Worksheets · Foundation and Higher

Equations of lines and parallel lines

8 exam-style questions, grades 4 to 6. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 3 marks
Answer each part.
(a) Write down the equation of the line that is parallel to $y = 3x + 2$ and passes through $(0, -5)$ . (2)
(b) Which line is parallel to $y = 4x - 1$ ? (1)
1. $y = 4x + 3$
2. $y = x - 1$
3. $y = -4x - 1$
4. $y = \frac{1}{4}x$
Question 2Non-calculator · 3 marks
A straight line has gradient $-1$ and passes through $(0, 2)$ .
(a) Write down the equation of the line. (1)
(b) Does the point $(5, -3)$ lie on the line? You must show how you get your answer. (2)
1. Yes
2. No
Question 3Non-calculator · 5 marks
A line passes through $(4,10)$ and $(7,13)$ .
(a) Find the equation of the line. (3)
(b) Find the equation of a parallel line through $(0,11)$ . (2)
Question 4Non-calculator · 3 marks
Line A has equation $y = 5 - 2x$ .
(a) Write down the gradient of line A. (1)
(b) Line B is parallel to line A and passes through (3, 1). Find the equation of line B. (2)
Question 5Non-calculator · 4 marks
Lines L1 and L2 have equations $y = 3x - 2$ and $6x - 2y = 7$ .
(a) Show that L1 and L2 are parallel. (2)
(b) Find the coordinates of the point where L2 crosses the x-axis. (2)
Question 6Non-calculator · 4 marks
A straight line passes through the points (−3, 11) and (2, 1).
(a) Find the gradient of the line. (2)
(b) Find the equation of the line in the form y = mx + c. (2)
Question 7Non-calculator · 3 marks
A straight line passes through the points $(2, 1)$ and $(4, 7)$ .
(a) Find the equation of the line. (3)
Question 8Non-calculator · 5 marks
Line $L_1$ has equation $2y - 6x = 8$ . Line $L_2$ is parallel to $L_1$ and passes through the point $(4, 5)$ .
(a) Find an equation of $L_2$ . (3)
(b) Find the coordinates of the point where $L_2$ crosses the $x$ -axis. (2)

Worked solutions and marks

Question 1

(a) $y = 3x - 5$

Parallel lines have the same gradient, 3. The line passes through $(0, -5)$ , so it crosses the $y$ -axis at $-5$ : $y = 3x - 5$ .

B1 Gradient 3 in an equation $y = 3x + c$ .
B1 The correct answer, $y = 3x - 5$ .

(b) $y = 4x + 3$

Parallel lines have equal gradients. Only $y = 4x + 3$ has gradient 4.

B1 The correct answer, $y = 4x + 3$ .

Question 2

(a) $y = -x + 2$

Gradient $-1$ and $y$ -intercept 2: $y = -x + 2$ .

B1 $y = -x + 2$ (or $y = 2 - x$ ).

(b) Yes: $-5 + 2 = -3$ .

When $x = 5$ : $y = -5 + 2 = -3$ . This matches, so the point is on the line.

M1 Substituting $x = 5$ .
C1 "Yes", supported by $y = -3$ when $x = 5$ .

Question 3

(a) $y=x+6$

Calculate the gradient as rise divided by run.
$\frac{13-10}{3}$
Substitute a known point to find the intercept.
$10-1\times 4$
Therefore $y=x+6$ .

M1 Calculate the gradient as rise divided by run.
M1 Substitute a known point to find the intercept.
A1 Correct answer: $y=x+6$

(b) $y=x+11$

Parallel lines have equal gradients.
$y=x+11$
Therefore $y=x+11$ .

M1 Parallel lines have equal gradients.
A1 Correct answer: $y=x+11$

Question 4

(a) $-2$

Therefore $-2$ .

B1 Correct answer: $-2$

(b) $y=-2x+7$

Substitute the point into y = −2x + c.
$1=-2\times 3+c$
Therefore $y=-2x+7$ .

M1 Substitute the point into y = −2x + c.
A1 Correct answer: $y=-2x+7$

Question 5

(a) L2 rearranges to y = 3x - 3.5, so both lines have gradient 3.

Rearrange L2 into the form y = mx + c.
$2y=6x-7$
Therefore L2 rearranges to y = 3x - 3.5, so both lines have gradient 3.

M1 Rearrange L2 into the form y = mx + c.
C1 Correct conclusion with supporting reasoning: L2 rearranges to y = 3x - 3.5, so both lines have gradient 3.

(b) $(7/6,0)$

Substitute y = 0 into the equation of L2.
$6x=7$
Therefore $(7/6,0)$ .

M1 Substitute y = 0 into the equation of L2.
A1 Correct answer: $(7/6,0)$

Question 6

(a) $-2$

Divide the change in y by the change in x.
$(1-11)/(2-(-3))$
Therefore $-2$ .

M1 Divide the change in y by the change in x.
A1 Correct answer: $-2$

(b) $y=-2x+5$

Substitute one point to find c.
$1=-2\times 2+c$
Therefore $y=-2x+5$ .

M1 Substitute one point to find c.
A1 Correct answer: $y=-2x+5$

Question 7

(a) $y = 3x - 5$

Find the gradient.
$m = \frac{7 - 1}{4 - 2} = 3$
Substitute a point into $y = 3x + c$ .
$1 = 3 \times 2 + c \Rightarrow c = -5$
So $y = 3x - 5$ . Check with $(4, 7)$ : $12 - 5 = 7$ .

M1 Finding the gradient, 3.
M1 Substituting a point to find $c$ : $1 = 6 + c$ .
A1 The correct answer, $y = 3x - 5$ .

Question 8

(a) $y = 3x - 7$

Rearrange $L_1$ into $y = mx + c$ .
$2y = 6x + 8 \Rightarrow y = 3x + 4$
$L_2$ has gradient 3. Substitute $(4, 5)$ .
$5 = 3 \times 4 + c \Rightarrow c = -7$

P1 Rearranging $L_1$ to find the gradient 3.
P1 Substituting $(4, 5)$ into $y = 3x + c$ .
A1 The correct answer, $y = 3x - 7$ .

(b) $\left(\frac{7}{3}, 0\right)$

On the $x$ -axis, $y = 0$ : $0 = 3x - 7$ , so $x = \frac{7}{3}$ .

M1 Setting $y = 0$ .
A1 $\left(\frac{7}{3}, 0\right)$ .

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Equations of lines and parallel lines

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