What is the general equation of a straight line in Pure Mathematics 1 for Edexcel International A Levels?

In Pure Mathematics 1 for Edexcel International A Levels, the general equation of a straight line is given by y = mx + c, where 'm' represents the gradient or slope of the line, and 'c' is the y-intercept, which is the point where the line crosses the y-axis.

How do you find the gradient of a straight line graph?

The gradient of a straight line graph can be found using the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two distinct points on the line. The gradient represents the rate of change of y with respect to x.

What is the significance of the y-intercept in the equation of a straight line?

The y-intercept in the equation of a straight line, represented by 'c' in y = mx + c, indicates the point where the line crosses the y-axis. It is significant because it provides a starting point for graphing the line and helps in understanding the line's position relative to the origin.

How can you determine if two straight lines are parallel or perpendicular?

Two straight lines are parallel if they have the same gradient, meaning their slopes are equal (m1 = m2). They are perpendicular if the product of their gradients is -1 (m1 * m2 = -1). This is a key concept in Pure Mathematics 1 for Edexcel International A Levels.

How do you find the equation of a straight line given two points?

To find the equation of a straight line given two points, first calculate the gradient using m = (y2 - y1) / (x2 - x1). Then, use one of the points (x1, y1) and the gradient in the point-slope form of the equation: y - y1 = m(x - x1). Rearrange this to the form y = mx + c to find the equation of the line.

Why is "Subtracting in the wrong order: $(y_{1} - y_{2})/(x_{2} - x_{1})$" a common mistake in Straight line graphs?

Why it happens: Mismatching subscripts in numerator and denominator. How to avoid it: Always subtract IN THE SAME ORDER for both: (y_2 - y_1)/(x_2 - x_1) or (y_1 - y_2)/(x_1 - x_2). If the subscripts match, the sign is right.

Why is "Forgetting the negative when taking perpendicular gradient" a common mistake in Straight line graphs?

Why it happens: Just flipping the fraction. How to avoid it: Perpendicular gradient is the NEGATIVE RECIPROCAL. Both flip AND change sign. m_1 = 3 m_ = -1/3. Verify: 3 × (-1/3) = -1. ✓ Source: Persistent issue in WMA11/01 examiner reports

Why is "Leaving the answer as $y = -\tfrac{2}{3}x + \tfrac{5}{3}$ when integer form is demanded" a common mistake in Straight line graphs?

Why it happens: Not reading the instruction. How to avoid it: If 'in the form ax + by + c = 0 where a, b, c are integers' is stated, MULTIPLY through to clear fractions: 3y = -2x + 5 2x + 3y - 5 = 0.

Why is "Computing distance as $(x_{2} - x_{1}) + (y_{2} - y_{1})$ (i.e. forgetting the square root)" a common mistake in Straight line graphs?

Why it happens: Confusing distance with midpoint or change-of-coords. How to avoid it: Distance comes from Pythagoras. ALWAYS: square both coordinate differences, sum, then take ROOT.

Why is "'Showing parallel' by computing gradients but not stating the conclusion" a common mistake in Straight line graphs?

Why it happens: Forgetting the verbal conclusion. How to avoid it: 'Show that' demands a clear conclusion. After computing m_1 = m_2 = 2/3, write: 'Since both gradients are equal and the y-intercepts differ, the lines are parallel.' Source: WMA11/01 examiner reports

Pure Mathematics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Straight line graphs

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Straight Line Graphs Study Notes — Edexcel IAL Mathematics P1 (WMA11/01, 2018 spec — 2026 onwards)

Gradients, intercepts, three forms of the line equation, parallel/perpendicular conditions, distance, midpoint, and linear modelling. Foundation for coordinate geometry across the IAL.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

5.1 — Understand and use the equation of a straight line, including the forms $y = mx + c$ , $y - y_{1} = m(x - x_{1})$ , and $ax + by + c = 0$ .
5.2 — Use gradient conditions for parallel and perpendicular lines.
5.3 — Find the length and midpoint of a line segment.
5.4 — Model real-world situations using linear equations.

Three forms of the equation of a line

Pick the form that matches what you have and what you want.

Form 1: gradient-intercept $y = mx + c$ .

$m$ = gradient (slope).
$c$ = $y$ -intercept (where the line crosses the $y$ -axis).
Use when you know $m$ and $c$ , or when the question wants 'gradient-intercept form'.

Form 2: point-gradient $y - y_{1} = m(x - x_{1})$ .

$(x_{1}, y_{1})$ = a known point on the line.
$m$ = gradient.
Use when you know one point and the gradient. This is the most useful form for most P1 problems.

Form 3: general $ax + by + c = 0$ (with $a, b, c$ integers and $a$ positive).

Use when the question demands 'integers' form, or when the line is vertical ( $b = 0$ , equation $x = k$ ).
Multiply through to clear fractions; rearrange so all terms are on one side.

Conversion example. Through $(1, -3)$ with $m = -2/3$ :

Point-gradient: $y + 3 = -\tfrac{2}{3}(x - 1)$ .
Gradient-intercept: $y = -\tfrac{2}{3}x + \tfrac{2}{3} - 3 = -\tfrac{2}{3}x - \tfrac{7}{3}$ .
General (multiply by 3): $3y = -2x - 7 \Rightarrow 2x + 3y + 7 = 0$ .

Special lines.

Line	Equation	Gradient
Horizontal through $(0, c)$	$y = c$	$m = 0$
Vertical through $(a, 0)$	$x = a$	Undefined (infinite)
Passes through origin	$y = mx$	Any $m$

Edexcel tip. When the question says 'integer form', clear all fractions first. Mark schemes typically credit any algebraically correct form unless integer form is explicitly demanded — but writing it as integers shows polish.

Gradient m = rise/run = Δy/Δx, with y-intercept c marking where the line meets the y-axis.

$y = mx + c$ for gradient-intercept.
$y - y_{1} = m(x - x_{1})$ for point-gradient.
$ax + by + c = 0$ for integer form.
Vertical lines are $x = a$ ; horizontal are $y = c$ .

Parallel and perpendicular gradients

Parallel: same gradient. Perpendicular: negative reciprocal.

Parallel lines. Same gradient, different $y$ -intercepts.

$m_{1} = m_{2}$ .
Geometrically: never meet, equidistant everywhere.

Perpendicular lines. Negative reciprocal gradients.

$m_{1} \cdot m_{2} = -1$ .
Equivalent: $m_{2} = -1/m_{1}$ .
Geometrically: meet at $90°$ .

Why $m_{1} m_{2} = -1$ ? Take a line $y = mx$ rotated by $90°$ . Original direction vector: $(1, m)$ . Rotated by $90°$ : $(-m, 1)$ . New gradient: $1/(-m) = -1/m$ . So $m \cdot (-1/m) = -1$ . ✓

Cheat sheet for finding perpendicular gradients:

If $m_{1} = \ldots$	Then $m_{\perp} = \ldots$
$3$	$-1/3$
$-5$	$1/5$
$2/3$	$-3/2$
$-7/4$	$4/7$
$0$ (horizontal)	undefined (vertical)
undefined (vertical)	$0$ (horizontal)

Worked example. Line $L_{1}$ has equation $3x - 2y + 5 = 0$ . Find an equation of $L_{2}$ perpendicular to $L_{1}$ through $(1, -3)$ .

Rearrange $L_{1}$ : $y = \tfrac{3}{2}x + \tfrac{5}{2}$ . So $m_{1} = 3/2$ .
Perpendicular gradient: $m_{2} = -2/3$ .
Use point-gradient with $(1, -3)$ : $y + 3 = -\tfrac{2}{3}(x - 1)$ .
Multiply by 3: $3y + 9 = -2x + 2 \Rightarrow 2x + 3y + 7 = 0$ .

Edexcel tip. Mark scheme awards 1 mark just for stating $m_{1} \cdot m_{2} = -1$ or equivalent. Always state the rule before applying it.

Parallel: same gradient.
Perpendicular: $m_{1} m_{2} = -1$ .
Perpendicular gradient = negative reciprocal.
Horizontal ⊥ vertical.

Distance and midpoint formulae

Distance from Pythagoras. Midpoint from averages.

Distance between two points. Pythagoras' theorem.

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

The $x$ -change and $y$ -change form the two legs of a right triangle; the line segment is the hypotenuse.

Example. $P(-1, 4)$ , $Q(5, -4)$ .

$\Delta x = 5 - (-1) = 6$ .
$\Delta y = -4 - 4 = -8$ .
$d = \sqrt{36 + 64} = \sqrt{100} = 10$ .

Midpoint. Average of the two coordinate pairs.

$M = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)$

For the same $P, Q$ : $M = ((-1 + 5)/2, (4 - 4)/2) = (2, 0)$ .

Perpendicular bisector. A line that:

passes through the midpoint of $PQ$ .
is perpendicular to $PQ$ .

Useful for circle centres (P2/P3) and locus questions.

Example. Perpendicular bisector of $PQ$ where $P(-1, 4)$ , $Q(5, -4)$ .

Midpoint: $(2, 0)$ .
Gradient of $PQ$ : $(-4 - 4)/(5 - (-1)) = -8/6 = -4/3$ .
Perpendicular gradient: $3/4$ .
Equation: $y - 0 = \tfrac{3}{4}(x - 2) \Rightarrow 4y = 3x - 6 \Rightarrow 3x - 4y - 6 = 0$ .

Distance from a point to a line (extension). Beyond P1, but useful: the perpendicular distance from $(x_{0}, y_{0})$ to $ax + by + c = 0$ is $|ax_{0} + by_{0} + c|/\sqrt{a^{2} + b^{2}}$ .

Distance: $\sqrt{\Delta x^{2} + \Delta y^{2}}$ .
Midpoint: average each coordinate.
Perpendicular bisector: through midpoint, perpendicular gradient.

Modelling with linear equations

Gradient = rate of change. Intercept = fixed/starting value.

Linear models. Real-world relationships modelled as $y = mx + c$ .

Interpretation.

$m$ (gradient): the rate of change of $y$ with respect to $x$ .
$c$ ( $y$ -intercept): the value of $y$ when $x = 0$ — often a fixed cost, starting amount, or initial value.

Example. A taxi charges a booking fee plus per-km cost. $C = mn + c$ where $n$ is km, $C$ is total cost. A 5 km journey costs $11; a 12 km journey costs $22.50.

Find $m, c$ : two points $(5, 11), (12, 22.50)$ . Gradient $m = (22.50 - 11)/(12 - 5) = 11.50/7 \approx 1.64$ .
$11 = 5(1.64) + c \Rightarrow c \approx 2.79$ .
Interpretation: $m \approx \$ 1.64 $per km (variable rate);$ c \approx $2.79$ fixed booking fee.

Limitations of linear models. Real systems are rarely linear over a wide range. The model:

assumes constant rate (no surge pricing, no quantity discounts).
breaks down outside the data range (e.g. for $n < 0$ , predicts negative distance).

Edexcel tip. Linear modelling questions usually have a verbal interpretation mark — make sure to ANSWER IT explicitly. Saying 'the gradient is the cost per km' or 'the intercept is the fixed booking fee' is enough.

Other common models.

Context	$m$ means	$c$ means
Taxi: cost vs distance	Cost per km	Booking fee
Plant height vs time	Growth rate per day	Initial height
Temperature vs altitude	°C drop per metre	Sea-level temperature
Spring extension vs load	Stiffness factor	Natural length (zero if calibrated)

$m$ = rate of change.
$c$ = initial/fixed value.
Always interpret in context.
State limitations if asked.

How it’s examined

Straight-line graphs appears in every WMA11/01 paper, usually as a 6-10 mark question early on. Typical questions: find the equation of a line through two points (then maybe rearrange to integer form); find an equation of a parallel/perpendicular line through a given point; show that two lines are parallel/perpendicular; modelling with linear interpretation. Mark schemes use Method-Accuracy-Bookwork awarding. The most-missed marks: forgetting the negative in perpendicular gradient, not stating the parallel/perpendicular CONCLUSION in 'show that', not writing the answer in the demanded form. A* candidates: full marks expected on this section.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA11/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Straight line graphs

Step-by-step solutions to past-paper-style questions on straight line graphs, written exactly the way a tutor would explain them at the board.

1Find the equation of a line through two points (4 marks, P1)
Core• Style: WMA11/01 January 2024 Q2• equation of line
Question
Find the equation of the line passing through $A(-2, 5)$ and $B(4, -1)$ . Give your answer in the form $ax + by + c = 0$ where $a$ , $b$ , $c$ are integers.
Step-by-step solution
1. Step 1
  Find the gradient:
  $m = \dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1$
2. Step 2
  Use $y - y_{1} = m(x - x_{1})$ with point $A(-2, 5)$ :
  $y - 5 = -1(x + 2)$
3. Step 3
  Expand: $y - 5 = -x - 2 \Rightarrow y = -x + 3 \Rightarrow x + y - 3 = 0$ .
Answer
$x + y - 3 = 0$
Examiner tip
M1 for the gradient with correct subtraction; A1 for $m = -1$ ; M1 for substituting into $y - y_{1} = m(x - x_{1})$ ; A1 for the answer in the demanded form. Reading 'integers' means $a, b, c$ all integers — so $y = -x + 3$ scores M marks only.
2Equation of perpendicular line (5 marks, P1)
Core• Style: WMA11/01 June 2024 Q4• perpendicular
Question
The line $L_{1}$ has equation $3x - 2y + 5 = 0$ . Find an equation of the line $L_{2}$ which is perpendicular to $L_{1}$ and passes through the point $(1, -3)$ .
Step-by-step solution
1. Step 1
  Rearrange $L_{1}$ to gradient-intercept form: $2y = 3x + 5 \Rightarrow y = \tfrac{3}{2}x + \tfrac{5}{2}$ . Gradient $m_{1} = 3/2$ .
2. Step 2
  Perpendicular gradient: $m_{2} = -1/m_{1} = -2/3$ .
3. Step 3
  Use point-gradient form with $(1, -3)$ :
  $y - (-3) = -\dfrac{2}{3}(x - 1)$
4. Step 4
  Multiply through by 3 to clear fractions: $3(y + 3) = -2(x - 1) \Rightarrow 3y + 9 = -2x + 2$ .
5. Step 5
  Rearrange: $2x + 3y + 7 = 0$ .
Answer
$2x + 3y + 7 = 0$
Examiner tip
M1 for rearranging $L_{1}$ to find $m_{1}$ ; A1 for $m_{1} = 3/2$ ; B1 for $m_{2} = -2/3$ (perpendicular condition); M1 for substituting point into line; A1 for the correct answer. The 'perpendicular condition' $m_{1}m_{2} = -1$ must be applied — students who use $m_{2} = 3/2$ (parallel by mistake) lose B1 and A1.
3Distance and midpoint (4 marks, P1)
Core• Style: WMA11/01 January 2023 Q3• distance, midpoint
Question
$P(-1, 4)$ and $Q(5, -4)$ are two points. Find (a) the distance $PQ$ , (b) the coordinates of the midpoint $M$ of $PQ$ .
Step-by-step solution
1. Step 1
  (a) Distance formula: $PQ = \sqrt{(5 - (-1))^{2} + (-4 - 4)^{2}}$ .
2. Step 2
  Compute: $\sqrt{6^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$ .
3. Step 3
  (b) Midpoint: $M = \left(\dfrac{-1 + 5}{2}, \dfrac{4 + (-4)}{2}\right) = (2, 0)$ .
Answer
(a) $PQ = 10$ . (b) $M = (2, 0)$ .
Examiner tip
(a) M1 for distance formula; A1 for the answer. (b) M1 for midpoint formula; A1 for both coordinates. These are foundation marks — A* candidates should not lose any.
4Show two lines are parallel and find distance (5 marks, P1)
Extended• Style: WMA11/01 June 2023 Q8• parallel, show that
Question
Lines $L_{1}: 2x - 3y + 6 = 0$ and $L_{2}: 4x - 6y + 9 = 0$ . (a) Show that $L_{1}$ and $L_{2}$ are parallel. (b) Find the $y$ -intercept of each, and hence the vertical distance between them.
Step-by-step solution
1. Step 1
  (a) Rearrange $L_{1}$ : $3y = 2x + 6 \Rightarrow y = \tfrac{2}{3}x + 2$ . Gradient $= 2/3$ .
2. Step 2
  Rearrange $L_{2}$ : $6y = 4x + 9 \Rightarrow y = \tfrac{2}{3}x + \tfrac{3}{2}$ . Gradient $= 2/3$ .
3. Step 3
  Same gradient ( $2/3$ ) but different intercepts ( $2$ vs $3/2$ ): so $L_{1}$ and $L_{2}$ are parallel and distinct.
4. Step 4
  (b) $y$ -intercepts: $L_{1}$ at $(0, 2)$ , $L_{2}$ at $(0, 3/2)$ .
5. Step 5
  Vertical distance between them: $2 - 3/2 = 1/2$ .
Answer
(a) Both have gradient $2/3$ and different intercepts, so parallel. (b) Vertical distance = $1/2$ .
Examiner tip
(a) M1 for finding both gradients, A1 for stating they are equal AND non-identical (must mention both for 'show that'). (b) B1 B1 for the intercepts, A1 for the distance. 'Show that' demands explicit conclusion — just calculating gradients isn't enough; you must state 'gradients equal so parallel'.
5Modelling with a linear equation (6 marks, P1)
Extended• Style: WMA11/01 January 2024 Q9• modelling
Question
A taxi charges a fixed booking fee plus a fee per km. The total cost $C$ (in $) is modelled as$ C = mn + c $, where$ n $is the distance in km. A 5 km journey costs$ $11 $; a 12 km journey costs$ $22.50 $. (a) Find$ m $and$ c $. (b) Interpret$ m $and$ c$ in the context. (c) Calculate the cost of a 20 km journey.
Step-by-step solution
1. Step 1
  (a) Two points: $(5, 11)$ and $(12, 22.50)$ . Gradient:
  $m = \dfrac{22.50 - 11}{12 - 5} = \dfrac{11.50}{7} = 1.642857... = 1.643 \text{ (3 sf)}$
2. Step 2
  Use $11 = m(5) + c$ : $c = 11 - 5(1.642857) = 11 - 8.2143 = 2.7857$ . To 3 sf: $c \approx \$ 2.79$.
3. Step 3
  (b) $m$ = cost per km ( $\$ 1.64 $/km).$ c $= fixed booking fee ($ $2.79$).
4. Step 4
  (c) $C = 1.643 \times 20 + 2.786 = 32.857 + 2.786 = 35.64$ . To the nearest cent: $\$ 35.64$.
Answer
(a) $m \approx 1.64$ , $c \approx 2.79$ . (b) $m$ = cost per km, $c$ = booking fee. (c) About $\$ 35.64$.
Examiner tip
(a) M1 for gradient calculation, A1 for $m$ , M1 for substitution, A1 for $c$ . (b) B1 for both interpretations. (c) M1 A1. Linear modelling questions appear regularly — the interpretation marks are easy but often missed.

Key Formulae — Straight line graphs

The formulae you need to memorise for straight line graphs on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Gradient of a line through two points
$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$
$(x_1, y_1), (x_2, y_2)$
two points on the line
When to use
Always for finding the gradient. NOT in the formula booklet.
Example
Through $(-2, 5)$ and $(4, -1)$ : $m = (-1 - 5)/(4 - (-2)) = -1$ .
Equation of a straight line (three forms)
$y = mx + c \quad \text{or} \quad y - y_{1} = m(x - x_{1}) \quad \text{or} \quad ax + by + c = 0$
$m$
gradient
$c$
$y$ -intercept
When to use
Pick the form matching the data. Through a point with gradient → point-gradient. Through two points → gradient first, then point-gradient. Required output form → rearrange. NOT in the formula booklet.
Example
Through $(1, -3)$ with $m = -2/3$ : $y + 3 = -\tfrac{2}{3}(x - 1)$ .
Parallel lines condition
$m_{1} = m_{2}$
$m_1, m_2$
gradients of the two lines
When to use
Two lines are parallel iff they have the same gradient.
Example
$y = 3x + 1$ and $y = 3x - 5$ are parallel (same gradient $3$ , different intercepts).
Perpendicular lines condition
$m_{1} m_{2} = -1 \quad \Leftrightarrow \quad m_{2} = -\dfrac{1}{m_{1}}$
$m_1, m_2$
gradients of the two lines
When to use
Two lines are perpendicular iff the product of their gradients is $-1$ . The perpendicular gradient is the NEGATIVE RECIPROCAL.
Example
If $m_{1} = 3/2$ , perpendicular gradient $m_{2} = -2/3$ .
Distance between two points
$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
$(x_1, y_1), (x_2, y_2)$
two points
When to use
Distance/length of a line segment. Comes from Pythagoras.
Example
From $(-1, 4)$ to $(5, -4)$ : $\sqrt{36 + 64} = 10$ .
Midpoint of a line segment
$M = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)$
$(x_1, y_1), (x_2, y_2)$
endpoints
When to use
Average of coordinates. Useful for finding centres, perpendicular bisectors.
Example
Midpoint of $(-1, 4)$ and $(5, -4)$ is $(2, 0)$ .

Key Definitions and Keywords — Straight line graphs

Definitions to memorise and the exact keywords mark schemes credit for straight line graphs answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Gradient (slope)
Examiner keyword
The rate of change of $y$ with respect to $x$ . 'Rise over run' — the $y$ -change divided by the $x$ -change between two points.
$y$ -intercept
Examiner keyword
The $y$ -value where the line crosses the $y$ -axis. For $y = mx + c$ , the $y$ -intercept is $c$ .
$x$ -intercept
The $x$ -value where the line crosses the $x$ -axis. Set $y = 0$ and solve.
Parallel lines
Examiner keyword
Two lines that never meet. Equivalent: same gradient, different intercepts.
Perpendicular lines
Examiner keyword
Two lines meeting at a right angle. Gradients satisfy $m_{1}m_{2} = -1$ .

Common Mistakes and Misconceptions — Straight line graphs

The traps other students keep falling into on straight line graphs questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Subtracting in the wrong order: $(y_{1} - y_{2})/(x_{2} - x_{1})$
Why it happens
Mismatching subscripts in numerator and denominator.
How to avoid it
Always subtract IN THE SAME ORDER for both: $(y_{2} - y_{1})/(x_{2} - x_{1})$ or $(y_{1} - y_{2})/(x_{1} - x_{2})$ . If the subscripts match, the sign is right.
✕Forgetting the negative when taking perpendicular gradient
Persistent issue in WMA11/01 examiner reports
Why it happens
Just flipping the fraction.
How to avoid it
Perpendicular gradient is the NEGATIVE RECIPROCAL. Both flip AND change sign. $m_{1} = 3 \Rightarrow m_{\perp} = -1/3$ . Verify: $3 \times (-1/3) = -1$ . ✓
✕Leaving the answer as $y = -\tfrac{2}{3}x + \tfrac{5}{3}$ when integer form is demanded
Why it happens
Not reading the instruction.
How to avoid it
If 'in the form $ax + by + c = 0$ where $a, b, c$ are integers' is stated, MULTIPLY through to clear fractions: $3y = -2x + 5 \Rightarrow 2x + 3y - 5 = 0$ .
✕Computing distance as $(x_{2} - x_{1}) + (y_{2} - y_{1})$ (i.e. forgetting the square root)
Why it happens
Confusing distance with midpoint or change-of-coords.
How to avoid it
Distance comes from Pythagoras. ALWAYS: square both coordinate differences, sum, then take ROOT.
✕'Showing parallel' by computing gradients but not stating the conclusion
WMA11/01 examiner reports
Why it happens
Forgetting the verbal conclusion.
How to avoid it
'Show that' demands a clear conclusion. After computing $m_{1} = m_{2} = 2/3$ , write: 'Since both gradients are equal and the $y$ -intercepts differ, the lines are parallel.'

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Straight line graphs — frequently asked questions

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Straight line graphs

Short Study Notes in the form of Slides

Short Study Notes — Straight line graphs

Detailed Notes

What you’ll learn

How it’s examined

1Find the equation of a line through two points (4 marks, P1)

2Equation of perpendicular line (5 marks, P1)

3Distance and midpoint (4 marks, P1)

4Show two lines are parallel and find distance (5 marks, P1)

5Modelling with a linear equation (6 marks, P1)

Gradient of a line through two points

Equation of a straight line (three forms)

Parallel lines condition

Perpendicular lines condition

Distance between two points

Midpoint of a line segment

Gradient (slope)

yyy-intercept

xxx-intercept

Parallel lines

Perpendicular lines

✕Subtracting in the wrong order: (y1−y2)/(x2−x1)(y_{1} - y_{2})/(x_{2} - x_{1})(y1​−y2​)/(x2​−x1​)

✕Forgetting the negative when taking perpendicular gradient

✕Leaving the answer as y=−23x+53y = -\tfrac{2}{3}x + \tfrac{5}{3}y=−32​x+35​ when integer form is demanded

✕Computing distance as (x2−x1)+(y2−y1)(x_{2} - x_{1}) + (y_{2} - y_{1})(x2​−x1​)+(y2​−y1​) (i.e. forgetting the square root)

✕'Showing parallel' by computing gradients but not stating the conclusion

Practice questions

Past paper style quiz

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Straight line graphs — frequently asked questions

$y$ -intercept

$x$ -intercept

✕Subtracting in the wrong order: $(y_{1} - y_{2})/(x_{2} - x_{1})$

✕Leaving the answer as $y = -\tfrac{2}{3}x + \tfrac{5}{3}$ when integer form is demanded

✕Computing distance as $(x_{2} - x_{1}) + (y_{2} - y_{1})$ (i.e. forgetting the square root)