Write down the gradient of the line with equation y = 5x + 1.
(Total for Question 1 is 1 mark)
2
A line has equation y = -2x + 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)
(Total for Question 2 is 2 marks)
3
A line has equation 2y = 6x - 4.
(a)Rearrange the equation into the form y = mx + c.(1)
(b)State the gradient of the line.(1)
(Total for Question 3 is 2 marks)
4
Line A has equation y = 4x + 3. Line B has equation y = 4x - 5. State, giving a reason, whether lines A and B are parallel.
(Total for Question 4 is 2 marks)
5
Find the equation of the line that is parallel to y = 3x + 2 and passes through the point (0, -4). Give your answer in the form y = mx + c.
(Total for Question 5 is 2 marks)
6
Find the equation of the line that is parallel to y = 2x - 1 and passes through the point (3, 10). Give your answer in the form y = mx + c.
(Total for Question 6 is 3 marks)
7
A line has gradient 3/4. Find the gradient of a line that is perpendicular to it.
(Total for Question 7 is 2 marks)
8
Find the equation of the line perpendicular to y = 2x + 5 that passes through the point (4, 1). Give your answer in the form y = mx + c.
(Total for Question 8 is 4 marks)
9
Line L has equation 3x + y = 6. Find the gradient of a line that is perpendicular to L.
(Total for Question 9 is 3 marks)
10
A line has equation 4x + 2y = 10.
(a)Rearrange the equation into the form y = mx + c.(2)
(b)State the gradient of a line that is parallel to this line.(1)
(c)State the gradient of a line that is perpendicular to this line.(1)
(Total for Question 10 is 4 marks)
11
Points A(1, 2) and B(5, 10) lie on a straight line. Find the gradient of the line AB.
(Total for Question 11 is 2 marks)
12
Find the equation of the straight line that passes through the points (2, -1) and (6, 7). Give your answer in the form y = mx + c.
(Total for Question 12 is 4 marks)
13
A line passes through the points (0, 1) and (4, 9). Determine, showing your working, whether this line is parallel to, perpendicular to, or neither parallel nor perpendicular to the line with equation y = -1/4 x + 3.
(Total for Question 13 is 3 marks)
14
C has coordinates (-2, 5) and D has coordinates (2, -3). Show that the line CD is perpendicular to the line with equation y = 1/2 x + 4.
(Total for Question 14 is 3 marks)
15
A(1, 4) and B(7, -2) are the endpoints of a line segment AB.
(a)Find the coordinates of the midpoint M of AB.(2)
(b)Find the gradient of AB.(2)
(c)Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.(3)
(Total for Question 15 is 7 marks)
16
The points W(0, 0), X(4, 2), Y(6, -2) and Z(2, -4) are the vertices of quadrilateral WXYZ. Show that WXYZ is a rectangle. You must show gradient calculations for at least two pairs of sides as part of your reasoning.
(Total for Question 16 is 5 marks)
17
Line L1 has equation y = 2x - 3. Line L2 is perpendicular to L1 and passes through the point where L1 crosses the y-axis. Find the equation of L2, giving your answer in the form y = mx + c.
(Total for Question 17 is 4 marks)
18
Line L1 has equation kx + 3y = 12. Line L2 has equation 2x - y = 7. Given that L1 is perpendicular to L2, find the value of k.
(Total for Question 18 is 5 marks)
19
The lines y = (2k - 1)x + 5 and y = 7x - 3 are parallel. Find the value of k.
(Total for Question 19 is 2 marks)
20
A circle has centre O(0, 0) and passes through the point P(6, 8). Find the equation of the tangent to the circle at the point P. Give your answer in the form y = mx + c.
(Total for Question 20 is 5 marks)
Mark scheme · 6.6D Parallel and Perpendicular Lines: Fluency and Exam Drill
Question 1
B1 5 cao
Answer: 5
Question 2
(a) B1 -2 cao
(a) Answer: -2
(b) B1 (0, 7) cao
(b) Answer: (0, 7)
Question 3
(a) B1 y = 3x - 2 oe
(a) Answer: y = 3x - 2
(b) B1 3 ft from part (a)
(b) Answer: 3
Question 4
B1 gradient of A = gradient of B (both = 4) identified
B1 correct conclusion: parallel, with valid reason (dep on first mark)
Answer: Yes, A and B are parallel because they both have gradient 4.
Question 5
M1 gradient = 3 stated or used (parallel lines have equal gradient)
A1 y = 3x - 4 oe cao
Answer: y = 3x - 4
Question 6
M1 gradient = 2 identified
M1 10 = 2(3) + c oe substituted
A1 y = 2x + 4 cao
Answer: y = 2x + 4
Question 7
M1 negative reciprocal method shown, e.g. -1 divided by (3/4)
M1 gradient of the line through the points = (9 - 1) / (4 - 0) oe
A1 gradient = 2
B1 correct conclusion 'neither', supported by comparison (2 does not equal -1/4, and 2 x (-1/4) = -1/2, not -1)
Answer: Neither parallel nor perpendicular.
Question 14
M1 gradient of CD = (-3 - 5) / (2 - (-2)) oe
A1 gradient of CD = -2
B1 (-2) x (1/2) = -1 stated, so lines are perpendicular (dep on correct gradient)
Answer: Shown: gradient of CD = -2; (-2) x (1/2) = -1, so CD is perpendicular to y = 1/2 x + 4.
Question 15
(a) M1 ((1 + 7)/2, (4 + (-2))/2) oe method shown
(a) A1 (4, 1) cao
(a) Answer: (4, 1)
(b) M1 (-2 - 4) / (7 - 1) oe
(b) A1 -1 cao
(b) Answer: -1
(c) M1 perpendicular gradient = 1 ft from part (b)
(c) M1 y - 1 = 1(x - 4) oe using midpoint from part (a), ft
(c) A1 y = x - 3 cao
(c) Answer: y = x - 3
Question 16
M1 gradient WX = (2 - 0) / (4 - 0) = 1/2 oe
M1 gradient XY = (-2 - 2) / (6 - 4) = -2 oe
B1 gradient YZ = 1/2 and gradient ZW = -2 both found, showing WX // YZ and XY // ZW
M1 (1/2) x (-2) = -1 identified, showing adjacent sides WX and XY are perpendicular
A1 correct conclusion: opposite sides parallel and adjacent sides meet at right angles, so WXYZ is a rectangle (dep on all previous marks)
Answer: WXYZ is a rectangle: gradient WX = gradient YZ = 1/2 and gradient XY = gradient ZW = -2, so opposite sides are parallel; and (1/2) x (-2) = -1, so adjacent sides are perpendicular.
Question 17
B1 L1 crosses the y-axis at (0, -3)
M1 perpendicular gradient = -1/2 oe
M1 -3 = -1/2 (0) + c oe, i.e. c = -3 identified
A1 y = -1/2 x - 3 oe cao
Answer: y = -0.5x - 3
Question 18
M1 rearrange L2 to y = 2x - 7 oe, gradient of L2 = 2 identified
M1 rearrange L1 to y = -k/3 x + 4 oe
M1 perpendicular condition set up: (-k/3) x 2 = -1 oe
M1 -2k/3 = -1 solved to 2k = 3 oe
A1 k = 3/2 oe (1.5) cao
Answer: k = 3/2
Question 19
M1 2k - 1 = 7 oe set up using equal gradients
A1 k = 4 cao
Answer: k = 4
Question 20
M1 gradient of OP = (8 - 0) / (6 - 0) oe
A1 gradient of OP = 4/3
M1 tangent gradient = -3/4 identified (radius is perpendicular to tangent)
M1 8 = -3/4 (6) + c oe substituted
A1 y = -3/4 x + 25/2 oe cao (accept y = -0.75x + 12.5)