Sec 3 G2 Maths Coordinate geometry worksheet with answers

A free Sec 3 G2 Maths worksheet on coordinate geometry: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 G2 Maths worksheet: Coordinate geometry

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Find the gradient of the line joining

A(-5, 1)

and

B(1, 6)

Answer[2]

(a)Find the coordinates of the point where the line

y = -2x - 4

crosses the

x

-axis.

Answer (a)[2]

(b)The point

(k, 11)

lies on the line

y = -4x - 1

. Find

k

Answer (b)[1]

(a)Find the equation of the line parallel to

y = -x - 5

that passes through

(1, 0)

Answer (a)[2]

(b)Find the equation of the line parallel to

y = 4x + 4

that passes through

(3, 5)

Answer (b)[2]

(a)The distance between

A(3, k)

and

B(7, 5)

5

units. Find the two possible values of

k

Answer (a)[2]

(b)The distance between

A(-2, k)

and

B(3, 5)

13

units. Find the two possible values of

k

Answer (b)[2]

Section B

The lines

y = -2x + 7

and

y = -x - 6

meet at the point

P

(a)Find the coordinates of

P

Answer (a)[3]

(b)The point

(k, -7)

lies on the line

y = -2x + 7

. Find the value of

k

Answer (b)[2]

The diagram shows the points

A(-2, 1)

B(4, 3)

and

C(6, -3)

(a)Find the gradient of

AB

Answer (a)[1]

(b)Show that triangle

ABC

is isosceles.

[2]

(c)Find the equation of the line through

C

that is parallel to

AB

Answer (c)[2]

(d)This line crosses the

y

-axis at

D

. Write down the coordinates of

D

Answer (d)[1]

The line

3y = 2x + 6

crosses the

x

-axis at

P

and the

y

-axis at

Q

(a)Find the coordinates of

P

Answer (a)[1]

(b)Find the coordinates of

Q

Answer (b)[1]

(c)Find the length of

PQ

, correct to 3 significant figures.

Answer (c)units[2]

(d)The point

R(k, 6)

lies on the line. Find the value of

k

Answer (d)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\phantom{m} = \dfrac{5}{6}

[2]

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{6 - 1}{1 - (-5)}

m = \dfrac{5}{6}

\phantom{m} = \dfrac{5}{6}

Question 2

(a)Answer: The point is

(-2, 0)

.[2]

Put

y = 0

into

y = -2x - 4

0 = -2x - 4

, so

x = -2

The point is

(-2, 0)

(b)Answer:

k = -3

[1]

11 = -4k - 1

k = -3

Question 3

(a)Answer:

y = -x + 1

[2]

m = -1

0 = -1(1) + c

c = 1

y = -x + 1

(b)Answer:

y = 4x - 7

[2]

m = 4

5 = 4(3) + c

c = -7

y = 4x - 7

Question 4

(a)Answer:

k = 2

k = 8

[2]

(7 - 3)^2 + (5 - k)^2 = 5^2

(5 - k)^2 = 25 - 16

= 9

5 - k = \pm 3

k = 2

k = 8

(b)Answer:

k = -7

k = 17

[2]

(3 - (-2))^2 + (5 - k)^2 = 13^2

(5 - k)^2 = 169 - 25

= 144

5 - k = \pm 12

k = -7

k = 17

Question 5

(a)Answer:

(13, -19)

[3]

-2x + 7 = -x - 6

x = 13

y = -19

The lines meet at

(13, -19)

(b)Answer:

k = 7

[2]

-7 = -2k + 7

k = 7

Question 6

(a)Answer:

\dfrac{1}{3}

[1]

m = \dfrac{3 - 1}{4 - (-2)}

\phantom{m} = \dfrac{2}{6}

\phantom{m} = \dfrac{1}{3}

(b)[2]

AB = \sqrt{6^2 + 2^2}

\phantom{AB} = \sqrt{40}

BC = \sqrt{2^2 + 6^2}

\phantom{BC} = \sqrt{40}

AB = BC

, so triangle

ABC

is isosceles (shown)

(c)Answer:

y = \dfrac{1}{3}x - 5

[2]

m = \dfrac{1}{3}

-3 = \dfrac{1}{3} \times 6 + c

, so

c = -5

y = \dfrac{1}{3}x - 5

(d)Answer:

(0, -5)

[1]

On the

y

-axis,

x = 0

y = -5

D(0, -5)

Question 7

(a)Answer:

(-3, 0)

[1]

On the

x

-axis,

y = 0

0 = 2x + 6

x = -3

, so

P(-3, 0)

(b)Answer:

(0, 2)

[1]

On the

y

-axis,

x = 0

3y = 6

y = 2

, so

Q(0, 2)

(c)Answer:

3.61

units[2]

PQ = \sqrt{3^2 + 2^2}

\phantom{PQ} = \sqrt{13}

PQ = 3.6055\ldots

\phantom{PQ} = 3.61

units (3 s.f.)

(d)Answer:

k = 6

[2]

3(6) = 2k + 6

18 = 2k + 6

k = 6

Sec 3 G2 Maths Coordinate geometry worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Coordinate geometry notes. Or practise it online with step-by-step help when you get one wrong.

Answers and worked solutions

Question 1

Answer:

\phantom{m} = \dfrac{5}{6}

[2]

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{6 - 1}{1 - (-5)}

m = \dfrac{5}{6}

\phantom{m} = \dfrac{5}{6}

Question 2

(a)Answer: The point is

(-2, 0)

.[2]

Put

y = 0

into

y = -2x - 4

0 = -2x - 4

, so

x = -2

The point is

(-2, 0)

(b)Answer:

k = -3

[1]

11 = -4k - 1

k = -3

Question 3

(a)Answer:

y = -x + 1

[2]

m = -1

0 = -1(1) + c

c = 1

y = -x + 1

(b)Answer:

y = 4x - 7

[2]

m = 4

5 = 4(3) + c

c = -7

y = 4x - 7

Question 4

(a)Answer:

k = 2

k = 8

[2]

(7 - 3)^2 + (5 - k)^2 = 5^2

(5 - k)^2 = 25 - 16

= 9

5 - k = \pm 3

k = 2

k = 8

(b)Answer:

k = -7

k = 17

[2]

(3 - (-2))^2 + (5 - k)^2 = 13^2

(5 - k)^2 = 169 - 25

= 144

5 - k = \pm 12

k = -7

k = 17

Question 5

(a)Answer:

(13, -19)

[3]

-2x + 7 = -x - 6

x = 13

y = -19

The lines meet at

(13, -19)

(b)Answer:

k = 7

[2]

-7 = -2k + 7

k = 7

Question 6

(a)Answer:

\dfrac{1}{3}

[1]

m = \dfrac{3 - 1}{4 - (-2)}

\phantom{m} = \dfrac{2}{6}

\phantom{m} = \dfrac{1}{3}

(b)[2]

AB = \sqrt{6^2 + 2^2}

\phantom{AB} = \sqrt{40}

BC = \sqrt{2^2 + 6^2}

\phantom{BC} = \sqrt{40}

AB = BC

, so triangle

ABC

is isosceles (shown)

(c)Answer:

y = \dfrac{1}{3}x - 5

[2]

m = \dfrac{1}{3}

-3 = \dfrac{1}{3} \times 6 + c

, so

c = -5

y = \dfrac{1}{3}x - 5

(d)Answer:

(0, -5)

[1]

On the

y

-axis,

x = 0

y = -5

D(0, -5)

Question 7

(a)Answer:

(-3, 0)

[1]

On the

x

-axis,

y = 0

0 = 2x + 6

x = -3

, so

P(-3, 0)

(b)Answer:

(0, 2)

[1]

On the

y

-axis,

x = 0

3y = 6

y = 2

, so

Q(0, 2)

(c)Answer:

3.61

units[2]

PQ = \sqrt{3^2 + 2^2}

\phantom{PQ} = \sqrt{13}

PQ = 3.6055\ldots

\phantom{PQ} = 3.61

units (3 s.f.)

(d)Answer:

k = 6

[2]

3(6) = 2k + 6

18 = 2k + 6

k = 6