Lesson 14 ~30 min Unit 2 · Linear Relationships +100 XP

Sketching Linear Graphs

Three powerful methods for sketching any straight line. Master the fastest approach for each equation form.

Today's hook: A straight line needs just two points. The art of sketching is choosing the easiest two points — and that depends on the equation form.

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Think First

warm-up

What's the minimum number of points you need to draw a straight line? What's the fastest way to find those points from an equation?

You only need 2 points. The fastest way depends on the equation form: if it's $y = mx + c$, the $y$-intercept gives one point instantly. If it's $ax + by = c$, the intercepts are quickest.

The Three Methods

+5 XP

Every linear graph can be sketched using just two points. Here's how the methods compare:

Method	Best for…	How it works
Two-point	Any equation	Pick $x \to$ find $y$ (×2) $\to$ plot $\to$ join
Gradient-intercept	$y = mx + c$	Plot $(0,c)$ $\to$ use $m$ $\to$ 2nd point $\to$ join
Intercept	$ax + by = c$	Find both intercepts $\to$ plot $\to$ join

Key insight

A sketch shows correct intercepts and gradient — understanding the method matters most.

Choosing your method

See $m$ and $c$? Gradient-intercept. See $ax+by=c$? Intercept. Else? Two-point.

Flowchart

Use the decision flowchart to pick the fastest method for any equation.

Method 1 — Two-Point Method

+5 XP

When to use: When the equation is in any form — this always works.

Choose two simple $x$-values (e.g., $x = 0$ and $x = 1$)
Substitute each into the equation to find $y$
Plot both points, join with a straight line

Worked Example — Sketch $y = 2x + 1$

Choose $x$-values and find $y$
$x = 0$: $y = 2(0) + 1 = 1$ → point $(0, 1)$

$x = 1$: $y = 2(1) + 1 = 3$ → point $(1, 3)$
Plot and join
Gradient $2$, $y$-intercept $(0, 1)$

$x = 0$ gives the $y$-intercept instantly. Pick a small second $x$ for easy calculations.

Tip

$x = 0$ gives the $y$-intercept instantly. Pick a small second $x$ for easy calculations.

Method 2 — Gradient-Intercept Method

+5 XP

When to use: Equation in $y = mx + c$ form — fastest method.

Identify $(0, c)$ — first point
Identify $m = \frac{\text{rise}}{\text{run}}$
From $(0, c)$, move run across and rise up/down
Plot second point, join

Worked Example — Sketch $y = -\tfrac{1}{2}x + 3$

Read $m$ and $c$
$m = -\frac{1}{2}$, $c = 3$. First point: $(0, 3)$
Use gradient: $\frac{-1}{2}$
From $(0, 3)$: run 2 right, rise 1 down → $(2, 2)$
Plot, join, label

Gradient $= \frac{\text{rise}}{\text{run}}$. Positive = rises right; negative = falls. $y$-intercept is at $x = 0$.

Method 3 — Intercept Method

+5 XP

When to use: Standard form $ax + by = c$.

$y$-intercept: set $x = 0$, solve for $y$
$x$-intercept: set $y = 0$, solve for $x$
Plot both, join

Worked Example — Sketch $2x + 3y = 6$

$y$-intercept ($x = 0$)
$3y = 6 \Rightarrow y = 2$. Intercept: $(0, 2)$
$x$-intercept ($y = 0$)
$2x = 6 \Rightarrow x = 3$. Intercept: $(3, 0)$
Plot and join

Cover up one variable to find the other intercept mentally.

Choosing the Best Method

strategy

Equation Form	Best Method	Why?
$y = mx + c$	Gradient-Intercept	$m$ and $c$ directly visible
$ax + by = c$	Intercept	Easy to find both intercepts
$ax + by + c = 0$	Intercept or Two-point	Set $x=0$ or pick two values
Unfamiliar form	Two-point	Always works

Decision flow

See $m$ and $c$? Gradient-intercept. See $ax+by=c$? Intercept. Else? Two-point.

Special Lines

short-cut

Recognising these saves time — no point-calculation needed!

$y = c$Horizontal, gradient 0, through $(0, c)$

$x = a$Vertical, undefined gradient, through $(a, 0)$

$y = x$Diagonal, gradient 1, $45°$

$y = -x$Diagonal, gradient $-1$, falling $45°$

Quick check

No $x$ → horizontal. No $y$ → vertical.

Checking Your Sketch

checklist

Run through this checklist before you move on:

Gradient looks right? Positive = rises; negative = falls; steep = large $|m|$

$y$-intercept correct? Should cross at $(0, c)$

$x$-intercept correct? Set $y = 0$ and check

Both plotted points on the line? If not, recalculate!

Axes labelled? Include $x$, $y$, origin, and the equation

Pro tip

Pick a third point as a check — it should lie on your line too.

Brain Trainer

speed drill

Identify methods and features fast. Go!

$y = 4x - 2$: what is the $y$-intercept?

Answer

$(0, -2)$

Best method for $y = -x + 7$?

Answer

Gradient-intercept — $m=-1$, $c=7$ visible

$x$-intercept of $2x + 5y = 10$?

Answer

$(5, 0)$

Best method for $3x + 4y = 24$?

Answer

Intercept method

Gradient of $y = \frac{2}{3}x + 1$?

Answer

$m = \frac{2}{3}$

Does $(2, 4)$ lie on $y = 3x - 2$?

Answer

Yes — $3(2)-2 = 4$ ✓

What type of line is $y = 5$?

Answer

Horizontal

Two points on $y = 2x - 1$ at $x = 0, 3$?

Answer

$(0, -1)$ and $(3, 5)$

Best method for $y = 0.5x + 10$?

Answer

Gradient-intercept

$y$-intercept of $4x - 3y = 12$?

Answer

$(0, -4)$

Best method for sketching $y = 3x + 2$?

10 XP

Sketch $y = -x + 4$ using the intercept method. What are the intercepts?

10 XP

A line has gradient 2 and $y$-intercept $-1$. Best sketching method?

10 XP

Which point does NOT lie on $y = 2x - 3$?

10 XP

Sketch $x + 2y = 4$. Two easiest points?

10 XP

SAQ1

Short Answer

15 XP

Sketch $y = 2x + 3$ using the gradient-intercept method. Show all working.

Sample solution

Step 1: $m = 2$, $c = 3$. $y$-intercept: $(0, 3)$.

Step 2: Plot $(0, 3)$.

Step 3: Gradient $\frac{2}{1}$: from $(0, 3)$ go right 1, up 2 to $(1, 5)$.

Step 4: Plot $(1, 5)$, join with straight line, label equation.

SAQ2

Short Answer

15 XP

Sketch $3x + 2y = 12$ using the intercept method. Label both intercepts.

Sample solution

Step 1: $x = 0$: $2y = 12$ → $y = 6$. $y$-intercept: $(0, 6)$.

Step 2: $y = 0$: $3x = 12$ → $x = 4$. $x$-intercept: $(4, 0)$.

Step 3: Plot both, join. Label $3x + 2y = 12$.

SAQ3

Short Answer

20 XP

Compare the three sketching methods. When would you use each? Give an example equation for each.

Sample solution

Gradient-intercept: For $y = mx + c$. Ex: $y = -\frac{1}{3}x + 4$ — read $m$ and $c$ directly.

Intercept: For $ax + by = c$. Ex: $5x + 2y = 10$ — easy intercepts $(0, 5)$ and $(2, 0)$.

Two-point: For unfamiliar forms. Ex: $2(y+1) = 3(x-1)$ — pick $x$ values, find $y$.

Stretch Challenges

extension

Stretch 1: A line through $(-2, 5)$ with gradient $-2$.

Find its equation in $y = mx + c$.
Sketch using gradient-intercept.
Does $(1, -1)$ lie on it?

Solution

(i) $5 = (-2)(-2) + c$ → $c = 1$. Equation: $y = -2x + 1$.

(ii) Plot $(0, 1)$, rise $-2$, run $1$ to $(1, -1)$, join.

(iii) $y = -2(1) + 1 = -1$ ✓

Stretch 2: Match equations to descriptions.

A. $y = 5$

B. $x = -3$

C. $y = -2x$

D. $2x + 5y = 10$

1. Vertical line

2. Horizontal line

3. Through origin

4. Best: intercept method

Solution

A→2, B→1, C→3, D→4

Stretch 3: $x$-intercept $(6, 0)$, $y$-intercept $(0, -4)$. Find the gradient and equation.

Solution

Gradient $= \frac{-4-0}{0-6} = \frac{2}{3}$. Equation: $y = \frac{2}{3}x - 4$

Key Takeaways

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Three methods(1) Two-point: pick $x$, find $y$ ×2, plot, join. (2) Gradient-intercept: plot $(0,c)$, use $m$ for 2nd point. (3) Intercept: set $x=0$, set $y=0$, plot, join.

Key ruleOnly 2 points needed. Choose the fastest method for the equation form.

Special lines$y=c$ = horizontal; $x=a$ = vertical; $y=x$ = gradient 1; $y=-x$ = gradient $-1$.

Gradient formula$m = \frac{\text{rise}}{\text{run}} = \frac{y_2-y_1}{x_2-x_1}$

Lesson Complete!

+100 XP

You've mastered three methods for sketching linear graphs. You can now choose the fastest approach for any equation form.

Mark lesson as complete