Coordinate Geometry: Straight Lines Worksheet with Answers

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GCSE · Coordinate Geometry

G1 Coordinate Geometry: Straight Lines

AQA 8365 · Calculator allowed · about 110 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

Points A(1, 2) and B(5, 10) are two points on a coordinate grid.

(a)Calculate the gradient of the line AB.(2)

(b)Find the coordinates of the midpoint of AB.(2)

(c)Calculate the length of AB, giving your answer as a surd in its simplest form.(2)

(Total for Question 1 is 6 marks)

2

A taxi firm in Preston models the cost, C pounds, of a journey of m miles using the formula C = 2.50 + 1.20m.

(a)State the fixed charge and the cost per mile given by this model.(2)

(b)On a graph of C against m, write down the gradient and the y-intercept of the line.(2)

(c)A journey with this firm costs GBP14.50 in total. Work out the distance travelled, in miles.(3)

(Total for Question 2 is 7 marks)

3

A straight line passes through the point (3, -1) and has gradient 4.

(a)Find the equation of the line, giving your answer in the form y = mx + c.(3)

(b)Write your equation from part (a) in the form ax + by + c = 0, where a, b and c are integers and a > 0.(2)

(Total for Question 3 is 5 marks)

4

Line L1 has equation y = 3x + 2. Line L2 is parallel to L1 and passes through the point (2, 5).

(a)Find the equation of L2, giving your answer in the form y = mx + c.(3)

(b)Determine, with justification, whether the point (4, 11) lies on L2.(2)

(Total for Question 4 is 5 marks)

5

Line L1 has gradient 2/3.

(a)State the gradient of a line perpendicular to L1.(1)

(b)Line L2 is perpendicular to L1 and passes through the point (6, -4). Find the equation of L2, giving your answer in the form y = mx + c.(3)

(Total for Question 5 is 4 marks)

6

Find the equation of the straight line passing through the points P(-2, 3) and Q(4, -9), giving your answer in the form y = mx + c.

(a)Find the equation of the line PQ.(4)

(Total for Question 6 is 4 marks)

7

A straight line l has equation 2x + 3y = 18.

(a)Find the coordinates of the points where l crosses the x-axis and the y-axis.(2)

(b)Find the area of the triangle enclosed by l and the two coordinate axes.(2)

(Total for Question 7 is 4 marks)

8

Find the coordinates of the point of intersection of the lines y = 2x - 3 and y = -x + 9.

(a)Find the point of intersection.(3)

(Total for Question 8 is 3 marks)

9

A(1, 6) and B(7, -2) are two points. Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

(a)Find the equation of the perpendicular bisector of AB.(5)

(Total for Question 9 is 5 marks)

10

Show that the points A(-3, -5), B(1, 3) and C(4, 9) are collinear.

(a)Show that A, B and C are collinear.(3)

(Total for Question 10 is 3 marks)

11

A circle has centre O, the origin, and passes through the point P(6, 8).

(a)Find the radius of the circle.(2)

(b)Find the gradient of OP.(1)

(c)Find the equation of the tangent to the circle at P, giving your answer in the form y = mx + c.(3)

(Total for Question 11 is 6 marks)

12

A circle has centre C(2, -1) and passes through the point Q(5, 3).

(a)Find the radius of the circle.(2)

(b)Find the equation of the tangent to the circle at Q, giving your answer in the form ax + by = c, where a, b and c are integers.(4)

(Total for Question 12 is 6 marks)

13

Triangle ABC has vertices A(1, 1), B(5, 2) and C(3, 6). Find the area of triangle ABC.

(a)Find the area of triangle ABC.(4)

(Total for Question 13 is 4 marks)

14

Points A(2, 3) and B(14, 15) are given. The point Q lies on the line segment AB such that AQ : QB = 1 : 3.

(a)Find the coordinates of Q.(3)

(Total for Question 14 is 3 marks)

15

ABCD is a parallelogram with vertices A(1, 2), B(5, 4) and C(9, 10), listed in order around the parallelogram.

(a)Find the coordinates of D.(3)

(b)Show that AB is parallel to, and the same length as, DC, confirming ABCD is a parallelogram.(3)

(Total for Question 15 is 6 marks)

16

The line l has equation 4x + 3y = 25. The point P has coordinates (12, 9).

(a)Find the gradient of l.(1)

(b)Find the equation of the line through P perpendicular to l.(3)

(c)Find the coordinates of the point where the perpendicular line from part (b) meets l.(3)

(d)Hence find the shortest distance from P to the line l.(2)

(Total for Question 16 is 9 marks)

17

The lines l1: y = 2x + 1, l2: y = -x + 10 and l3: x = 0 (the y-axis) enclose a triangle.

(a)Find the coordinates of the point of intersection of l1 and l2.(3)

(b)State the coordinates of the points where l1 and l2 cross l3.(2)

(c)Hence find the area of the triangle enclosed by l1, l2 and l3.(3)

(Total for Question 17 is 8 marks)

18

A(0, 0) and B(8, 6) are two points. The perpendicular bisector of AB meets the x-axis at the point C. Find the coordinates of C.

(a)Find the coordinates of C.(5)

(Total for Question 18 is 5 marks)

19

A(-3, 1) and B(5, 7) are the endpoints of a diameter of a circle C.

(a)Find the coordinates of the centre of C.(2)

(b)Find the radius of C.(2)

(c)Determine, with justification, whether the point (6, 2) lies inside, on, or outside circle C.(3)

(Total for Question 19 is 7 marks)

20

Triangle PQR has vertices P(1, 1), Q(7, 13) and R(11, 11).

(a)Show that PQ is perpendicular to QR.(3)

(b)Find the area of triangle PQR.(3)

(c)Find the length of PR and hence verify Pythagoras' theorem for triangle PQR.(3)

(Total for Question 20 is 9 marks)

Mark scheme · G1 Coordinate Geometry: Straight Lines

Question 1

(a) M1 correct substitution, e.g. (10 - 2) / (5 - 1)
(a) A1 gradient = 2 cao
(a) Answer: 2
(b) M1 ((1+5)/2, (2+10)/2) oe
(b) A1 (3, 6) cao
(b) Answer: (3, 6)
(c) M1 √(5-1)² + (10-2)² oe
(c) A1 4sqrt(5) cao (accept √80)
(c) Answer: 4sqrt(5)

Question 2

(a) B1 fixed charge = GBP2.50 oe
(a) B1 cost per mile = GBP1.20 oe
(a) Answer: Fixed charge GBP2.50; cost per mile GBP1.20
(b) B1 gradient = 1.20 oe
(b) B1 y-intercept = 2.50 oe (accept (0, 2.50))
(b) Answer: Gradient 1.20, y-intercept 2.50
(c) M1 14.50 - 2.50 (= 12) oe
(c) M1 12 / 1.20 (dependent on previous method)
(c) A1 10 miles cao
(c) Answer: 10 miles

Question 3

(a) M1 y - (-1) = 4(x - 3) oe
(a) M1 correct expansion, e.g. y + 1 = 4x - 12
(a) A1 y = 4x - 13 cao
(a) Answer: y = 4x - 13
(b) M1 correct rearrangement of terms, ft from part (a)
(b) A1 4x - y - 13 = 0 oe cao
(b) Answer: 4x - y - 13 = 0

Question 4

(a) B1 gradient of L2 = 3 (parallel lines have equal gradient)
(a) M1 y - 5 = 3(x - 2) oe
(a) A1 y = 3x - 1 cao
(a) Answer: y = 3x - 1
(b) M1 substitute x = 4 into equation of L2, ft from part (a): 3(4) - 1
(b) A1 = 11, which matches the y-coordinate, so (4, 11) lies on L2 cso
(b) Answer: Yes, (4, 11) lies on L2

Question 5

(a) B1 -3/2 oe (accept -1.5)
(a) Answer: -3/2
(b) M1 y - (-4) = -3/2(x - 6) oe, ft gradient from part (a)
(b) M1 correct expansion, e.g. y + 4 = -1.5x + 9
(b) A1 y = -1.5x + 5 oe cao
(b) Answer: y = -1.5x + 5

Question 6

(a) M1 gradient = (-9 - 3) / (4 - (-2)) oe
(a) A1 gradient = -2
(a) M1 y - 3 = -2(x + 2) oe
(a) A1 y = -2x - 1 cao
(a) Answer: y = -2x - 1

Question 7

(a) M1 set y = 0 or x = 0 to find one intercept correctly
(a) A1 (9, 0) and (0, 6) cao (both required)
(a) Answer: (9, 0) and (0, 6)
(b) M1 Area = 1/2 x 9 x 6 oe, ft from part (a)
(b) A1 27 (units²) cao
(b) Answer: 27 square units

Question 8

(a) M1 2x - 3 = -x + 9 oe
(a) M1 3x = 12, x = 4
(a) A1 (4, 5) cao (both coordinates required)
(a) Answer: (4, 5)

Question 9

(a) M1 midpoint of AB = (4, 2) oe
(a) M1 gradient AB = (-2-6)/(7-1) = -4/3 oe
(a) M1 perpendicular gradient = 3/4
(a) M1 y - 2 = 3/4(x - 4) oe
(a) A1 y = 0.75x - 1 oe cao
(a) Answer: y = 0.75x - 1

Question 10

(a) M1 gradient AB = (3-(-5))/(1-(-3)) = 2
(a) M1 gradient BC = (9-3)/(4-1) = 2
(a) A1 gradient AB = gradient BC and B is a common point, so A, B, C lie on the same straight line cso
(a) Answer: A, B and C are collinear (both gradients equal 2)

Question 11

(a) M1 √6² + 8² oe
(a) A1 10 cao
(a) Answer: 10
(b) B1 4/3 oe
(b) Answer: 4/3
(c) M1 perpendicular gradient = -3/4, ft from part (b)
(c) M1 y - 8 = -3/4(x - 6) oe
(c) A1 y = -0.75x + 12.5 oe cao
(c) Answer: y = -0.75x + 12.5

Question 12

(a) M1 √(5-2)² + (3-(-1))² oe
(a) A1 5 cao
(a) Answer: 5
(b) M1 gradient CQ = 4/3 oe
(b) M1 perpendicular gradient = -3/4
(b) M1 y - 3 = -3/4(x - 5) oe
(b) A1 3x + 4y = 27 oe cao
(b) Answer: 3x + 4y = 27

Question 13

(a) M1 correct use of Area = 1/2 |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|
(a) M1 correct substitution: 1/2 |1(2-6) + 5(6-1) + 3(1-2)|
(a) A1 simplifies to 1/2 |18|
(a) A1 Area = 9 (units²) cao
(a) Answer: 9 square units

Question 14

(a) M1 vector AB = (12, 12) oe
(a) M1 Q = A + 1/4(vector AB) oe
(a) A1 (5, 6) cao
(a) Answer: (5, 6)

Question 15

(a) M1 midpoint of AC = midpoint of BD (diagonals of a parallelogram bisect each other) oe
(a) M1 midpoint of AC = (5, 6)
(a) A1 D = (5, 8) cao
(a) Answer: D = (5, 8)
(b) M1 vector AB = (4, 2)
(b) M1 vector DC = (9-5, 10-8) = (4, 2), ft from part (a)
(b) A1 AB = DC, so AB is parallel to DC and equal in length cso
(b) Answer: AB = DC = (4, 2), confirming ABCD is a parallelogram

Question 16

(a) B1 -4/3 oe
(a) Answer: -4/3
(b) M1 perpendicular gradient = 3/4, ft from part (a)
(b) M1 y - 9 = 3/4(x - 12) oe
(b) A1 y = 0.75x oe cao
(b) Answer: y = 0.75x
(c) M1 substitute y = 0.75x into 4x + 3y = 25, ft from part (b)
(c) M1 solve to give x = 4
(c) A1 (4, 3) cao
(c) Answer: (4, 3)
(d) dM1 √(12-4)² + (9-3)², dependent on part (c)
(d) A1 10 cao
(d) Answer: 10

Question 17

(a) M1 2x + 1 = -x + 10 oe
(a) M1 3x = 9, x = 3
(a) A1 (3, 7) cao
(a) Answer: (3, 7)
(b) B1 (0, 1)
(b) B1 (0, 10)
(b) Answer: (0, 1) and (0, 10)
(c) M1 base along y-axis = 10 - 1 = 9, ft from part (b)
(c) M1 Area = 1/2 x 9 x 3 oe, using horizontal distance from y-axis to their part (a) point
(c) A1 13.5 (units²) cao
(c) Answer: 13.5 square units

Question 18

(a) M1 midpoint AB = (4, 3) oe
(a) M1 gradient AB = 3/4 oe
(a) M1 perpendicular gradient = -4/3
(a) M1 y - 3 = -4/3(x - 4) oe, set y = 0 and solve
(a) A1 C = (6.25, 0) oe cao
(a) Answer: C = (6.25, 0)

Question 19

(a) M1 midpoint formula applied to A and B
(a) A1 (1, 4) cao
(a) Answer: (1, 4)
(b) M1 AB = √8² + 6² = 10 oe
(b) A1 radius = 5 cao
(b) Answer: 5
(c) M1 distance from centre (1, 4) to (6, 2): √5² + (-2)² oe, ft from part (a)
(c) A1 √29, awrt 5.39
(c) A1 since √29 > 5 (the radius), the point lies outside C cso
(c) Answer: Outside C (distance = √29, approx 5.39 > radius 5)

Question 20

(a) M1 gradient PQ = (13-1)/(7-1) = 2
(a) M1 gradient QR = (11-13)/(11-7) = -1/2
(a) A1 product of gradients = 2 x (-1/2) = -1, so PQ is perpendicular to QR cso
(a) Answer: Product of gradients = -1, so PQ is perpendicular to QR
(b) M1 PQ = √6² + 12² = √180 = 6sqrt(5) oe
(b) M1 QR = √4² + 2² = √20 = 2sqrt(5) oe
(b) A1 Area = 1/2 x 6sqrt(5) x 2sqrt(5) = 30 (units²) cao, using the right angle at Q from part (a)
(b) Answer: 30 square units
(c) M1 PR = √(11-1)² + (11-1)² = √200 oe
(c) M1 PQ² + QR² = 180 + 20 = 200, ft from part (b)
(c) A1 PR² = 200 = PQ² + QR², verifying Pythagoras' theorem cso
(c) Answer: PR = 10sqrt(2); PQ² + QR² = PR² = 200, verifying Pythagoras' theorem

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