Graphing Lines and Regions

Not every relation arrives in the form $y = mx + b$. Some look like $x + y = 10$, $3x + 2y = 12$ or $xy = 6$. This page shows you a quick way to graph each kind. Then you’ll graph inequalities like $x + y \le 10$, where the answer isn’t just a line but a whole shaded region of points. Regions are perfect for questions like “what can I buy with $20?”

Key ideas

Vertical and horizontal lines

• The graph of $x = k$ is a vertical line. Every point on it has the same $x$-coordinate, $k$. For example, $x = 3$ goes through $(3, 0)$, $(3, 5)$ and $(3, -2)$.
• The graph of $y = k$ is a horizontal line. Every point on it has the same $y$-coordinate, $k$. For example, $y = -2$ goes through $(0, -2)$ and $(4, -2)$.

It feels backwards at first: $x = 3$ is the line that goes up and down. Think “all the points where $x$ is $3$”.

Lines like x + y = k and x − y = k

For $x + y = 5$, find pairs of numbers that add to $5$:

$x$	$0$	$1$	$2$	$5$
$y$	$5$	$4$	$3$	$0$

Plot them and join them: the line falls $1$ for every $1$ step right.

For $x - y = 2$, find pairs where $x$ is $2$ more than $y$: $(2, 0)$, $(3, 1)$, $(4, 2)$, $(0, -2)$. This line rises $1$ for every $1$ step right.

Lines like ax + by = k: the intercept method

For an equation like $3x + 2y = 12$, the easiest two points to find are the intercepts:

• $x$-intercept: set $y = 0$. Then $3x = 12$, so $x = 4$. The point is $(4, 0)$.
• $y$-intercept: set $x = 0$. Then $2y = 12$, so $y = 6$. The point is $(0, 6)$.

Plot both points and draw the line through them. It’s smart to check a third point: $(2, 3)$ gives $3(2) + 2(3) = 12$. ✓ (The equations of lines page shows how to rewrite these equations in $y = mx + b$ form.)

The relation xy = k

$xy = k$ means ”$x$ times $y$ equals $k$”. For $xy = 6$:

$x$	$-6$	$-3$	$-2$	$-1$	$1$	$2$	$3$	$6$
$y$	$-1$	$-2$	$-3$	$-6$	$6$	$3$	$2$	$1$

This relation is non-linear. Its graph has two separate curved pieces, called branches. Each branch gets closer and closer to the axes but never touches them, because $x = 0$ or $y = 0$ would make the product $0$, not $6$. When $k$ is positive, the branches are in the first and third quadrants. When $k$ is negative, they are in the second and fourth quadrants.

Left: $x = 3$, $y = -2$, $x + y = 5$ and $x - y = 2$. Right: the two branches of $xy = 6$.

Inequalities and regions

An inequality like $x + y \le 4$ is true for a whole region of points, not just a line. To graph it:

1. Draw the boundary line $x + y = 4$.
   • Use a solid line for $\le$ or $\ge$ (points on the line count).
   • Use a dashed line for $\lt$ or $\gt$ (points on the line don’t count).
2. Test a point that’s not on the line. $(0, 0)$ is easiest, unless the line goes through it.
3. Shade the side that contains the test point if it makes the inequality true. Otherwise, shade the other side.

Symbol	Means	Boundary
$\lt$	less than	dashed
$\le$	less than or equal to	solid
$\gt$	greater than	dashed
$\ge$	greater than or equal to	solid

For $x + y \le 4$: test $(0, 0)$, which gives $0 + 0 = 0$, and $0 \le 4$ is true. So shade the side with the origin.

For $x - y \gt 2$: test $(0, 0)$, which gives $0 - 0 = 0$, and $0 \gt 2$ is false. So shade the side without the origin. Notice that this shaded region is below the line even though the symbol is “greater than”. That’s why testing a point is safer than guessing.

A solid boundary for $\le$ and a dashed boundary for $\gt$. The test point $(0, 0)$ decides which side to shade.

For simple inequalities, you can read the region directly: $y \gt 3$ is everything above the dashed line $y = 3$, and $x \le -1$ is everything to the left of the solid line $x = -1$, including the line itself.

Points and regions in context

In a real problem, every point in the shaded region is a combination that works, and every point outside it doesn’t. Points on a solid boundary line are the combinations that use up the limit exactly. Watch out for context, though: you can’t buy $-2$ muffins or $3.5$ tickets, so often only points with whole-number coordinates in the first quadrant make sense.

Worked examples

Example 1: Special lines

Graph $x = -4$, $y = 1$ and $x + y = 3$ on the same grid. Where does $y = 1$ cross each of the other two lines?

Solution.

• $x = -4$ is a vertical line through $(-4, 0)$.
• $y = 1$ is a horizontal line through $(0, 1)$.
• For $x + y = 3$, use pairs that add to $3$: $(0, 3)$, $(1, 2)$, $(3, 0)$. Join them with a straight line.

$y = 1$ crosses $x = -4$ at $(-4, 1)$: that point has $x = -4$ and $y = 1$.

$y = 1$ crosses $x + y = 3$ where $x + 1 = 3$, so $x = 2$. The point is $(2, 1)$.

Check: $2 + 1 = 3$. ✓

Example 2: The intercept method

Graph $2x + 5y = 20$.

Solution.

$x$-intercept: set $y = 0$. Then $2x = 20$, so $x = 10$. The point is $(10, 0)$.

$y$-intercept: set $x = 0$. Then $5y = 20$, so $y = 4$. The point is $(0, 4)$.

Plot $(10, 0)$ and $(0, 4)$ and draw the line through them.

Check a third point: $(5, 2)$ gives $2(5) + 5(2) = 10 + 10 = 20$. ✓ So $(5, 2)$ should be on your line, halfway between the intercepts.

Example 3: Graphing xy = −8

Make a table of values for $xy = -8$ and describe its graph.

Solution. Pick values of $x$ that divide evenly into $-8$, and find $y = \dfrac{-8}{x}$:

$x$	$-8$	$-4$	$-2$	$-1$	$1$	$2$	$4$	$8$
$y$	$1$	$2$	$4$	$8$	$-8$	$-4$	$-2$	$-1$

When $x$ is negative, $y$ is positive (second quadrant). When $x$ is positive, $y$ is negative (fourth quadrant). So the graph has two branches, one in the second quadrant and one in the fourth. Neither branch touches an axis, and there’s no point at $x = 0$.

It’s non-linear. From $x = 1$ to $x = 2$, $y$ goes from $-8$ to $-4$: up $4$ for one step, a rate of change of $4$. From $x = 2$ to $x = 4$, $y$ goes from $-4$ to $-2$: up $2$ over two steps, a rate of change of $1$. The rates are different, so the graph isn’t a straight line.

Example 4: A snack budget

Ari has $10 to spend at a bake sale. Muffins cost $2 each and cookies cost $1 each. Let $x$ be the number of muffins and $y$ the number of cookies.

• (a) Write an inequality for the combinations Ari can afford, and describe its graph.
• (b) Can Ari buy $3$ muffins and $4$ cookies? $4$ muffins and $3$ cookies?

Solution.

(a) Muffins cost $2x$ dollars and cookies cost $y$ dollars, and the total can be at most $10:

$$2x + y \le 10$$

Boundary line: $2x + y = 10$. The $x$-intercept is $(5, 0)$ (only muffins) and the $y$-intercept is $(0, 10)$ (only cookies). Use a solid line, because spending exactly $10 is allowed.

Test $(0, 0)$: $2(0) + 0 = 0$, and $0 \le 10$ is true. So shade the side with the origin.

Ari can’t buy negative amounts, so only the part of the region in the first quadrant (including the axes) makes sense, and only the points with whole-number coordinates.

(b) $(3, 4)$: $2(3) + 4 = 10$, and $10 \le 10$ is true. Yes: the point is on the boundary line, so Ari spends exactly $10.

$(4, 3)$: $2(4) + 3 = 11$, and $11 \le 10$ is false. No: that costs $11, and the point is outside the shaded region.

Common mistakes

Mixing up $x = k$ and $y = k$. $x = 3$ is vertical (every point has $x$-coordinate $3$), and $y = 3$ is horizontal. If you’re unsure, list two points, like $(3, 0)$ and $(3, 4)$, and see which way they line up.

Using the wrong kind of boundary line. Solid for $\le$ and $\ge$, dashed for $\lt$ and $\gt$. A dashed line tells the reader “the points on this line are not included”.

Shading by the symbol instead of testing. “Greater than means shade above” doesn’t always work. In $x - y \gt 2$, the correct region is below the line. Always test a point.

Testing a point that’s on the boundary line. For $y \lt 2x$, the line goes through $(0, 0)$, so testing the origin gives $0 \lt 0$, which tells you nothing. Pick a point clearly off the line, like $(1, 0)$: $0 \lt 2$ is true, so shade the side containing $(1, 0)$.

Joining the two branches of xy = k. The branches never touch the axes and never meet. Don’t draw a curve through the origin, and don’t connect the two pieces.

Ignoring the context. In a budget problem, a point like $(-1, 12)$ might be in the shaded region, but you can’t buy $-1$ muffins. Only use points that make sense in the situation.

Practice

1. (Warm-up) Which of $x = 5$ and $y = -3$ is a vertical line? Where do the two lines cross?

Solution

$x = 5$ is vertical (every point has $x$-coordinate $5$) and $y = -3$ is horizontal.

They cross at the point that has $x = 5$ and $y = -3$: $(5, -3)$.

2. (Warm-up) Find the intercepts of $4x + 3y = 24$.

Solution

$x$-intercept: set $y = 0$, so $4x = 24$ and $x = 6$. The point is $(6, 0)$.

$y$-intercept: set $x = 0$, so $3y = 24$ and $y = 8$. The point is $(0, 8)$.

3. (Warm-up) Is each point in the region $x + y \ge 6$?

• (a) $(2, 5)$
• (b) $(1, 4)$
• (c) $(3, 3)$

Solution

(a) $2 + 5 = 7$, and $7 \ge 6$ is true. Yes.

(b) $1 + 4 = 5$, and $5 \ge 6$ is false. No.

(c) $3 + 3 = 6$, and $6 \ge 6$ is true. Yes: the point is on the solid boundary line, so it’s included.

4. (Core) Make a table of values for $x - y = 4$ and graph it. Which of the points $(1, -3)$ and $(3, 1)$ is on the line?

Solution

Find pairs where $x$ is $4$ more than $y$:

$x$	$0$	$2$	$4$	$6$
$y$	$-4$	$-2$	$0$	$2$

Plot the points and join them with a straight line. It rises $1$ for every $1$ step right.

$(1, -3)$: $1 - (-3) = 4$. ✓ It’s on the line.

$(3, 1)$: $3 - 1 = 2$, not $4$. It’s not on the line.

5. (Core) A rectangle has an area of $18$ cm². Let $l$ be its length and $w$ its width, both in centimetres.

• (a) Write an equation relating $l$ and $w$.
• (b) Make a table of whole-number lengths and widths.
• (c) Why does only one branch of the graph make sense here? What happens to the width as the length gets bigger?

Solution

(a) Area is length times width, so $lw = 18$.

(b)

$l$ (cm)	$1$	$2$	$3$	$6$	$9$	$18$
$w$ (cm)	$18$	$9$	$6$	$3$	$2$	$1$

(c) Lengths and widths must be positive, so only the first-quadrant branch makes sense. As the length gets bigger, the width gets smaller, but it never reaches $0$. This relation is non-linear: doubling the length from $3$ to $6$ halves the width from $6$ to $3$.

6. (Core) Explain how to graph $3x + 2y \lt 12$. Is the point $(4, 0)$ part of the region?

Solution

Boundary line: $3x + 2y = 12$, with intercepts $(4, 0)$ and $(0, 6)$. Draw it dashed, because the symbol is $\lt$.

Test $(0, 0)$: $3(0) + 2(0) = 0$, and $0 \lt 12$ is true. Shade the side containing the origin.

$(4, 0)$ gives $3(4) + 2(0) = 12$, and $12 \lt 12$ is false. So $(4, 0)$ is not in the region: it sits on the dashed boundary.

7. (Core) A school play charges $15 for adult tickets and $10 for student tickets. The drama club needs to raise at least $600. Let $x$ be the number of adult tickets and $y$ the number of student tickets.

• (a) Write an inequality and find the intercepts of its boundary line.
• (b) Which side of the line should be shaded?
• (c) Do $20$ adult and $30$ student tickets raise enough?

Solution

(a) $15x + 10y \ge 600$. Intercepts: $y = 0$ gives $15x = 600$, so $(40, 0)$; $x = 0$ gives $10y = 600$, so $(0, 60)$. Draw a solid line.

(b) Test $(0, 0)$: $0 \ge 600$ is false. So shade the side away from the origin (more tickets sold).

(c) $15(20) + 10(30) = 300 + 300 = 600$, and $600 \ge 600$ is true. Yes: they raise exactly $600, so the point is on the boundary line.

8. (Challenge) A region has a dashed boundary line through $(0, 3)$ and $(3, 0)$, and the shaded side does not contain the origin. Write the inequality.

Solution

The line through $(0, 3)$ and $(3, 0)$ contains points whose coordinates add to $3$, so its equation is $x + y = 3$.

Dashed means the symbol is $\lt$ or $\gt$. The origin gives $0 + 0 = 0$, which is less than $3$, and the origin is not shaded. So the shaded points must have $x + y$ greater than $3$:

$$x + y \gt 3$$

Check with a point on the shaded side, like $(3, 3)$: $6 \gt 3$. ✓

9. (Challenge) Mei has at most $10$ hours a week for guitar practice ($x$ hours) and studying ($y$ hours), and she wants to study more than $3$ hours. So $x + y \le 10$ and $y \gt 3$.

• (a) Which of these combinations work: $(5, 4)$, $(7, 3)$, $(2, 8)$, $(6, 5)$?
• (b) If she only plans whole numbers of hours, what is the most guitar practice she can fit in?

Solution

(a) A combination must make both inequalities true.

• $(5, 4)$: $5 + 4 = 9 \le 10$ ✓ and $4 \gt 3$ ✓. Works.
• $(7, 3)$: $7 + 3 = 10 \le 10$ ✓, but $3 \gt 3$ is false. Doesn’t work.
• $(2, 8)$: $2 + 8 = 10 \le 10$ ✓ and $8 \gt 3$ ✓. Works.
• $(6, 5)$: $6 + 5 = 11$, which is more than $10$. Doesn’t work.

(b) With whole numbers, $y \gt 3$ means she studies at least $4$ hours. Then $x + 4 \le 10$, so $x \le 6$. The most guitar practice is $6$ hours (with exactly $4$ hours of studying).

On a graph, the allowed points are in the region below the solid line $x + y = 10$ and above the dashed line $y = 3$.