Look at the point P(-6, 3). State which quadrant P lies in.
(1)
2
Reflect the point R(5, -3) in the x-axis. Write down the coordinates of the image.
(1)
3
Reflect the point S(-4, 7) in the y-axis. Write down the coordinates of the image.
(1)
4
Write down the coordinates of the origin.
(1)
5
State which axis the point (0, -8) lies on.
(1)
6
State which axis the point (9, 0) lies on.
(1)
7
Write down the coordinates of the point that is 3 units to the right and 2 units up from (1, 1).
(1)
8
Find the midpoint of the line joining the points A(4, 9) and B(10, 3).
(2)
9
Find the midpoint of C(-5, -2) and D(3, 6).
(2)
10
Calculate the distance between E(6, 4) and F(6, -9).
(2)
11
Calculate the distance between G(-8, 5) and H(7, 5).
(2)
12
Work out the gradient of the line through J(2, 3) and K(6, 15).
(2)
13
Work out the gradient of the line through L(-1, 8) and M(5, 2).
(2)
14
A straight line passes through the points P(0, 5) and Q(3, 11). Work out the equation of the line in the form y = mx + c.
(3)
15
A straight line passes through the points R(2, 7) and S(6, 15). Work out the equation of the line in the form y = mx + c.
(3)
16
A line has equation y = 3x - 4. (a) State the gradient of the line. (b) Determine, showing your working, whether the point (5, 11) lies on the line.
(3)
17
A triangle has vertices A(1, 1), B(7, 1) and C(1, 9). Using the coordinates, find the lengths of AB and AC, and hence work out the area of triangle ABC.
(3)
18
Points A(-3, 1) and B(5, 7) are given.
(a)Find the midpoint M of AB.(2)
(b)Find the distance AB, using the horizontal and vertical differences and Pythagoras' theorem.(2)
19
A line passes through P(-2, -1) and Q(4, 11).
(a)Find the gradient of the line PQ.(1)
(b)Find the equation of the line PQ in the form y = mx + c.(2)
(c)M is the midpoint of PQ. Find the coordinates of M, and verify that M lies on the line found in part (b).(2)
Mark scheme · KS3.M-G13D Coordinates and Line Geometry: Fluency and Exam Drill
Question 1
B1 second quadrant cao
Answer: second quadrant
Question 2
B1 (5, 3) cao
Answer: (5, 3)
Question 3
B1 (4, 7) cao
Answer: (4, 7)
Question 4
B1 (0, 0) cao
Answer: (0, 0)
Question 5
B1 y-axis cao
Answer: y-axis
Question 6
B1 x-axis cao
Answer: x-axis
Question 7
B1 (4, 3) cao
Answer: (4, 3)
Question 8
M1 average the x-coordinates and the y-coordinates
A1 (7, 6) cao
Answer: (7, 6)
Question 9
M1 average the x-coordinates and the y-coordinates
A1 (-1, 2) cao
Answer: (-1, 2)
Question 10
M1 recognise the points share an x-coordinate, so subtract the y-coordinates
A1 13 cao
Answer: 13
Question 11
M1 recognise the points share a y-coordinate, so subtract the x-coordinates
A1 15 cao
Answer: 15
Question 12
M1 (15-3)/(6-2) or equivalent method
A1 3 cao
Answer: 3
Question 13
M1 (2-8)/(5-(-1)) or equivalent method
A1 -1 cao
Answer: -1
Question 14
M1 find the gradient, (11-5)/(3-0) = 2
M1 identify c = 5, since P is on the y-axis
A1 y = 2x + 5 cao
Answer: y = 2x + 5
Question 15
M1 find the gradient, (15-7)/(6-2) = 2
M1 substitute into y = 2x + c using R(2,7) to find c = 3
A1 y = 2x + 3 cao
Answer: y = 2x + 3
Question 16
B1 gradient = 3
M1 substitute x = 5 into 3x - 4
A1 yes, (5,11) lies on the line, since 3(5)-4 = 11 cao
Answer: gradient = 3; yes, (5, 11) lies on the line
Question 17
M1 AB = 6 (horizontal) and AC = 8 (vertical)
M1 recognise the right angle at A, and use area = 0.5 x base x height
A1 24 cao
Answer: 24
Question 18
(a) M1 average the x-coordinates and the y-coordinates
(a) A1 (1, 4) cao
(a) Answer: (1, 4)
(b) M1 horizontal difference 8, vertical difference 6, then √82+62
(b) A1 10 cao
(b) Answer: 10
Question 19
(a) B1 2 cao
(a) Answer: 2
(b) M1 substitute P(-2,-1) into y = 2x + c to find c