Maths · 4 Coordinate geometry

4.5 Parallel lines

Before you start

Finding the gradient of a line from two points on it, as the change in $y$ divided by the change in $x$ (subtopic 4.2).
In $y = m x + c$ , $m$ is the gradient and $c$ is the $y$ -intercept, where the line crosses the $y$ -axis (subtopic 4.4).
Rearranging a linear equation by doing the same operation to every term on both sides.
Substituting numbers into an equation and solving a one-step equation such as $- 3 = 4 + c$ .
Arithmetic with negative numbers and fractions, e.g. $- \frac{3}{4} \times 8 = - 6$ .

Syllabus objectives covered (1)

Find the gradient and equation of a straight line parallel to a given line.

Part 1 of 7

Parallel lines have the same gradient

Two straight lines are parallel if they point in exactly the same direction. They stay the same distance apart and never meet, however far you extend them.
The gradient of a line measures its direction: how far it goes up (or down) for every 1 unit across. If two lines go up by the same amount for every unit across, they point the same way. So parallel lines have equal gradients.
It works the other way round too: if two different lines have equal gradients, they are parallel. This one fact is what the whole subtopic rests on.
In $y = m x + c$ , parallel lines share the same $m$ . What makes them different lines is $c$ , the point where each one crosses the $y$ -axis.
graph diagram — not drawn yet
Axes from -5 to 5 on both axes with equal scales, showing the three lines y = 2x - 3, y = 2x and y = 2x + 4, each labelled with its equation, and on each line a small right-angled triangle 1 unit across and 2 units up.
Each line goes up 2 for every 1 across, so all three have gradient 2 and never meet. Only the y-intercepts, -3, 0 and 4, are different.
If two equations have the same gradient and the same $y$ -intercept, they are not two parallel lines. They are one line written two ways. $y = 2 x + 4$ and $2 y = 4 x + 8$ are the same line, because dividing every term of the second by 2 gives the first.
Parallel means you keep the gradient exactly as it is: the same number with the same sign. Changing the sign, or turning the gradient upside down, gives a line in a different direction. Doing both together is the rule for perpendicular lines (subtopic 4.6), not for parallel ones.

Part 2 of 7

Reading the gradient from the given line

Every question on this objective starts the same way: find the gradient of the line you are given. If the equation is in the form $y = m x + c$ , with $y$ on its own on the left, the gradient is the number multiplying $x$ .
$y = 4 x - 1$ has gradient $4$ . $y = - \frac{2}{3} x + 5$ has gradient $- \frac{2}{3}$ . $y = x + 6$ has gradient $1$ , because $x$ means $1 x$ .
The order of the terms does not matter. In $y = 5 - 3 x$ the term with $x$ in it is $- 3 x$ , so the gradient is $- 3$ . The $5$ is the $y$ -intercept. Always take the sign in front of the $x$ term along with the number.
A fraction of $x$ is still a coefficient of $x$ . $y = \frac{x}{2} + 7$ is the same as $y = \frac{1}{2} x + 7$ , so the gradient is $\frac{1}{2}$ .
So any line parallel to $y = 5 - 3 x$ has gradient $- 3$ , and any line parallel to $y = \frac{x}{2} + 7$ has gradient $\frac{1}{2}$ . The first half of the objective, finding the gradient of the parallel line, is done as soon as you have read this number correctly.

Part 3 of 7

When y is not on its own, rearrange first

The given line is not always in $y = m x + c$ form. It might be $2 y = 6 x + 5$ or $3 x + 4 y = 12$ . Then the number next to $x$ is not the gradient, because $y$ is not on its own.
Rearrange until the equation starts $y = \dots$ . For $2 y = 6 x + 5$ , divide every term by 2. You divide by 2 because 2 is multiplying $y$ , and dividing every term keeps the equation balanced. The gradient is $3$ , not $6$ .
$2 y = 6 x + 5 \Rightarrow y = 3 x + \frac{5}{2}$
For $3 x + 4 y = 12$ , first subtract $3 x$ from both sides so that only the $y$ term is on the left. Then divide every term by 4. The gradient is $- \frac{3}{4}$ . The minus sign appears when $3 x$ moves across, and it is the easiest thing to drop.
$4 y = - 3 x + 12 \Rightarrow y = - \frac{3}{4} x + 3$
Watch a negative $y$ term. In $6 x - 2 y = 7$ , subtract $6 x$ to get $- 2 y = - 6 x + 7$ . Then divide every term by $- 2$ . Dividing by a negative number changes the sign of every term, so the gradient is $+ 3$ .
$- 2 y = - 6 x + 7 \Rightarrow y = 3 x - \frac{7}{2}$
Divide every term, including the number on its own. The gradient comes out right either way, but if a later part asks where the line crosses the $y$ -axis, an undivided constant gives the wrong intercept.
A check for a line written as $a x + b y = c$ (here $c$ is just the number on the right, not the intercept): the gradient is $- \frac{a}{b}$ . For $3 x + 4 y = 12$ that gives $- \frac{3}{4}$ , which agrees with the rearrangement. Use it to check your working, and still show the rearrangement.

Part 4 of 7

Finding c: make the new line pass through the point

Once you have the gradient, the new line is $y = m x + c$ with $m$ known and $c$ unknown. There is a whole family of lines with that gradient. The point in the question picks out exactly one of them.
A point lies on a line exactly when its coordinates make the equation true. So put the point's $x$ -coordinate in for $x$ and its $y$ -coordinate in for $y$ , then solve for $c$ .
Take the line parallel to $y = 4 x - 1$ that passes through $(1, - 3)$ . The gradient is $4$ , so the new line is $y = 4 x + c$ .
Substitute $x = 1$ and $y = - 3$ . Then subtract 4 from both sides to get $c$ on its own. $- 3 - 4$ means start at $- 3$ and go 4 further down, which is $- 7$ , not $- 1$ .
$- 3 = 4 \times 1 + c \Rightarrow - 3 = 4 + c \Rightarrow c = - 7$
Now write the full equation: $y = 4 x - 7$ . The question asked for an equation, so $c = - 7$ on its own is not the answer.
graph diagram — not drawn yet
Axes with x from -2 to 3 and y from -9 to 9, showing the line y = 4x - 1 and the parallel line y = 4x - 7, each labelled with its equation, with the point (1, -3) marked with a cross on y = 4x - 7 and the y-intercepts (0, -1) and (0, -7) marked.
The new line keeps the gradient 4 and sits lower, so that it passes through (1, -3). That puts its y-intercept at -7.
The first coordinate goes in for $x$ and the second for $y$ . If you swap them for $(1, - 3)$ you get $1 = 4 \times (- 3) + c$ , so $c = 13$ . The line $y = 4 x + 13$ passes through $(- 3, 1)$ , which is a different point.
If you have met the form $y - y_{1} = m (x - x_{1})$ , where $(x_{1}, y_{1})$ is the point, it gives the same line: $y + 3 = 4 (x - 1)$ simplifies to $y = 4 x - 7$ . Either method is fine, as long as you end with the simplified equation.

Part 5 of 7

Check it, then give it in the form asked for

Two checks take seconds. First, is the number in front of $x$ in your answer the gradient of the given line? Second, does the point fit? For $y = 4 x - 7$ and $(1, - 3)$ : $4 \times 1 - 7 = - 3$ , so it does.
If the point does not fit, look first at the step where you solved for $c$ . A sign slip there changes the answer without changing the gradient.
When the gradient is a fraction, keep it as a fraction. For the line parallel to $3 x + 4 y = 12$ through $(8, - 5)$ , the gradient is $- \frac{3}{4}$ . Then $- 5 = - \frac{3}{4} \times 8 + c = - 6 + c$ , so $c = 1$ and the line is $y = - \frac{3}{4} x + 1$ .
Leave the answer in $y = m x + c$ form unless the question names another form. If it asks for $a x + b y = c$ with integers, multiply every term by the denominator to clear the fraction, then collect $x$ and $y$ on one side.
$4 y = - 3 x + 4 \Rightarrow 3 x + 4 y = 4$
Notice what happened. The given line was $3 x + 4 y = 12$ and the parallel line is $3 x + 4 y = 4$ . In this form, parallel lines can keep the same $x$ and $y$ terms, and only the number on the right changes. That gives a fast check: $3 \times 8 + 4 \times (- 5) = 4$ , so $(8, - 5)$ fits.
The syllabus expects the equation of a line in a fully simplified form. Write $3 x + 4 y = 4$ , not $6 x + 8 y = 8$ , and write $y = 4 x - 7$ , not $y = 4 x + (- 7)$ .

Part 6 of 7

Other starting points: two points, or a horizontal or vertical line

Sometimes you are given two points on the line instead of its equation, such as $A (- 1, 2)$ and $B (3, 10)$ . Find the gradient of $A B$ from the two points first. Subtract in the same order on the top and the bottom.
$m = \frac{10 - 2}{3 - (- 1)} = \frac{8}{4} = 2$
The line parallel to $A B$ through $(0, - 5)$ is then $y = 2 x - 5$ . When the point is on the $y$ -axis, its $y$ -coordinate is $c$ straight away, because $x = 0$ makes $m x$ zero.
A horizontal line has equation $y = k$ and gradient $0$ . A line parallel to $y = 6$ is also horizontal, so every point on it has the same $y$ -coordinate. Through $(2, - 1)$ it is $y = - 1$ .
A vertical line has equation $x = k$ . Its gradient cannot be calculated, so $y = m x + c$ cannot describe it. A line parallel to $x = 3$ is also vertical. Through $(2, - 1)$ it is $x = 2$ .
graph diagram — not drawn yet
Axes from -3 to 7 on both axes with the horizontal lines y = 6 and y = -1 and the vertical lines x = 3 and x = 2, each labelled with its equation, and the point (2, -1) marked with a cross where y = -1 meets x = 2.
Through (2, -1), the line parallel to y = 6 is y = -1 and the line parallel to x = 3 is x = 2. Each new line copies one coordinate of the point.
The trap is the wording. A line parallel to the $x$ -axis is horizontal, so its equation is $y = \dots$ . A line parallel to the $y$ -axis is vertical, so its equation is $x = \dots$ . The axis named in the question is not the letter in the answer.

Part 7 of 7

Deciding whether two lines are parallel

To test two lines, put each one into $y = m x + c$ form and compare the gradients. You cannot compare the numbers in front of $x$ until both equations start $y = \dots$ .
$y = 3 x + 1$ and $6 x - 2 y = 7$ : you rearranged the second one earlier to $y = 3 x - \frac{7}{2}$ . Both gradients are $3$ and the intercepts, $1$ and $- \frac{7}{2}$ , are different, so the lines are parallel.
$y = 3 x + 1$ and $3 x + y = 4$ : the second rearranges to $y = - 3 x + 4$ , with gradient $- 3$ . The gradients are not equal, so the lines are not parallel. Both equations have a 3 in them, but one line goes up and the other goes down.
So there are three outcomes. Equal gradients and different intercepts: parallel. Equal gradients and equal intercepts: the same line. Different gradients: the lines are not parallel, and they cross at one point.
When a question says Show that two lines are parallel, the evidence is both gradients, each from a correct rearrangement. Finish with the sentence that links them: the gradients are equal, so the lines are parallel.

Worked examples

Example 1. Find the equation of the line that is parallel to $y = 3 - 2 x$ and passes through $(4, 1)$ .

Show the working

Read the gradient of the given line. The term with $x$ in it is $- 2 x$ , so the gradient is $- 2$ . The $3$ is the intercept, not the gradient.
The parallel line has the same gradient, so it is $y = - 2 x + c$ .
Substitute the point: $x = 4$ and $y = 1$ . Then add 8 to both sides to get $c$ on its own.
$1 = - 2 \times 4 + c \Rightarrow 1 = - 8 + c \Rightarrow c = 9$
Check that the point fits the answer: $- 2 \times 4 + 9 = 1$ . It does.

Answer: $y = - 2 x + 9$ (or $y = 9 - 2 x$ )

Example 2. Line $L$ has equation $3 x + 5 y = 10$ . (a) Find the gradient of $L$ . (b) Find the equation of the line that is parallel to $L$ and passes through $(- 5, 4)$ . Give your answer in the form $a x + b y = c$ , where $a$ , $b$ and $c$ are integers.

Show the working

(a) $y$ is not on its own, so rearrange. Subtract $3 x$ from both sides, then divide every term by 5.
$5 y = - 3 x + 10 \Rightarrow y = - \frac{3}{5} x + 2$
The gradient is the number multiplying $x$ now that $y$ is on its own: $- \frac{3}{5}$ .
(b) The parallel line is $y = - \frac{3}{5} x + c$ . Substitute $x = - 5$ and $y = 4$ . A negative times a negative is positive, so $- \frac{3}{5} \times (- 5) = 3$ .
$4 = - \frac{3}{5} \times (- 5) + c \Rightarrow 4 = 3 + c \Rightarrow c = 1$
So the line is $y = - \frac{3}{5} x + 1$ . The question asks for integers, so multiply every term by 5 to clear the fraction, then add $3 x$ to both sides.
$5 y = - 3 x + 5 \Rightarrow 3 x + 5 y = 5$
Check with the point: $3 \times (- 5) + 5 \times 4 = - 15 + 20 = 5$ . Notice that $L$ is $3 x + 5 y = 10$ and the new line is $3 x + 5 y = 5$ : same $x$ and $y$ terms, different number on the right, as parallel lines in this form can be.

Answer: (a) $- \frac{3}{5}$ (b) $3 x + 5 y = 5$

Example 3. $A$ is the point $(- 2, 7)$ and $B$ is the point $(4, - 2)$ . Line $L$ is parallel to $A B$ and passes through $C (4, 3)$ . (a) Find the equation of $L$ . (b) Find the coordinates of the point where $L$ crosses the $x$ -axis.

Show the working

(a) You are not given an equation for $A B$ , so find its gradient from the two points. Take $B$ minus $A$ on the top and on the bottom, keeping the signs.
$m = \frac{- 2 - 7}{4 - (- 2)} = \frac{- 9}{6} = - \frac{3}{2}$
$L$ is parallel to $A B$ , so $L$ is $y = - \frac{3}{2} x + c$ .
Substitute $C$ : $x = 4$ and $y = 3$ .
$3 = - \frac{3}{2} \times 4 + c \Rightarrow 3 = - 6 + c \Rightarrow c = 9$
So $L$ is $y = - \frac{3}{2} x + 9$ . Check: $- \frac{3}{2} \times 4 + 9 = - 6 + 9 = 3$ , so $C$ fits.
(b) Every point on the $x$ -axis has $y = 0$ . Put $y = 0$ into the equation of $L$ and solve for $x$ .
$0 = - \frac{3}{2} x + 9 \Rightarrow \frac{3}{2} x = 9 \Rightarrow x = 6$
Check: the gradient from $(4, 3)$ to $(6, 0)$ is $\frac{0 - 3}{6 - 4} = - \frac{3}{2}$ , the same as $A B$ .

Answer: (a) $y = - \frac{3}{2} x + 9$ (b) $(6, 0)$

Example 4. Show that the lines $y = 4 x - 3$ and $8 x - 2 y = 5$ are parallel.

Show the working

The first line is already in $y = m x + c$ form. Its gradient is $4$ .
Rearrange the second line. Subtract $8 x$ from both sides, then divide every term by $- 2$ . Dividing by a negative changes every sign.
$- 2 y = - 8 x + 5 \Rightarrow y = 4 x - \frac{5}{2}$
Its gradient is $4$ . The intercepts, $- 3$ and $- \frac{5}{2}$ , are different, so these are two different lines, not one line written twice.
Write the conclusion as a sentence. A Show that is finished only when you have said what the equal gradients mean.

Answer: Both lines have gradient $4$ , so they are parallel.

Where people go wrong

Mistake: Taking the gradient of $2 y = 6 x + 5$ as $6$ .

Instead: Divide every term by 2 first: $y = 3 x + \frac{5}{2}$ , so the gradient is $3$ .

The number in front of $x$ is the gradient only when $y$ is on its own. In $2 y = 6 x + 5$ , the $6$ is how fast $2 y$ changes, which is twice as fast as $y$ .

Mistake: Reading the gradient of $y = 5 - 3 x$ as $5$ , or as $3$ .

Instead: The gradient is the number multiplying $x$ , with its sign: $- 3$ .

Writing the terms in a different order does not change which number multiplies $x$ . The $5$ is the $y$ -intercept, and dropping the minus sign turns a line that goes down into one that goes up.

Mistake: Solving $- 3 = 4 + c$ as $c = 1$ or $c = 7$ .

Instead: Subtract 4 from both sides: $c = - 3 - 4 = - 7$ .

Substitute back to see it: with $c = 1$ , the point $x = 1$ gives $4 + 1 = 5$ , not $- 3$ . Checking the point in your final equation catches this every time.

Mistake: Putting the coordinates in the wrong way round, e.g. $x = - 3$ and $y = 1$ for the point $(1, - 3)$ .

Instead: The first coordinate is $x$ and the second is $y$ : $- 3 = 4 \times 1 + c$ .

Swapping them gives $c = 13$ , and $y = 4 x + 13$ passes through $(- 3, 1)$ , which is not the point in the question.

Mistake: Changing the gradient for the parallel line, e.g. using $- \frac{1}{4}$ for a line parallel to $y = 4 x - 1$ .

Instead: A parallel line has exactly the same gradient: $4$ .

Turning the gradient upside down and changing its sign is the rule for perpendicular lines. A gradient of $- \frac{1}{4}$ gives a line that crosses $y = 4 x - 1$ at a right angle.

Check yourself

Write down the gradient of any line parallel to $y = 7 - 2 x$ .
Answer
$- 2$ . The $x$ term is $- 2 x$ ; the $7$ is the intercept.
Find the gradient of a line parallel to $4 y = 3 x + 8$ .
Answer
Divide every term by 4: $y = \frac{3}{4} x + 2$ , so the gradient is $\frac{3}{4}$ .
Find the equation of the line parallel to $y = 5 x + 2$ that passes through $(0, - 4)$ .
Answer
$y = 5 x - 4$ . The point is on the $y$ -axis, so $c = - 4$ straight away.
Find the equation of the line parallel to $y = - 3 x + 1$ that passes through $(2, 5)$ .
Answer
$5 = - 3 \times 2 + c = - 6 + c$ , so $c = 11$ and the line is $y = - 3 x + 11$ .
Find the equation of the line parallel to $2 x + y = 9$ that passes through $(- 1, 6)$ .
Answer
$y = - 2 x + 9$ , so the gradient is $- 2$ . Then $6 = - 2 \times (- 1) + c = 2 + c$ , so $c = 4$ : $y = - 2 x + 4$ (or $2 x + y = 4$ ).
Find the equation of the line parallel to $y = \frac{1}{2} x - 3$ that passes through $(6, 1)$ .
Answer
$1 = \frac{1}{2} \times 6 + c = 3 + c$ , so $c = - 2$ and the line is $y = \frac{1}{2} x - 2$ .
Write down the equation of the line through $(3, - 2)$ that is (a) parallel to the $x$ -axis, (b) parallel to $x = - 1$ .
Answer
(a) $y = - 2$ , a horizontal line. (b) $x = 3$ , a vertical line.
Are the lines $y = 2 x + 3$ and $4 x - 2 y = 1$ parallel? Explain how you know.
Answer
Rearrange the second: $- 2 y = - 4 x + 1$ , so $y = 2 x - \frac{1}{2}$ . Both gradients are $2$ and the intercepts are different, so yes, they are parallel.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Line $L_{1}$ has equation $y = 3 x - 5$ .

(a) Write down the gradient of $L_{1}$ . 1 mark No calculator

Mark scheme

B1 3

(b) Line $L_{2}$ is parallel to $L_{1}$ and passes through the point $(2, 9)$ . Find the equation of $L_{2}$ . 3 marks No calculator

Mark scheme

B1 gradient 3 used for L2, e.g. y = 3x + c
M1 (2, 9) substituted: 9 = 3 × 2 + c oe
A1 y = 3x + 3

(c) Find the coordinates of the point where $L_{2}$ crosses the $x$ -axis. 2 marks No calculator

Mark scheme

M1 3x + 3 = 0 oe, FT their equation from (b)
A1 (-1, 0)

Question 2. Point $A$ is $(- 3, 4)$ and point $B$ is $(5, - 2)$ .

(a) Calculate the gradient of the line $A B$ . 2 marks No calculator

Mark scheme

M1 (-2 - 4) ÷ (5 - (-3)) oe
A1 -3/4 oe

(b) Line $M$ is parallel to $A B$ and passes through the point $(4, - 1)$ . Find the equation of $M$ in the form $y = m x + c$ . 3 marks No calculator

Mark scheme

B1 their gradient from (a) used for M, e.g. y = -3/4 x + c, FT
M1 (4, -1) substituted: -1 = -3/4 × 4 + c oe, FT their gradient
A1 y = -3/4 x + 2 oe

(c) Write the equation of $M$ in the form $p x + q y = r$ , where $p$ , $q$ and $r$ are integers. 1 mark No calculator

Mark scheme

B1 3x + 4y = 8 oe with integer values, FT their (b)

Question 3. Line $P$ has equation $2 x + 4 y = 9$ . Line $Q$ has equation $y = 3 - \frac{1}{2} x$ .

(a) Show that lines $P$ and $Q$ are parallel. 2 marks No calculator

Mark scheme

B1 P rearranged to y = -1/2 x + 9/4 oe, or gradient of P = -1/2 from a correct rearrangement
B1 gradient of Q = -1/2 and a statement that the gradients are equal, so the lines are parallel

(b) Line $R$ is parallel to $P$ and passes through the point $(6, 1)$ . Find the equation of $R$ . 2 marks No calculator

Mark scheme

M1 1 = -1/2 × 6 + c oe
A1 y = -1/2 x + 4 oe

(c) Line $S$ passes through the point $(6, 1)$ and is parallel to the $x$ -axis. Write down the equation of $S$ . 1 mark No calculator

Mark scheme

B1 y = 1

How this is examined

A line parallel to $A B$ when you are given only the points $A$ and $B$ : A two-part question: Calculate the gradient of $A B$ , then Find the equation of the line parallel to $A B$ through a third point, about 5 marks across the two parts The gradient as difference in $y$ over difference in $x$ , subtracting in the same order with signs kept: $\frac{- 2 - 7}{4 - (- 2)} = - \frac{3}{2}$ . Then that same gradient in $y = m x + c$ with the third point substituted, ending with the full equation. If your gradient is wrong, using it correctly in the next part still shows a correct method (follow through).
The equation of the line parallel to a given line through a given point: A 3-mark Find the equation, with the given line's equation and one point stated The given line's gradient used unchanged, the point substituted with $x$ and $y$ the right way round ( $- 3 = 4 \times 1 + c$ ), and the full equation as the last line ( $y = 4 x - 7$ ). The written substitution line is your method; $c = - 7$ on its own is not the equation.
Getting the gradient out of an equation that is not in $y = m x + c$ form: A 1- or 2-mark Find the gradient, or the first step of a longer part, with the line given as $a x + b y = c$ or as $2 y = \dots$ The equation rearranged to $y = \dots$ with every term divided ( $y = - \frac{3}{5} x + 2$ ), and the gradient read as the coefficient of $x$ with its sign: $- \frac{3}{5}$ , not $3$ or $- 3$ .
Proving that two lines are parallel: A 2-mark Show (that), with at least one line that needs rearranging Both gradients shown, each from a correct rearrangement where one is needed, then the conclusion in words: 'the gradients are both 4, so the lines are parallel'. Writing only '4 and 4' leaves the conclusion unsaid.
Giving the equation in the form the question names: A 1-mark Write the equation in the form $a x + b y = c$ with integers, after you have found it as $y = m x + c$ Every term multiplied by the denominator, $x$ and $y$ on the same side, integers with no common factor: $3 x + 5 y = 5$ , not $5 y = - 3 x + 5$ and not $6 x + 10 y = 10$ .
Lines parallel to an axis: A 1-mark Write down the equation of the line through a given point that is parallel to the $x$ -axis, the $y$ -axis, $y = k$ or $x = k$ Parallel to the $x$ -axis: $y =$ the point's $y$ -coordinate. Parallel to the $y$ -axis: $x =$ the point's $x$ -coordinate. For $(2, - 1)$ that is $y = - 1$ or $x = 2$ , a full equation, not just a number.

Words to know

Parallel lines: Straight lines that point in the same direction and never meet. Parallel lines have equal gradients.
Gradient: How far a line goes up (or down) for every 1 unit across. In $y = m x + c$ it is $m$ .
$y$ -intercept: The $y$ -coordinate of the point where a line crosses the $y$ -axis. In $y = m x + c$ it is $c$ .
Coefficient: The number multiplying a letter. In $y = - 3 x + 4$ the coefficient of $x$ is $- 3$ .
Horizontal line: A line with gradient $0$ and an equation of the form $y = k$ . It is parallel to the $x$ -axis.
Vertical line: A line with an equation of the form $x = k$ , whose gradient cannot be calculated. It is parallel to the $y$ -axis.

All Maths lessons Next: 4.6 Perpendicular lines