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GCSE · Algebra: straight-line graphs

5.15D Equation of a Line: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A straight line has equation y = 5x - 3.

(a)Write down the gradient of the line.(1)

(b)Write down the coordinates of the point where the line crosses the y-axis.(1)

(Total for Question 1 is 2 marks)

Find the gradient of the line joining the points A(2, 5) and B(6, 13).

(Total for Question 2 is 2 marks)

Find the coordinates of the midpoint of the line segment joining the points (-3, 7) and (5, -1).

(Total for Question 3 is 2 marks)

A straight line has gradient -2 and passes through the point (3, 1). Find the equation of the line, giving your answer in the form y = mx + c.

(Total for Question 4 is 3 marks)

A straight line passes through the points (2, 7) and (5, -2). Find the equation of the line, giving your answer in the form y = mx + c.

(Total for Question 5 is 3 marks)

A straight line has equation 3x + 2y = 16. Write the equation in the form y = mx + c, and state the gradient and the y-intercept of the line.

(Total for Question 6 is 3 marks)

A straight line has equation 2x + 5y = 20.

(a)Find the coordinates of the point where the line crosses the x-axis.(2)

(b)Find the coordinates of the point where the line crosses the y-axis.(1)

(Total for Question 7 is 3 marks)

A candle is lit and burns down at a constant rate. The height, h cm, of the candle t minutes after being lit is modelled by h = 20 - 0.4t.

(a)State what the value 20 represents in this context.(1)

(b)State what the gradient of the line represents in this context.(1)

(c)Find the height of the candle 25 minutes after it is lit.(2)

(Total for Question 8 is 4 marks)

Line L1 has equation y = 5x - 2. Line L2 has equation 10x - 2y + 4 = 0. Show that L1 and L2 are parallel but are not the same line.

(Total for Question 9 is 3 marks)

Find the equation of the straight line that is parallel to y = 3x - 5 and passes through the point (-2, 4). Give your answer in the form y = mx + c.

(Total for Question 10 is 3 marks)

A is the point (1, 2) and B is the point (6, 9). Calculate the length of AB, giving your answer correct to 3 significant figures.

(Total for Question 11 is 3 marks)

Determine whether the point (5, 12) lies on the line with equation y = 2x + 3. You must show your working.

(Total for Question 12 is 3 marks)

A line has gradient 2/5. Find the gradient of a line that is perpendicular to it.

(Total for Question 13 is 2 marks)

Line L has equation y = 2x - 1. Find the equation of the line that is perpendicular to L and passes through the point (4, 9). Give your answer in the form y = mx + c.

(Total for Question 14 is 4 marks)

Show that the points A(1, 1), B(3, 5) and C(6, 11) are collinear.

(Total for Question 15 is 4 marks)

R is the point (-4, 1) and S is the point (2, 9). Find the equation of the perpendicular bisector of RS. Give your answer in the form y = mx + c.

(Total for Question 16 is 5 marks)

The line L1 passes through the points A(-1, -2) and B(3, 6). The line L2 passes through the points B(3, 6) and D(7, 4).

(a)Find the gradient of L1.(2)

(b)Find the gradient of L2.(2)

(c)Hence show that L1 and L2 are perpendicular at the point B.(2)

(Total for Question 17 is 6 marks)

A quadrilateral has vertices A(0, 1), B(4, 3), C(6, -1) and D(2, -3).

(a)Show that AB is parallel to DC.(3)

(b)Show that AD is perpendicular to AB.(3)

(Total for Question 18 is 6 marks)

Mark scheme · 5.15D Equation of a Line: Fluency and Exam Drill

Question 1

(a) B1 5 cao
(a) Answer: 5
(b) B1 (0, -3) cao
(b) Answer: (0, -3)

Question 2

M1 correct substitution into (y2 - y1)/(x2 - x1), e.g. (13-5)/(6-2)
A1 2 cao
Answer: 2

Question 3

M1 correct method, e.g. ((-3+5)/2, (7+(-1))/2)
A1 (1, 3) cao
Answer: (1, 3)

Question 4

M1 uses y - 1 = -2(x - 3), or substitutes (3, 1) into y = -2x + c
A1 c = 7 found (or equivalent correct intermediate step)
A1 y = -2x + 7 oe cao
Answer: y = -2x + 7

Question 5

M1 finds the gradient, m = (-2-7)/(5-2) = -3
M1 substitutes a point into y = -3x + c to find c (or uses y - y1 = m(x - x1))
A1 y = -3x + 13 oe cao
Answer: y = -3x + 13

Question 6

M1 rearranges to make y the subject, e.g. 2y = -3x + 16
A1 y = -3/2 x + 8 oe cao
B1 gradient = -3/2 (or -1.5) and y-intercept = (0, 8) stated, ft their equation
Answer: y = -1.5x + 8; gradient = -1.5; y-intercept = (0, 8)

Question 7

(a) M1 sets y = 0 and solves, 2x = 20
(a) A1 (10, 0) cao
(a) Answer: (10, 0)
(b) B1 (0, 4) cao
(b) Answer: (0, 4)

Question 8

(a) B1 the height of the candle when it is lit (at t = 0) oe
(a) Answer: The starting height of the candle, 20 cm, at the moment it is lit.
(b) B1 the rate at which the candle burns down, 0.4 cm per minute oe
(b) Answer: The candle burns down at 0.4 cm per minute.
(c) M1 substitutes t = 25 into h = 20 - 0.4t
(c) A1 10 cm cao
(c) Answer: 10 cm

Question 9

M1 rearranges L2 into the form y = mx + c
A1 y = 5x + 2
B1 (dep on both marks) correct conclusion: both lines have gradient 5, so they are parallel; the y-intercepts are different (-2 and 2), so they are not the same line
Answer: L2 rearranges to y = 5x + 2. Both lines have gradient 5, so they are parallel, but the y-intercepts (-2 and 2) differ, so they are not the same line.

Question 10

B1 gradient = 3 stated (parallel lines have equal gradients)
M1 substitutes (-2, 4) into y = 3x + c (or uses y - 4 = 3(x + 2))
A1 y = 3x + 10 oe cao
Answer: y = 3x + 10

Question 11

M1 correct substitution into the distance formula, √(6-1)² + (9-2)²
A1 √74 oe seen
A1 awrt 8.60
Answer: 8.60

Question 12

M1 substitutes x = 5 into y = 2x + 3
A1 y = 13 found
B1 (dep) correct conclusion: since 13 is not equal to 12, the point (5, 12) does not lie on the line
Answer: The point (5, 12) does not lie on the line, since y = 2(5) + 3 = 13, not 12.

Question 13

M1 uses the negative reciprocal, -1 divided by (2/5)
A1 -5/2 oe (accept -2.5) cao
Answer: -5/2

Question 14

B1 gradient of L = 2 identified
M1 perpendicular gradient = -1/2 (negative reciprocal)
M1 substitutes (4, 9) into y = -1/2 x + c (or uses y - 9 = -1/2(x - 4))
A1 y = -1/2 x + 11 oe cao
Answer: y = -1/2 x + 11

Question 15

M1 finds the gradient of AB
A1 gradient of AB = 2
M1 finds the gradient of BC
A1 (dep on all previous marks) gradient of BC = 2; since gradient AB = gradient BC = 2 and B is a common point to both segments, A, B and C are collinear
Answer: Gradient AB = gradient BC = 2, and B is common to both, so A, B and C are collinear.

Question 16

M1 finds the midpoint of RS, (-1, 5)
M1 finds the gradient of RS, 4/3
M1 finds the perpendicular gradient, -3/4
M1 substitutes the midpoint into y = -3/4 x + c (or uses y - 5 = -3/4(x + 1))
A1 y = -3/4 x + 17/4 oe cao
Answer: y = -3/4 x + 17/4 (or y = -0.75x + 4.25)

Question 17

(a) M1 correct substitution, (6-(-2))/(3-(-1))
(a) A1 2 cao
(a) Answer: 2
(b) M1 correct substitution, (4-6)/(7-3)
(b) A1 -1/2 cao
(b) Answer: -1/2
(c) M1 multiplies their two gradients together, ft their answers to (a) and (b)
(c) A1 product = -1, so L1 and L2 are perpendicular at B, cao
(c) Answer: 2 x (-1/2) = -1, so L1 is perpendicular to L2 at B.

Question 18

(a) M1 finds gradient of AB
(a) M1 finds gradient of DC
(a) A1 both gradients equal 1/2, so AB is parallel to DC
(a) Answer: Gradient AB = gradient DC = 1/2, so AB is parallel to DC.
(b) M1 finds gradient of AD
(b) M1 multiplies gradient of AD by gradient of AB (ft their part a gradient)
(b) A1 product = -1, so AD is perpendicular to AB
(b) Answer: Gradient AD = -2. -2 x 1/2 = -1, so AD is perpendicular to AB.

Equation of a Line: Fluency and Exam Drill - Worksheets, Questions and Revision