Find the gradient of the line joining A(2, 3) and B(6, 11).
(Total for Question 1 is 1 mark)
2
Find the midpoint of the points P(-3, 5) and Q(7, 1).
(Total for Question 2 is 1 mark)
3
Find the distance between the points C(0, 0) and D(6, 8).
(Total for Question 3 is 1 mark)
4
Find the equation of the line with gradient 3 that passes through the point (2, 7). Give your answer in the form y = mx + c.
(Total for Question 4 is 2 marks)
5
Find the equation of the line passing through the points (0, 2) and (4, 14). Give your answer in the form y = mx + c.
(Total for Question 5 is 2 marks)
6
A line has equation y = 4x - 5. Write down (a) the gradient of the line, (b) the y-intercept of the line.
(a)Write down the gradient of the line.(1)
(b)Write down the y-intercept of the line.(1)
(Total for Question 6 is 2 marks)
7
Write the equation 2x + 3y = 12 in the form y = mx + c.
(Total for Question 7 is 1 mark)
8
Determine whether the lines y = 2x + 3 and y = 2x - 7 are parallel, perpendicular, or neither.
(Total for Question 8 is 2 marks)
9
Determine whether the lines y = 3x + 1 and y = -x/3 + 4 are parallel, perpendicular, or neither.
(Total for Question 9 is 2 marks)
10
Find the equation of the line parallel to y = 5x - 2 that passes through the point (1, 9). Give your answer in the form y = mx + c.
(Total for Question 10 is 2 marks)
11
Two points are given: E(-1, 4) and F(3, 4). Find (a) the midpoint of EF, (b) the length of EF.
(a)Find the midpoint of EF.(1)
(b)Find the length of EF.(1)
(Total for Question 11 is 2 marks)
12
Find the midpoint and the length of the line segment joining (-2, -1) and (4, 7).
(a)Find the midpoint of the segment.(1)
(b)Find the length of the segment.(1)
(Total for Question 12 is 2 marks)
13
The line L1 passes through the points A(-1, 2) and B(5, 20). Find the equation of L1 in the form y = mx + c.
(Total for Question 13 is 3 marks)
14
Line L has equation 4x + 2y = 10. A second line M is perpendicular to L and passes through the point (3, -1). Find the equation of M in the form y = mx + c.
(Total for Question 14 is 3 marks)
15
Points P(2, 5), Q(8, 5) and R(8, 13) form a right-angled triangle, right-angled at Q. Find the length of the hypotenuse PR.
(Total for Question 15 is 3 marks)
16
The line y = 2x - 3 crosses the line x + y = 12 at the point T. Find the coordinates of T.
(Total for Question 16 is 3 marks)
17
A line passes through A(-3, 4) and is perpendicular to the line joining B(1, 2) and C(5, -6). Find the equation of the line through A, giving your answer in the form ax + by + c = 0.
(Total for Question 17 is 4 marks)
18
Points C(2, 1), V(10, 7) and L(18, 14) are marked on a coordinate grid. Determine, showing your working, whether the three points lie on a single straight line.
(Total for Question 18 is 4 marks)
Mark scheme · G1D Coordinate Geometry: Straight Lines: Fluency and Exam Drill
Question 1
B1 gradient = 2 cao
Answer: gradient = 2
Question 2
B1 (2, 3) cao
Answer: (2, 3)
Question 3
B1 10 cao
Answer: 10
Question 4
M1 uses y - 7 = 3(x - 2) or substitutes into y = 3x + c
A1 y = 3x + 1 cao
Answer: y = 3x + 1
Question 5
M1 finds the gradient = (14-2)/(4-0) = 3
A1 y = 3x + 2 cao
Answer: y = 3x + 2
Question 6
(a) B1 4 cao
(a) Answer: 4
(b) B1 (0, -5) oe -5 cao
(b) Answer: (0, -5)
Question 7
B1 y = -(2/3)x + 4 oe cao
Answer: y = -(2/3)x + 4
Question 8
M1 identifies that both lines have gradient 2
A1 parallel (equal gradients, different y-intercepts, so distinct lines) cao
Answer: Parallel
Question 9
M1 identifies gradients 3 and -1/3 and computes their product = -1
A1 perpendicular cao
Answer: Perpendicular
Question 10
M1 uses gradient 5 (parallel lines have equal gradients) with the point (1, 9)
A1 y = 5x + 4 cao
Answer: y = 5x + 4
Question 11
(a) B1 (1, 4) cao
(a) Answer: (1, 4)
(b) B1 4 cao
(b) Answer: 4
Question 12
(a) B1 (1, 3) cao
(a) Answer: (1, 3)
(b) B1 10 cao
(b) Answer: 10
Question 13
M1 finds the gradient = (20-2)/(5-(-1)) = 3
M1 uses the gradient with either point in a point-gradient form
A1 y = 3x + 5 oe cao
Answer: y = 3x + 5
Question 14
M1 rearranges L to y = 5 - 2x and identifies its gradient as -2
M1 finds the perpendicular gradient 1/2 and uses it with the point (3, -1)
A1 y = x/2 - 5/2 oe (e.g. y = 0.5x - 2.5) cao
Answer: y = x/2 - 5/2
Question 15
M1 finds PQ = 6 and QR = 8 (or sets up the distance formula for PR directly)
M1 applies Pythagoras' theorem or the distance formula: √62 + 82
A1 PR = 10 cao
Answer: PR = 10
Question 16
M1 substitutes y = 2x - 3 into x + y = 12
M1 solves the resulting equation to find x = 5, dependent on previous M1 (dM1)
A1 T = (5, 7) cao
Answer: T = (5, 7)
Question 17
M1 finds the gradient of BC = (-6-2)/(5-1) = -2
M1 finds the perpendicular gradient = 1/2
M1 forms the line through A using y - 4 = (1/2)(x + 3), dependent on both previous marks (dM1)
A1 x - 2y + 11 = 0 oe cao
Answer: x - 2y + 11 = 0
Question 18
M1 finds the gradient of CV = (7-1)/(10-2) = 3/4
M1 finds the gradient of VL = (14-7)/(18-10) = 7/8
A1 compares the two gradients and states that they are different (3/4 not equal to 7/8)
A1 concludes that the three points do not lie on a single straight line (cso)
Answer: The points are not collinear (gradient CV = 3/4, gradient VL = 7/8, which are different)