Straight-Line Graphs

Straight-line graphs represent linear relationships. This lesson covers the equation y = mx + c, finding gradients and intercepts, plotting lines, parallel and perpendicular lines, and finding the equation of a line from its graph or from given information.

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Key Vocabulary

Term	Meaning
Gradient (m)	The steepness of a line — $rise \div run$rise÷run
y-intercept (c)	Where the line crosses the y-axis
x-intercept	Where the line crosses the x-axis (set y = 0)
y = mx + c	The equation of a straight line in gradient-intercept form
Parallel lines	Lines with the same gradient (they never meet)
Perpendicular lines	Lines that meet at right angles; their gradients multiply to −1

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The Equation y = mx + c

Every straight line can be written in the form y = mx + c, where:

• m is the gradient (slope)
• c is the y-intercept (the value of y when x = 0)

Worked Example 1

For the line y = 3x − 5:

• Gradient = 3
• y-intercept = −5 (the line crosses the y-axis at (0, −5))

Worked Example 2

Rearrange 2x + 3y = 12 into the form y = mx + c.

3y = −2x + 12

y = −(2/3)x + 4

Gradient = −2/3, y-intercept = 4

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Finding the Gradient

From Two Points

Gradient $= (y_{2} - y_{1}) / (x_{2} - x_{1})$=(y2​−y1​)/(x2​−x1​)

Worked Example 3

Find the gradient of the line through (2, 3) and (6, 11).

m = (11 − 3) / (6 − 2) = 8/4 = 2

Worked Example 4

Find the gradient of the line through (−1, 5) and (3, −3).

m = (−3 − 5) / (3 − (−1)) = −8/4 = −2

From a Graph

Pick two points where the line passes through grid intersections. Calculate rise/run.

• A positive gradient slopes upward from left to right (/).
• A negative gradient slopes downward from left to right ().
• A horizontal line has gradient 0.
• A vertical line has an undefined gradient.

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Plotting a Straight Line

Method 1: Table of Values

Choose at least 3 x-values, calculate y, plot the points, join with a ruler.

Worked Example 5

Plot y = 2x − 1.

x	−1	0	1	2	3
y	−3	−1	1	3	5

Plot these points and draw a straight line through them.

Method 2: Gradient and y-intercept

1. Plot the y-intercept.
2. Use the gradient to find another point (gradient = rise/run).
3. Draw the line.

Worked Example 6

Plot y = −(1/2)x + 3.

• y-intercept = 3. Plot (0, 3).
• Gradient = −1/2. From (0, 3), go right 2 and down 1 to reach (2, 2). Plot this point.
• Draw the line through (0, 3) and (2, 2).

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Finding the Equation of a Line

From the Gradient and y-intercept

Substitute m and c into y = mx + c.

From the Gradient and a Point

Use $y - y_{1} = m(x - x_{1})$y−y1​=m(x−x1​), then rearrange to y = mx + c.

Worked Example 7

Find the equation of the line with gradient 4 passing through (2, 11).

y − 11 = 4(x − 2)

y − 11 = 4x − 8

y = 4x + 3

Answer: y = 4x + 3

From Two Points

1. Find the gradient.
2. Substitute one point into $y - y_{1} = m(x - x_{1})$y−y1​=m(x−x1​).

Worked Example 8

Find the equation of the line through (1, 2) and (4, 11).

m = (11 − 2)/(4 − 1) = 9/3 = 3

y − 2 = 3(x − 1)

y = 3x − 1

Answer: y = 3x − 1

Check with (4, 11): 3(4) − 1 = 11 ✓

Finding the Equation of a Line — Decision Flow

flowchart TD
    START[Find equation of a line] --> WHAT{What is given?}
    WHAT -->|Gradient + y-intercept| MC[Substitute into y = mx + c]
    WHAT -->|Gradient + one point| PG[Use y minus y1 = m times x minus x1]
    WHAT -->|Two points| TP[Step 1: m = y2 minus y1 over x2 minus x1]
    WHAT -->|Parallel to known line| PAR[Same gradient as known line]
    WHAT -->|Perpendicular to known line| PERP[Gradient = minus 1 over m]
    TP --> PG
    PAR --> PG
    PERP --> PG
    MC --> FORM[Rearrange to required form]
    PG --> FORM
    FORM --> CHECK[Verify by substituting given point]

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Parallel and Perpendicular Lines

Parallel Lines

Two lines are parallel if they have the same gradient.

y = 3x + 1 and y = 3x − 7 are parallel (both have m = 3).

Perpendicular Lines

Two lines are perpendicular if the product of their gradients is −1.

If one line has gradient m, the perpendicular gradient is −1/m.

Worked Example 9

Line L has equation y = 2x + 5. Find the equation of the line perpendicular to L passing through (4, 1).

Gradient of L = 2. Perpendicular gradient = −1/2.

y − 1 = −(1/2)(x − 4)

y − 1 = −(1/2)x + 2

y = −(1/2)x + 3

Answer: y = −(1/2)x + 3

Worked Example 10

Show that the lines 3x + 4y = 12 and 4x − 3y = 6 are perpendicular.

Rearrange: y = −(3/4)x + 3, so $m_{1} = -3/4$m1​=−3/4.

Rearrange: y = (4/3)x − 2, so $m_{2} = 4/3$m2​=4/3.

Product: $(-3/4) \times (4/3) = -12/12 =$(−3/4)×(4/3)=−12/12= −1 ✓

Therefore the lines are perpendicular. ∎

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Special Lines

Equation	Description
y = c (e.g. y = 4)	Horizontal line through (0, c); gradient = 0
x = a (e.g. x = −2)	Vertical line through (a, 0); gradient undefined
y = x	Diagonal through the origin at $45°$45°; gradient = 1
y = −x	Diagonal through the origin; gradient = −1

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The Midpoint and Length of a Line Segment

Midpoint

Midpoint $= ((x_{1} + x_{2})/2$=((x1​+x2​)/2, $(y_{1} + y_{2})/2)$(y1​+y2​)/2)

Worked Example 11

Find the midpoint of (2, 3) and (8, 7).

Midpoint = ((2 + 8)/2, (3 + 7)/2) = (5, 5)

Length of a Line Segment [H]

Using Pythagoras: length $= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$=(x2​−x1​)2+(y2​−y1​)2​

Worked Example 12

Find the distance between (1, 2) and (4, 6).

$= \sqrt{(4 - 1)^{2} + (6 - 2)^{2}} = \sqrt{9 + 16} = \sqrt{25} =$=(4−1)2+(6−2)2​=9+16​=25​= 5

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Common Mistakes and Misconceptions