gradient and c is the y-intercept.
Parallel lines: Lines with the same gradient that never intersect.
Perpendicular lines: Lines that meet at right angles (90°); their gradients multiply to -1.
Negative reciprocal: For a gradient m, the negative reciprocal is -1/m. Used to find perpendicular gradients.
Perpendicular bisector: A line that cuts a segment into two equal parts at a right angle; passes through the midpoint and has gradient negative reciprocal of the segment.

Slides

Equation of a Straight Line

The general equation is $y = mx + c$ .
m is the gradient (steepness).
c is the y-intercept (where the line crosses the y-axis).
Example: $y = 5x + 2$ has gradient 5 and y-intercept 2.

What is the gradient of the line $y = 5x + 12$ ?

5
12
-5
-12

Finding the y-intercept

The y-intercept is the point where the line crosses the $y-axis (x=0)$ .
For $y = mx + c$ , the y-intercept is (0, c).
Example: $y = 3x - 2$ crosses the y-axis at (0, -2).

The line $y = 3x - 8$ crosses the y-axis at $(0, ____)$ .

Finding the Equation from a Graph

Find the gradient using rise/run between two points.
Read off the y-intercept from the graph.
Substitute into $y = mx + c$ .
If y-intercept is not visible, substitute a point to solve for c.

A line passes through (2, 15) with gradient 6. What is its equation?

$y = 6x + 3$
$y = 6x + 15$
$y = 2x + 6$
$y = 6x - 3$

Drawing Straight Line Graphs

Use a table of values with at least 3 points.
Plot points and join with a straight line.
Alternatively, start at y-intercept and use gradient to find other points.
If equation is not in $y=mx+c$ , rearrange first.

Which of these are true about drawing straight lines?

You should plot at least 3 points.
The gradient tells you how many units up for 1 unit right.
You never need to rearrange the equation.

Parallel Lines

Parallel lines have the same gradient.
They never meet.
Example: $y = 2x + 1$ and $y = 2x + 5$ are parallel.
To find a parallel line through a point, use the same gradient and solve for c.

The line parallel to $y = 3x + 7$ passing through (2, 1) has equation $y = 3x + ____$ .

Perpendicular Lines

Perpendicular lines meet at right angles (90°).
Gradients m₁ and m₂ are perpendicular if m₁ × m₂ $= -1$ .
They are negative reciprocals: m₂ = -1/m₁.
Example: gradients $\frac{1}{3}$ and -3 are perpendicular.

Which gradient is perpendicular to 2?

$-\frac{1}{2}$
$\frac{1}{2}$
-2
2

Finding a Perpendicular Line

Find the gradient of the given line.
Take the negative reciprocal for the perpendicular gradient.
Use $y = mx + c$ with the new gradient.
Substitute the given point to find c.

A line perpendicular to $y = 2x - 3$ through (0, 2) has equation $y = ____ x + 2$ .

Perpendicular Bisector

A perpendicular bisector cuts a line segment in half at a right angle.
Find the midpoint of the segment.
Find the gradient of the segment, then its negative reciprocal.
Write the equation using the midpoint.

The perpendicular bisector of (4,-6) and (8,6) passes through which point?

(6,0)
(12,0)
(4,6)
(0,0)

Horizontal and Vertical Lines

Horizontal line: $y = c (constant y-value)$ .
Vertical line: $x = k (constant x-value)$ .
Example: $y = 4$ is horizontal; $x = -2$ is vertical.

Which of these are true?

A horizontal line has gradient 0.
A vertical line has equation $y = k$ .
The line $x = 3$ is vertical.

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