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6.6 Parallel and Perpendicular Lines

EDEXCEL 1MA1 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A = (2, 3), B = (6, 11)
C = (-1, 5), D = (3, -3)

(a)Work out the gradient of the line through A and B.(2)

(b)Work out the gradient of the line through C and D.(2)

(Total for Question 1 is 4 marks)

A straight line has equation y = 4x - 7

(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)

(c)Show that the point (3, 5) lies on the line.(2)

(Total for Question 2 is 4 marks)

Which of these lines is parallel to y = 5 - 3x?
A) y = 3x + 5
B) y = -3x + 2
C) y = (1/3)x - 5
D) y = 3x - 5

A) y = 3x + 5
B) y = -3x + 2
C) y = (1/3)x - 5
D) y = 3x - 5

(Total for Question 3 is 1 mark)

A line has gradient -3/4. What is the gradient of a line perpendicular to it?
A) -3/4
B) 3/4
C) -4/3
D) 4/3

A) -3/4
B) 3/4
C) -4/3
D) 4/3

(Total for Question 4 is 1 mark)

A straight line has equation 3x + 2y = 12

(a)Rearrange the equation into the form y = mx + c(2)

(b)State the gradient of the line and the coordinates of the point where it crosses the y-axis.(2)

(Total for Question 5 is 4 marks)

For each pair of lines below, state whether the lines are parallel, perpendicular or neither, giving a reason based on their gradients.

(a)Line P has gradient 5. Line Q has gradient 5.(1)

(b)Line R has gradient 2. Line S has gradient -1/2.(1)

(c)Line T has gradient 3. Line U has gradient 1/3.(1)

(Total for Question 6 is 3 marks)

A straight line passes through the points (-3, 7) and (5, -1).
Find the equation of the line, giving your answer in the form y = mx + c

(Total for Question 7 is 3 marks)

Show that the line y = 2x - 5 is parallel to the line with equation 6x - 3y + 9 = 0

(Total for Question 8 is 3 marks)

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

(Total for Question 9 is 3 marks)

A straight line has equation y = -(2/5)x + 3

(a)State the gradient of a line that is perpendicular to this line.(1)

(b)Find the equation of the line perpendicular to y = -(2/5)x + 3 that passes through the point (-4, 1). Give your answer in the form y = mx + c(3)

(Total for Question 10 is 4 marks)

Three straight lines are given below.
L1: y = 4x + 1
L2: 8x - 2y = 5
L3: x + 4y = 8

(a)Show that L1 and L2 are parallel.(2)

(b)Show that L1 and L3 are perpendicular.(2)

(c)State the relationship between L2 and L3, giving a reason for your answer.(2)

(Total for Question 11 is 6 marks)

P and Q are the points (1, 2) and (7, 6).

(a)Find the midpoint of PQ.(2)

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

(c)Find the equation of the perpendicular bisector of PQ. Give your answer in the form ax + by = c, where a, b and c are integers.(3)

(Total for Question 12 is 7 marks)

Triangle ABC has vertices A(-2, -1), B(4, 1) and C(2, 7).
Show that triangle ABC has a right angle at B.

(Total for Question 13 is 4 marks)

Quadrilateral WXYZ has vertices W(0, 1), X(4, 3), Y(6, 0) and Z(2, -2).
Show that WXYZ is a parallelogram.

(Total for Question 14 is 5 marks)

On a coordinate grid, distances are measured in kilometres from a fixed reference point. Road A passes through the villages at J(1, 2) and K(5, 10).

(a)Find the gradient of Road A.(2)

(b)A new road, Road B, is to be built through the junction at M(3, 4), parallel to Road A. Find the equation of Road B.(3)

(c)A third road, Road C, must be perpendicular to Road A and pass through the point N(9, 1). Find the equation of Road C, giving your answer in the form ax + by + c = 0, where a, b and c are integers.(4)

(Total for Question 15 is 9 marks)

The line joining the points A(k, 3) and B(5, 11) is parallel to the line with equation y = 4x + 2
Find the value of k.

(Total for Question 16 is 3 marks)

Line l has equation 5x - 2y = 8
Point A has coordinates (4, -3).
Find the equation of the line perpendicular to l that passes through A. Give your answer in the form y = mx + c

(Total for Question 17 is 4 marks)

Show that the points A(1, 1), B(4, 7) and C(6, 11) lie on the same straight line.

(Total for Question 18 is 3 marks)

Line l1 has equation y = 2x - 1
Line l2 is perpendicular to l1 and passes through the point (6, 5).

(a)Find the equation of l2.(3)

(b)Find the coordinates of the point where l1 and l2 intersect.(2)

(Total for Question 19 is 5 marks)

Mark scheme · 6.6 Parallel and Perpendicular Lines

Question 1

(a) M1 correct substitution into the gradient formula, e.g. (11 - 3)/(6 - 2) (oe)
(a) A1 2 (cao)
(a) Answer: 2
(b) M1 correct substitution into the gradient formula, e.g. (-3 - 5)/(3 - (-1)) (oe)
(b) A1 -2 (cao)
(b) Answer: -2

Question 2

(a) B1 4 (cao)
(a) Answer: 4
(b) B1 (0, -7) (cao)
(b) Answer: (0, -7)
(c) M1 substitutes x = 3 into y = 4x - 7
(c) A1 correctly shows y = 5, matching the given point (cso)
(c) Answer: (3, 5) lies on the line since y = 4(3) - 7 = 5

Question 3

B1 B selected (cao)
Answer: B (y = -3x + 2)

Question 4

B1 D selected (cao)
Answer: D (4/3)

Question 5

(a) M1 correct first step, e.g. 2y = -3x + 12 (oe)
(a) A1 y = -1.5x + 6 (oe, e.g. y = -(3/2)x + 6) (cao)
(a) Answer: y = -1.5x + 6
(b) B1 gradient = -1.5 (oe -3/2), ft their part (a)
(b) B1 y-intercept at (0, 6), ft their part (a)
(b) Answer: gradient = -3/2, y-intercept = (0, 6)

Question 6

(a) B1 parallel, since the gradients are equal (cao)
(a) Answer: Parallel (equal gradients)
(b) B1 perpendicular, since 2 x (-1/2) = -1 (cao)
(b) Answer: Perpendicular (gradients multiply to -1)
(c) B1 neither, since the gradients are not equal and 3 x (1/3) = 1, not -1 (cao)
(c) Answer: Neither (gradients not equal, and product is 1, not -1)

Question 7

M1 correct gradient found, e.g. (-1 - 7)/(5 - (-3)) = -1 (oe)
M1 correct method to find c using their gradient and one of the given points
A1 y = -x + 4 (oe, e.g. y = 4 - x) (cao)
Answer: y = -x + 4

Question 8

M1 rearranges 6x - 3y + 9 = 0 into the form y = mx + c
M1 states both gradients are 2
C1 correct conclusion with full reasoning: both lines have gradient 2 and different y-intercepts (-5 and 3), so they are parallel and distinct (cso)
Answer: Parallel, since both lines have gradient 2

Question 9

M1 identifies the gradient as 3 (parallel lines have equal gradients)
M1 correct method, e.g. y - 5 = 3(x - 2)
A1 y = 3x - 1 (cao)
Answer: y = 3x - 1

Question 10

(a) B1 5/2 (oe 2.5) (cao)
(a) Answer: 5/2
(b) M1 uses the correct perpendicular gradient of 5/2, ft their part (a)
(b) M1 correct method, e.g. y - 1 = (5/2)(x - (-4))
(b) A1 y = (5/2)x + 11 (oe, e.g. y = 2.5x + 11) (cao)
(b) Answer: y = (5/2)x + 11

Question 11

(a) M1 rearranges L2 into the form y = mx + c, y = 4x - 2.5 (oe)
(a) C1 correct conclusion: both lines have gradient 4 and different y-intercepts, so L1 and L2 are parallel (cso)
(a) Answer: L1 and L2 are parallel (both have gradient 4)
(b) M1 rearranges L3 into the form y = mx + c, y = -0.25x + 2 (oe), and multiplies gradients
(b) C1 correct conclusion: 4 x (-1/4) = -1, so L1 and L3 are perpendicular (cso)
(b) Answer: L1 and L3 are perpendicular (gradients multiply to -1)
(c) M1 identifies gradient of L2 as 4 and gradient of L3 as -1/4, ft their parts (a) and (b)
(c) A1 perpendicular, since 4 x (-1/4) = -1 (cao)
(c) Answer: L2 and L3 are perpendicular

Question 12

(a) M1 correct method, e.g. ((1 + 7)/2, (2 + 6)/2)
(a) A1 (4, 4) (cao)
(a) Answer: (4, 4)
(b) M1 correct substitution, e.g. (6 - 2)/(7 - 1)
(b) A1 2/3 (oe) (cao)
(b) Answer: 2/3
(c) M1 finds the perpendicular gradient -3/2, ft their part (b)
(c) M1 correct method using the midpoint from part (a), e.g. y - 4 = -(3/2)(x - 4)
(c) A1 3x + 2y = 20 (oe, integer coefficients) (cao)
(c) Answer: 3x + 2y = 20

Question 13

M1 correct gradient of AB, e.g. (1 - (-1))/(4 - (-2)) = 1/3 (oe)
M1 correct gradient of BC, e.g. (7 - 1)/(2 - 4) = -3 (oe)
M1 multiplies the two gradients together
C1 correct conclusion with full reasoning: (1/3) x (-3) = -1, so AB is perpendicular to BC, giving a right angle at B (cso)
Answer: Right angle at B confirmed, since gradient AB x gradient BC = -1

Question 14

M1 correct gradient of WX, e.g. (3 - 1)/(4 - 0) = 1/2 (oe)
A1 correct gradient of ZY = 1/2 (oe), with statement that WX is parallel to ZY
M1 correct gradient of XY, e.g. (0 - 3)/(6 - 4) = -3/2 (oe)
A1 correct gradient of WZ = -3/2 (oe), with statement that XY is parallel to WZ
C1 correct overall conclusion: since both pairs of opposite sides are parallel, WXYZ is a parallelogram (cso)
Answer: WXYZ is a parallelogram, since WX is parallel to ZY and XY is parallel to WZ

Question 15

(a) M1 correct substitution, e.g. (10 - 2)/(5 - 1)
(a) A1 2 (cao)
(a) Answer: 2
(b) M1 uses gradient 2 for Road B (parallel to Road A), ft their part (a)
(b) M1 correct method, e.g. y - 4 = 2(x - 3)
(b) A1 y = 2x - 2 (cao)
(b) Answer: y = 2x - 2
(c) P1 identifies the perpendicular gradient as -1/2, ft their part (a)
(c) M1 correct method using point N(9, 1), e.g. y - 1 = -(1/2)(x - 9)
(c) M1 rearranges into integer form with all terms on one side
(c) A1 x + 2y - 11 = 0 (oe, integer coefficients) (cao)
(c) Answer: x + 2y - 11 = 0

Question 16

M1 sets up the gradient of AB in terms of k, e.g. (11 - 3)/(5 - k), and equates to 4
M1 correct algebraic method to solve for k, e.g. 8 = 4(5 - k)
A1 k = 3 (cao)
Answer: k = 3

Question 17

M1 rearranges l into the form y = mx + c and identifies its gradient as 5/2
M1 finds the perpendicular gradient as -2/5
M1 correct method using point A, e.g. y - (-3) = -(2/5)(x - 4)
A1 y = -(2/5)x - 7/5 (oe, e.g. y = -0.4x - 1.4) (cao)
Answer: y = -(2/5)x - 7/5

Question 18

M1 correct gradient of AB, e.g. (7 - 1)/(4 - 1) = 2 (oe)
M1 correct gradient of BC, e.g. (11 - 7)/(6 - 4) = 2 (oe)
C1 correct conclusion with full reasoning: gradient AB = gradient BC and B is a common point, so A, B and C are collinear (cso)
Answer: A, B and C are collinear, since gradient AB = gradient BC = 2

Question 19

(a) M1 finds the perpendicular gradient as -1/2
(a) M1 correct method using the point (6, 5), e.g. y - 5 = -(1/2)(x - 6)
(a) A1 y = -(1/2)x + 8 (oe, e.g. y = -0.5x + 8) (cao)
(a) Answer: y = -(1/2)x + 8
(b) P1 sets the two equations equal to each other, e.g. 2x - 1 = -(1/2)x + 8, ft their part (a)
(b) A1 (3.6, 6.2) (oe, e.g. (18/5, 31/5)), both coordinates correct (cao)
(b) Answer: (3.6, 6.2)

Parallel and Perpendicular Lines - Worksheets, Questions and Revision