Worksheets · Foundation and Higher

Plotting lines, gradient and intercept

8 exam-style questions, grades 2 to 6. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 2 marks
(a) A straight line passes through (1, 3) and (5, 11). Work out its gradient. (2)
Question 2Non-calculator · 3 marks
A straight line has equation $y = 2x - 1$ .
(a) Work out the value of $y$ when $x = -1$ . (1)
(b) Write down the gradient of the line. (1)
(c) Write down the $y$ -coordinate of the point where the line crosses the $y$ -axis. (1)
Question 3Non-calculator · 3 marks
Answer each part.
(a) A line passes through $(1, 4)$ and $(3, 10)$ . Work out its gradient. (2)
(b) Which of these lines is the steepest: $y = 2x + 5$ , $y = 4x - 1$ or $y = x + 10$ ? (1)
1. $y = 2x + 5$
2. $y = 4x - 1$
3. $y = x + 10$
Question 4Non-calculator · 5 marks
A line passes through $(4,5)$ and $(7,8)$ .
(a) Find the equation of the line. (3)
(b) Find the equation of a parallel line through $(0,6)$ . (2)
Question 5Non-calculator · 6 marks
The line L has equation $2y = 6x - 8$ .
(a) Find the gradient of L. (2)
(b) Find the coordinates of the point where L crosses the y-axis. (2)
(c) Does the point (5, 11) lie on L? Show how you decide. (2)
Question 6Non-calculator · 4 marks
A table of values for a straight line gives y = 7, 4, 1 and −2 when x = −1, 0, 1 and 2.
(a) Write down the y-intercept of the line. (1)
(b) Find the gradient of the line. (2)
(c) Write down the equation of the line. (1)
Question 7Non-calculator · 6 marks
The line $L$ passes through the points $(-2, 7)$ and $(4, -5)$ .
(a) Find the gradient of $L$ . (2)
(b) Find an equation of $L$ . (2)
(c) Does the point $(10, -17)$ lie on $L$ ? You must show how you get your answer. (2)
1. Yes
2. No
Question 8Calculator · 2 marks
(a) Find the coordinates of the point where the line 4x + 3y = 18 crosses the x-axis. (2)

Worked solutions and marks

Question 1

(a) $2$

$\frac{11-3}{5-1}$
Gradient = change in y $\div$ change in x.
The gradient is (11 $-$ 3)/(5 $-$ 1) = 8/4 = 2.

P1 Establishing $\frac{11-3}{5-1}$ or an equivalent valid method.
A1 Correct answer: $2$

Question 2

(a) $-3$

$y = 2 \times (-1) - 1 = -2 - 1 = -3$ .

B1 The correct answer, $-3$ .

(b) $2$

In $y = mx + c$ , the gradient is $m$ , the number multiplying $x$ : 2.

B1 The correct answer, $2$ .

(c) $-1$

In $y = mx + c$ , the line crosses the $y$ -axis at $c$ : here $-1$ .

B1 The correct answer, $-1$ .

Question 3

(a) $3$

Change in $y$ : $10 - 4 = 6$ . Change in $x$ : $3 - 1 = 2$ . Gradient $= \frac{6}{2} = 3$ .

M1 Finding both changes, 6 and 2.
A1 The correct answer, $3$ .

(b) $y = 4x - 1$

Steepness is the gradient: 2, 4 and 1. The steepest is $y = 4x - 1$ .

B1 The correct answer, $y = 4x - 1$ .

Question 4

(a) $y=x+1$

Calculate the gradient as rise divided by run.
$\frac{8-5}{3}$
Substitute a known point to find the intercept.
$5-1\times 4$
Therefore $y=x+1$ .

M1 Calculate the gradient as rise divided by run.
M1 Substitute a known point to find the intercept.
A1 Correct answer: $y=x+1$

(b) $y=x+6$

Parallel lines have equal gradients.
$y=x+6$
Therefore $y=x+6$ .

M1 Parallel lines have equal gradients.
A1 Correct answer: $y=x+6$

Question 5

(a) $3$

Divide every term by 2 to make y the subject.
$6/2$
Therefore $3$ .

M1 Divide every term by 2 to make y the subject.
A1 Correct answer: $3$

(b) $(0,-4)$

Substitute x = 0.
$2y=-8$
Therefore $(0,-4)$ .

M1 Substitute x = 0.
A1 Correct answer: $(0,-4)$

(c) Yes: when x = 5, y = 3 × 5 − 4 = 11.

Substitute x = 5 into y = 3x - 4.
$3\times 5-4$
Yes: when x = 5, y = 3 × 5 − 4 = 11.

M1 Substitute x = 5 into y = 3x - 4.
C1 Correct conclusion with supporting reasoning: Yes: when x = 5, y = 3 × 5 − 4 = 11.

Question 6

(a) $4$

Therefore $4$ .

B1 Correct answer: $4$

(b) $-3$

Divide the change in y by the change in x.
$\frac{1-4}{1-0}$
Therefore $-3$ .

M1 Divide the change in y by the change in x.
A1 Correct answer: $-3$

(c) $y=-3x+4$

Therefore $y=-3x+4$ .

B1 Correct answer: $y=-3x+4$

Question 7

(a) $-2$

$\frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2$

M1 Writing $\frac{-5 - 7}{4 - (-2)}$ or equivalent.
A1 The correct answer, $-2$ .

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

Substitute a point into $y = -2x + c$ .
$7 = -2(-2) + c = 4 + c \Rightarrow c = 3$
So $y = -2x + 3$ .

M1 Substituting a point into $y = -2x + c$ .
A1 The correct answer, $y = -2x + 3$ .

(c) Yes: $-2(10) + 3 = -17$ .

When $x = 10$ : $y = -2 \times 10 + 3 = -17$ . This matches, so the point lies on $L$ .

M1 Substituting $x = 10$ into their equation.
C1 "Yes", supported by $y = -17$ when $x = 10$ .

Question 8

(a) $(4.5,0)$

$4x=18$
On the x-axis, y = 0.
The equation becomes 4x = 18, so x = 18/4 = 4.5.
The crossing point is (4.5, 0).

P1 Establishing $4x=18$ or an equivalent valid method.
A1 Correct answer: $(4.5,0)$

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Plotting lines, gradient and intercept

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