Horizontal and Vertical Lines

The lines $y = k$ and $x = h$ : what they look like, their gradients, the axes as lines, where they meet other lines, and the areas they enclose.

Before you start

A horizontal line has gradient $0$ : going along it, the vertical change is $0$ (Secondary 1).
The graph of $ax + by = c$ is a straight line. Its $x$ -intercept is found by putting $y = 0$ and its $y$ -intercept by putting $x = 0$ (lesson 2.2).

Horizontal Lines: y = k

In $ax + by = c$ , $a$ may be $0$ . Then the equation has no $x$ term: for example $0x + 3y = 6$ , which is $3y = 6$ , or $y = 2$ .

The equation $y = 2$ says nothing about $x$ , so $x$ can be any number, and $y$ is always $2$ : $(-3, 2)$ , $(0, 2)$ and $(5, 2)$ are all solutions. These points are all at the same height, so they lie on a horizontal line.

The line y = k

The graph of $y = k$ , where $k$ is a number, is the horizontal line through $(0, k)$ . Every point on it has $y$ -coordinate $k$ .

It is parallel to the $x$ -axis. Its gradient is $0$ : between any two of its points the vertical change is $0$ .

Vertical Lines: x = h

If $b = 0$ , the equation has no $y$ term: for example $2x + 0y = -6$ , which is $2x = -6$ , or $x = -3$ . Now $y$ can be any number and $x$ is always $-3$ : $(-3, -4)$ , $(-3, 0)$ and $(-3, 1.5)$ all lie on it, one above another, on a vertical line.

The line x = h

The graph of $x = h$ , where $h$ is a number, is the vertical line through $(h, 0)$ . Every point on it has $x$ -coordinate $h$ .

It is parallel to the $y$ -axis. Its gradient is undefined: between any two of its points the horizontal change is $0$ , and a vertical change cannot be divided by $0$ .

A vertical line cannot be written in the form $y = mx + c$ (lesson 2.3): on it one value of $x$ goes with every value of $y$ , so $y$ cannot be found from $x$ . (Its equation has no $y$ term, so $y$ cannot be made the subject.)

Example 1: Drawing horizontal and vertical lines

On one pair of axes, draw the lines $y = 2$ , $y = -3$ , $x = 4$ and $x = -1$ . State the gradient of each.

Solution

$y = 2$ : the horizontal line through $(0, 2)$ , $2$ units above the $x$ -axis.

$y = -3$ : the horizontal line through $(0, -3)$ , $3$ units below the $x$ -axis.

$x = 4$ : the vertical line through $(4, 0)$ , $4$ units to the right of the $y$ -axis.

$x = -1$ : the vertical line through $(-1, 0)$ , $1$ unit to the left of the $y$ -axis.

$y = 2$ and $y = -3$ are horizontal, with gradient $0$ ; $x = 4$ and $x = -1$ are vertical, with gradient undefined.

The Axes, Lines Through a Point, and Tidying an Equation

The axes as lines

Every point on the $x$ -axis has $y$ -coordinate $0$ , so the $x$ -axis is the line $y = 0$ .

Every point on the $y$ -axis has $x$ -coordinate $0$ , so the $y$ -axis is the line $x = 0$ .

Example 2: Lines through a given point

Write down the equation of the line through $(3, -2)$ that is (a) parallel to the $x$ -axis, (b) parallel to the $y$ -axis.

Solution

(a) A line parallel to the $x$ -axis is horizontal. Every point on it has the same $y$ -coordinate as $(3, -2)$ , which is $-2$ : $y = -2$ .

(b) A line parallel to the $y$ -axis is vertical. Every point on it has the same $x$ -coordinate, $3$ : $x = 3$ .

(a) $y = -2$ (b) $x = 3$

An equation with only one variable in it may need tidying before you can see which line it is. Solve it for that variable.

Example 3: Tidying an equation

Write each equation in the form $y = k$ or $x = h$ and say whether its graph is horizontal or vertical. (a) $2y - 6 = 0$ (b) $3x + 12 = 0$ (c) $5 - y = 1$ (d) $4x = 0$

Solution

(a) Add $6$ : $2y = 6$ . Divide by $2$ : $y = 3$ . Horizontal.

(b) Subtract $12$ : $3x = -12$ . Divide by $3$ : $x = -4$ . Vertical.

(d) Divide by $4$ : $x = 0$ . Vertical: it is the $y$ -axis itself.

(a) $y = 3$ , horizontal (b) $x = -4$ , vertical (c) $y = 4$ , horizontal (d) $x = 0$ , vertical (the $y$ -axis)

Example 4: Which points lie on the line?

Which of the points $A(-2, 5)$ , $B(2, -2)$ , $C(-2, 0)$ and $D(0, -2)$ lie on (a) $x = -2$ , (b) $y = -2$ ?

Solution

(a) A point lies on $x = -2$ when its $x$ -coordinate is $-2$ , whatever its $y$ -coordinate: $A$ and $C$ .

(b) A point lies on $y = -2$ when its $y$ -coordinate is $-2$ : $B$ and $D$ .

Take care with $B(2, -2)$ and $D(0, -2)$ : they contain the number $-2$ , but as the $y$ -coordinate, so they are not on $x = -2$ .

(a) $A$ and $C$ (b) $B$ and $D$

Where Lines Meet

The point where two lines meet lies on both, so its coordinates make both equations true.

A vertical line $x = h$ and a horizontal line $y = k$ always meet, at $(h, k)$ . To find where $x = h$ meets a sloping line, put $x = h$ into the sloping line's equation and solve for $y$ ; for $y = k$ , put $y = k$ in and solve for $x$ .

Example 5: Meeting points

Find the point where (a) $x = 3$ meets $y = -1$ , (b) $x = 2$ meets $2x + y = 7$ , (c) $y = -1$ meets $2x - y = 5$ .

Solution

(a) The point has $x = 3$ and $y = -1$ : $(3, -1)$ .

(b) The point has $x = 2$ . Put it into $2x + y = 7$ : $4 + y = 7$ , so $y = 3$ . The point is $(2, 3)$ . Check: $2(2) + 3 = 7$ .

(c) The point has $y = -1$ . Put it into $2x - y = 5$ : $2x - (-1) = 5$ , so $2x + 1 = 5$ , $2x = 4$ and $x = 2$ . The point is $(2, -1)$ . Check: $2(2) - (-1) = 4 + 1 = 5\text{.}$

(a) $(3, -1)$ (b) $(2, 3)$ (c) $(2, -1)$

Areas Enclosed by Lines

Horizontal and vertical lines, the axes and a sloping line can enclose a rectangle or a triangle. Find the corners as meeting points first. A length along a horizontal or vertical side is the difference of the coordinates at its ends, and it is never negative.

Example 6: A rectangle

The lines $x = 4$ and $y = 3$ and the two axes enclose a rectangle. Find its area and its perimeter.

Solution

The axes are $x = 0$ and $y = 0$ . The corners are where the lines meet: $(0, 0)$ , $(4, 0)$ , $(4, 3)$ and $(0, 3)$ .

Width: from $x = 0$ to $x = 4$ , $4$ units. Height: from $y = 0$ to $y = 3$ , $3$ units.

Area $= 4 \times 3 = 12$ square units. Perimeter $= 2 \times (4 + 3) = 14$ units.

The area is $12$ square units and the perimeter is $14$ units.

Example 7: A triangle with a sloping side

The lines $y = 1$ , $x = 1$ and $x + y = 5$ enclose a triangle. Find the coordinates of its corners and its area.

Solution

$x = 1$ meets $y = 1$ at $(1, 1)$ .

$x = 1$ meets $x + y = 5$ : $1 + y = 5$ , so $y = 4$ , at $(1, 4)$ .

$y = 1$ meets $x + y = 5$ : $x + 1 = 5$ , so $x = 4$ , at $(4, 1)$ .

The triangle has a right angle at $(1, 1)$ . Its base along $y = 1$ is $4 - 1 = 3$ units and its height along $x = 1$ is $4 - 1 = 3$ units.

Area $= \frac{1}{2} \times 3 \times 3 = 4.5$ square units.

The corners are $(1, 1)$ , $(1, 4)$ and $(4, 1)$ , and the area is $4.5$ square units.

Equations from a Graph

Example 8: Reading the equations

Write down the equations of the lines $A$ , $B$ , $C$ and $D$ .

Solution

$A$ is horizontal and crosses the $y$ -axis at $3$ : every point on it has $y = 3$ .

$B$ is vertical and crosses the $x$ -axis at $-2$ : every point on it has $x = -2$ .

$C$ is horizontal, halfway between $-2$ and $-3$ on the $y$ -axis: $y = -2.5$ .

$D$ is vertical through $4$ on the $x$ -axis: $x = 4$ .

$A$ : $y = 3$ ; $B$ : $x = -2$ ; $C$ : $y = -2.5$ ; $D$ : $x = 4$ .

Common Mistakes

Drawing $x = 3$ across the page. $x = 3$ is the line where $x$ is always $3$ , so it runs up and down through $(3, 0)$ . $y = 3$ runs across.
Calling the $x$ -axis $x = 0$ . On the $x$ -axis it is $y$ that is $0$ : the $x$ -axis is $y = 0$ and the $y$ -axis is $x = 0$ .
Saying a vertical line has gradient $0$ . Gradient $0$ belongs to a horizontal line; the gradient of a vertical line is undefined.
Losing a minus sign in a length. The side from $x = -4$ to $x = 2$ is $2 - (-4) = 6$ units long, not $2$ .

Practice Questions

Work each question on paper before you open its solution. The questions get harder as you go.

Question 1

For each equation, say whether its graph is horizontal or vertical, and give its gradient. (a) $y = -4$ (b) $x = 7$ (c) $0x + 5y = 15$ (d) $4x + 0y = -2$

ShowHide solution

(a) Horizontal; gradient $0$ .

(b) Vertical; gradient undefined.

(d) $4x = -2$ , so $x = -0.5$ : vertical; gradient undefined.

(a) horizontal, $0$ (b) vertical, undefined (c) $y = 3$ , horizontal, $0$ (d) $x = -0.5$ , vertical, undefined

Question 2

On one pair of axes, draw the lines $x = -3$ , $x = 2.5$ , $y = 4$ and $y = -1$ .

ShowHide solution

$x = -3$ and $x = 2.5$ are vertical, through $(-3, 0)$ and $(2.5, 0)$ . $y = 4$ and $y = -1$ are horizontal, through $(0, 4)$ and $(0, -1)$ .

Two vertical lines through $-3$ and $2.5$ on the $x$ -axis, and two horizontal lines through $4$ and $-1$ on the $y$ -axis.

Question 3

Write down the equation of (a) the $x$ -axis, (b) the line through $(-5, 6)$ parallel to the $y$ -axis, (c) the line through $(-5, 6)$ parallel to the $x$ -axis, (d) the line through $(2, 7)$ and $(2, -3)$ , (e) the line through $(-1, 4)$ and $(6, 4)$ .

ShowHide solution

(a) Every point on the $x$ -axis has $y = 0$ : $y = 0$ .

(b) Vertical, with the $x$ -coordinate of $(-5, 6)$ : $x = -5$ .

(d) Both points have $x$ -coordinate $2$ , so the line is vertical: $x = 2$ .

(e) Both points have $y$ -coordinate $4$ , so the line is horizontal: $y = 4$ .

(a) $y = 0$ (b) $x = -5$ (c) $y = 6$ (d) $x = 2$ (e) $y = 4$

Question 4

Write each equation in the form $y = k$ or $x = h$ . (a) $4y + 8 = 0$ (b) $2x - 7 = 0$ (c) $3(y - 1) = 6$ (d) $5x = 2x + 9$

ShowHide solution

(a) $4y = -8$ , so $y = -2$ .

(b) $2x = 7$ , so $x = 3.5$ .

(d) Subtract $2x$ : $3x = 9$ , so $x = 3$ .

(a) $y = -2$ (b) $x = 3.5$ (c) $y = 3$ (d) $x = 3$

Question 5

Which of the points $P(3, -3)$ , $Q(-3, 3)$ , $R(3, 0)$ and $S(0, 3)$ lie on (a) $x = 3$ , (b) $y = 3$ ?

ShowHide solution

(a) $x$ -coordinate $3$ : $P$ and $R$ .

(b) $y$ -coordinate $3$ : $Q$ and $S$ .

(a) $P$ and $R$ (b) $Q$ and $S$

Question 6

Find the point where (a) $x = -4$ meets $y = 5$ , (b) $y = 6$ meets $3x - 2y = 0$ , (c) $x = -1$ meets $x + 2y = 9$ , (d) the $x$ -axis meets $5x + 2y = 15$ .

ShowHide solution

(a) $(-4, 5)$ .

(b) Put $y = 6$ : $3x - 12 = 0$ , so $3x = 12$ and $x = 4$ : $(4, 6)$ . Check: $12 - 12 = 0$ .

(d) The $x$ -axis is $y = 0$ : $5x = 15$ , so $x = 3$ : $(3, 0)$ .

(a) $(-4, 5)$ (b) $(4, 6)$ (c) $(-1, 5)$ (d) $(3, 0)$

Question 7

Write down the equations of the lines $E$ , $F$ and $G$ , and the coordinates of the point where $E$ and $F$ meet.

ShowHide solution

$E$ is vertical through $3$ on the $x$ -axis: $x = 3$ .

$F$ is horizontal through $-3$ on the $y$ -axis: $y = -3$ .

$G$ is vertical through $-4$ on the $x$ -axis: $x = -4$ .

$E$ and $F$ meet where $x = 3$ and $y = -3$ : $(3, -3)$ .

$E$ : $x = 3$ ; $F$ : $y = -3$ ; $G$ : $x = -4$ . $E$ and $F$ meet at $(3, -3)$ .

Question 8

The lines $x = -2$ , $x = 3$ , $y = -1$ and $y = 3$ enclose a rectangle. Find its area and its perimeter.

ShowHide solution

Width: from $x = -2$ to $x = 3$ is $3 - (-2) = 5$ units. Height: from $y = -1$ to $y = 3$ is $3 - (-1) = 4$ units.

Area $= 5 \times 4 = 20$ square units. Perimeter $= 2 \times (5 + 4) = 18$ units.

Area $20$ square units, perimeter $18$ units.

Question 9

The $y$ -axis and the lines $y = 1$ and $x + 2y = 8$ enclose a triangle. Find its corners and its area.

ShowHide solution

The $y$ -axis is $x = 0$ . It meets $y = 1$ at $(0, 1)$ .

$x = 0$ in $x + 2y = 8$ : $2y = 8$ , $y = 4$ , at $(0, 4)$ .

$y = 1$ in $x + 2y = 8$ : $x + 2 = 8$ , $x = 6$ , at $(6, 1)$ .

The right angle is at $(0, 1)$ . Base $6 - 0 = 6$ units, height $4 - 1 = 3$ units. Area $= \frac{1}{2} \times 6 \times 3 = 9$ square units.

Corners $(0, 1)$ , $(0, 4)$ and $(6, 1)$ ; area $9$ square units.

Question 10

The lines $x = 1$ , $x = 5$ , $y = 2$ and $y = k$ enclose a rectangle of area $20$ square units. Find the two possible values of $k$ .

ShowHide solution

The width is $5 - 1 = 4$ units, so the height is $20 \div 4 = 5$ units.

The line $y = k$ is $5$ units above or below $y = 2$ : $k = 2 + 5 = 7$ or $k = 2 - 5 = -3$ .

$k = 7$ or $k = -3$

Question 11

A car park charges a flat $ $3$ for any stay of up to 4 hours. (a) Draw the graph of the charge, $ $C$ , against the time, $t$ hours, for $0 < t \le 4$ . (b) Write down its equation and its gradient, and say what the gradient means.

ShowHide solution

(a) Whatever the time, the charge is $ $3$ : a horizontal line at $C = 3$ from $t = 0$ to $t = 4$ , with an open circle at $(0, 3)$ , since $t = 0$ is not included, and a filled dot at $(4, 3)$ , since $t = 4$ is.

(b) The equation is $C = 3$ , with gradient $0$ : the charge does not change as the time goes on.

(b) $C = 3$ ; gradient $0$ , because the charge stays at $ $3$ however long the stay (up to 4 hours).

Question 12

A rectangular garden plot is marked out by the lines $x = 0$ , $y = 0$ , $x = 6$ and $y = k$ , where $k > 0$ and 1 unit stands for 1 m. Fencing all round it costs $ $15$ per metre, and the fence costs $ $300$ . Find $k$ and the area of the plot.

ShowHide solution

Length of fence $= 300 \div 15 = 20$ m.

The plot is $6$ m by $k$ m, so its perimeter is $2(6 + k) = 20$ . Divide by $2$ : $6 + k = 10$ , so $k = 4$ .

Area $= 6 \times 4 = 24$ m $^{2}$ .

$k = 4$ ; the area is $24$ m $^{2}$ .

Key Points

$y = k$ is the horizontal line through $(0, k)$ ; its gradient is $0$ . It is $ax + by = c$ with $a = 0$ .
$x = h$ is the vertical line through $(h, 0)$ ; its gradient is undefined. It is $ax + by = c$ with $b = 0$ .
The $x$ -axis is $y = 0$ and the $y$ -axis is $x = 0$ .
The line through $(p, q)$ parallel to the $x$ -axis is $y = q$ ; parallel to the $y$ -axis, $x = p$ .
Tidy an equation in one variable by solving it: $2y - 6 = 0$ is $y = 3$ .
$x = h$ and $y = k$ meet at $(h, k)$ . To meet a sloping line, substitute and solve.
For an enclosed area, find the corners first; lengths are differences of coordinates and never negative.