Sec 3 A-Math Coordinate geometry worksheet with answers

A free Sec 3 A-Math worksheet on coordinate geometry: 8 questions, 40 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 A-Math worksheet: Coordinate geometry

Total: 40 marksSuggested time: 1 hour

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 coordinates of the mid-point of the line joining

A(-2, -4)

and

B(0, 7)

Answer[2]

Find the area of quadrilateral

ABCD

with vertices

A(0, 1)

B(4, -3)

C(7, 2)

and

D(2, 5)

Answerunits²[2]

The point

P(-1, 3)

lies on the circle

(x + 2)^2 + (y - 1)^2 = 5

. Find the equation of the tangent to the circle at

P

Answer[2]

(a)A circle has centre

(4, 3)

and passes through the point

(2, 6)

. Find the equation of the circle.

Answer (a)[2]

(b)A circle has centre

(2, 1)

and passes through the point

(-1, 2)

. Find the equation of the circle.

Answer (b)[2]

(a)Find the equation of the line which passes through the point

(4, -1)

and is perpendicular to the line

4x + 3y = 12

Answer (a)[3]

(b)Find the equation of the line which passes through the point

(-1, -2)

and is perpendicular to the line

x + 4y = -4

Answer (b)[3]

Section B

(a)Find the gradient of a line perpendicular to the line

3x - 2y = 4

Answer (a)[3]

(b)Find the radius of the circle

x^2 + y^2 + 4x - 2y - 8 = 0

, leaving your answer in surd form.

Answer (b)[3]

The points

A(4, 1)

and

C(8, 7)

are opposite vertices of a rhombus

ABCD

. The vertex

B

lies on the

y

-axis. The diagram is not drawn to scale.

(a)A point

P

moves so that

PA = PC

. Find the equation of the locus of

P

Answer (a)[3]

(b)Explain why

B

lies on this locus, and find the coordinates of

B

Answer (b)[2]

(c)Find the coordinates of

D

Answer (c)[2]

(d)Find the area of the rhombus

ABCD

Answer (d)units²[2]

A circle, with centre

C

, has equation

{x^{2} + y^{2} + 2x + 8y + 14 = 0}

. The line

L

has equation

{y = -2x - 16}

(a)Find the coordinates of

C

Answer (a)[1]

(b)Find the radius of the circle, leaving your answer in surd form where necessary.

Answer (b)[1]

(c)Find the equation of the line through

C

perpendicular to

L

Answer (c)[2]

(d)Find the coordinates of

N

, the foot of the perpendicular from

C

L

Answer (d)[2]

(e)Find the shortest distance from

C

L

, leaving your answer in surd form where necessary.

Answer (e)[2]

(f)Hence state whether

L

cuts the circle at two points, touches the circle or does not meet the circle.

Answer (f)[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

y = \dfrac{3}{2}

[2]

\mathstrut x = \dfrac{-2 + 0}{2}

x = -1

\mathstrut y = \dfrac{-4 + 7}{2}

y = \dfrac{3}{2}

Question 2

Answer:

= 29

units²[2]

\text{Area} = \dfrac{1}{2}\left| \begin{matrix} 0 & 4 & 7 & 2 & 0 \\[8pt] 1 & -3 & 2 & 5 & 1 \end{matrix} \right|

(0)(-3) + 4(2) + 7(5) + 2(1)

= 45

(4)(1) + 7(-3) + 2(2) + 0(5)

= -13

\text{Area} = \dfrac{1}{2}|45 - (-13)|

= 29

Question 3

Answer:

y = -\dfrac{1}{2}x + \dfrac{5}{2}

[2]

m_{\text{radius}} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{\text{radius}} = \dfrac{3 - 1}{-1 - (-2)}

= 2

m_{\text{tangent}} = -\dfrac{1}{2}

y = -\dfrac{1}{2}x + \dfrac{5}{2}

Question 4

(a)Answer:

(x - 4)^2 + (y - 3)^2 = 13

[2]

\text{Centre } (4, 3)

r^2 = (2 - 4)^2 + (6 - 3)^2

r^2 = 13

(x - 4)^2 + (y - 3)^2 = 13

(b)Answer:

(x - 2)^2 + (y - 1)^2 = 10

[2]

\text{Centre } (2, 1)

r^2 = (-1 - 2)^2 + (2 - 1)^2

r^2 = 10

(x - 2)^2 + (y - 1)^2 = 10

Question 5

(a)Answer:

\text{y-intercept } c = -4

[3]

3y = -4x + 12

y = \dfrac{-4}{3}x + 4

\text{Gradient} = -\dfrac{4}{3}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = \dfrac{3}{4}

y - (-1) = \dfrac{3}{4}(x - 4)

\text{y-intercept } c = -4

(b)Answer:

\text{y-intercept } c = 2

[3]

4y = -x - 4

y = \dfrac{-1}{4}x - 1

\text{Gradient} = -\dfrac{1}{4}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = 4

y - (-2) = 4(x - (-1))

\text{y-intercept } c = 2

Question 6

(a)Answer:

\text{Perpendicular gradient} = -\dfrac{2}{3}

[3]

-2y = -3x + 4

y = \dfrac{3}{2}x - 2

\text{Gradient} = \dfrac{3}{2}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = -\dfrac{2}{3}

(b)Answer:

r = \sqrt{13}

[3]

x^2 + 4x + y^2 - 2y = 8

(x + 2)^2 - 4 + (y - 1)^2 - 1 = 8

(x + 2)^2 + (y - 1)^2 = 13

r = \sqrt{13}

Question 7

(a)Answer:

y = -\dfrac{2}{3}x + 8

[3]

The locus of

P

is the perpendicular bisector of

AC

.Mid-point of

AC

= \left(\dfrac{4 + 8}{2}, \dfrac{1 + 7}{2}\right)

= (6, 4)

Gradient of

AC

= \dfrac{7 - 1}{8 - 4}

= \dfrac{3}{2}

Gradient of the perpendicular

= -\dfrac{2}{3}

y - 4 = -\dfrac{2}{3}(x - 6)

y = -\dfrac{2}{3}x + 8

(b)Answer:

B(0, 8)

[2]

BA = BC

because the sides of a rhombus are equal, so

B

lies on the locus.On the

y

-axis,

x = 0

y = 8

B(0, 8)

(c)Answer:

D(12, 0)

[2]

The diagonals of a rhombus bisect each other, so the mid-point of

BD

(6, 4)

D = (2 \times 6 - 0, 2 \times 4 - 8)

D(12, 0)

(d)Answer:

52

units²[2]

Area

= \dfrac{1}{2}\left|4(8) + 0(7) + 8(0) + 12(1) - 1(0) - 8(8) - 7(12) - 0(4)\right|

= 52

units²

Question 8

(a)Answer:

C(-1, -4)

[1]

(x + 1)^{2} + (y + 4)^{2} = 3

C(-1, -4)

(b)Answer:

\sqrt{3}

units[1]

r^{2} = 1 + 16 - 14

r = \sqrt{3}

(c)Answer:

y = \dfrac{1}{2}x - \dfrac{7}{2}

[2]

Gradient of

L

= -2

, so the perpendicular gradient

= \dfrac{1}{2}

y - (-4) = \dfrac{1}{2}(x - (-1))

y = \dfrac{1}{2}x - \dfrac{7}{2}

(d)Answer:

N(-5, -6)

[2]

-2x - 16 = \dfrac{1}{2}x - \dfrac{7}{2}

x = -5

y = -2(-5) - 16

y = -6

(e)Answer:

2\sqrt{5}

units[2]

CN = \sqrt{(-5 - (-1))^{2} + (-6 - (-4))^{2}}

= \sqrt{20}

= 2\sqrt{5}

(f)Answer: It does not meet the circle[1]

CN = 2\sqrt{5}

and

r = \sqrt{3}

CN^{2} = 20

r^{2} = 3

, so

CN > r

.So

L

does not meet the circle.

Practice questions by skill

Sec 3 A-Math 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:

y = \dfrac{3}{2}

[2]

\mathstrut x = \dfrac{-2 + 0}{2}

x = -1

\mathstrut y = \dfrac{-4 + 7}{2}

y = \dfrac{3}{2}

Question 2

Answer:

= 29

units²[2]

\text{Area} = \dfrac{1}{2}\left| \begin{matrix} 0 & 4 & 7 & 2 & 0 \\[8pt] 1 & -3 & 2 & 5 & 1 \end{matrix} \right|

(0)(-3) + 4(2) + 7(5) + 2(1)

= 45

(4)(1) + 7(-3) + 2(2) + 0(5)

= -13

\text{Area} = \dfrac{1}{2}|45 - (-13)|

= 29

Question 3

Answer:

y = -\dfrac{1}{2}x + \dfrac{5}{2}

[2]

m_{\text{radius}} = \dfrac{y_2 - y_1}{x_2 - x_1}

m_{\text{radius}} = \dfrac{3 - 1}{-1 - (-2)}

= 2

m_{\text{tangent}} = -\dfrac{1}{2}

y = -\dfrac{1}{2}x + \dfrac{5}{2}

Question 4

(a)Answer:

(x - 4)^2 + (y - 3)^2 = 13

[2]

\text{Centre } (4, 3)

r^2 = (2 - 4)^2 + (6 - 3)^2

r^2 = 13

(x - 4)^2 + (y - 3)^2 = 13

(b)Answer:

(x - 2)^2 + (y - 1)^2 = 10

[2]

\text{Centre } (2, 1)

r^2 = (-1 - 2)^2 + (2 - 1)^2

r^2 = 10

(x - 2)^2 + (y - 1)^2 = 10

Question 5

(a)Answer:

\text{y-intercept } c = -4

[3]

3y = -4x + 12

y = \dfrac{-4}{3}x + 4

\text{Gradient} = -\dfrac{4}{3}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = \dfrac{3}{4}

y - (-1) = \dfrac{3}{4}(x - 4)

\text{y-intercept } c = -4

(b)Answer:

\text{y-intercept } c = 2

[3]

4y = -x - 4

y = \dfrac{-1}{4}x - 1

\text{Gradient} = -\dfrac{1}{4}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = 4

y - (-2) = 4(x - (-1))

\text{y-intercept } c = 2

Question 6

(a)Answer:

\text{Perpendicular gradient} = -\dfrac{2}{3}

[3]

-2y = -3x + 4

y = \dfrac{3}{2}x - 2

\text{Gradient} = \dfrac{3}{2}

m_1 \times m_2 = -1

\text{Perpendicular gradient} = -\dfrac{2}{3}

(b)Answer:

r = \sqrt{13}

[3]

x^2 + 4x + y^2 - 2y = 8

(x + 2)^2 - 4 + (y - 1)^2 - 1 = 8

(x + 2)^2 + (y - 1)^2 = 13

r = \sqrt{13}

Question 7

(a)Answer:

y = -\dfrac{2}{3}x + 8

[3]

The locus of

P

is the perpendicular bisector of

AC

.Mid-point of

AC

= \left(\dfrac{4 + 8}{2}, \dfrac{1 + 7}{2}\right)

= (6, 4)

Gradient of

AC

= \dfrac{7 - 1}{8 - 4}

= \dfrac{3}{2}

Gradient of the perpendicular

= -\dfrac{2}{3}

y - 4 = -\dfrac{2}{3}(x - 6)

y = -\dfrac{2}{3}x + 8

(b)Answer:

B(0, 8)

[2]

BA = BC

because the sides of a rhombus are equal, so

B

lies on the locus.On the

y

-axis,

x = 0

y = 8

B(0, 8)

(c)Answer:

D(12, 0)

[2]

The diagonals of a rhombus bisect each other, so the mid-point of

BD

(6, 4)

D = (2 \times 6 - 0, 2 \times 4 - 8)

D(12, 0)

(d)Answer:

52

units²[2]

Area

= \dfrac{1}{2}\left|4(8) + 0(7) + 8(0) + 12(1) - 1(0) - 8(8) - 7(12) - 0(4)\right|

= 52

units²

Question 8

(a)Answer:

C(-1, -4)

[1]

(x + 1)^{2} + (y + 4)^{2} = 3

C(-1, -4)

(b)Answer:

\sqrt{3}

units[1]

r^{2} = 1 + 16 - 14

r = \sqrt{3}

(c)Answer:

y = \dfrac{1}{2}x - \dfrac{7}{2}

[2]

Gradient of

L

= -2

, so the perpendicular gradient

= \dfrac{1}{2}

y - (-4) = \dfrac{1}{2}(x - (-1))

y = \dfrac{1}{2}x - \dfrac{7}{2}

(d)Answer:

N(-5, -6)

[2]

-2x - 16 = \dfrac{1}{2}x - \dfrac{7}{2}

x = -5

y = -2(-5) - 16

y = -6

(e)Answer:

2\sqrt{5}

units[2]

CN = \sqrt{(-5 - (-1))^{2} + (-6 - (-4))^{2}}

= \sqrt{20}

= 2\sqrt{5}

(f)Answer: It does not meet the circle[1]

CN = 2\sqrt{5}

and

r = \sqrt{3}

CN^{2} = 20

r^{2} = 3

, so

CN > r

.So

L

does not meet the circle.