mathsbooks.netmathsbooks.net/Maths Quest 11 General Mathematics...516 General Mathematics Distance between two points Coordinate geometry is a branch of Mathematics with many practical

12In this chapter12A Distance between two

points12B Midpoint of a line segment12C Dividing a line segment

internally in the ratio a:b12D Dividing a line segment

externally in the ratio a:b12E Parallel lines12F Perpendicular lines12G Applications

VCEVCEcocovverageerageArea of studyUnit 1 and 2 • Geometry

Coordinategeometry

Chapter 12 GM Page 515 Wednesday, December 15, 1999 9:41 PM

516 G e n e r a l M a t h e m a t i c s

Distance between two pointsCoordinate geometry is a branch of Mathematicswith many practical applications. The distancebetween two points can be calculated easily usingPythagoras’ Theorem. It is particularly usefulwhen trying to find a distance that is difficult tomeasure directly. For example, finding thedistance from a point on one side of a lake to apoint on the other side.

Let A (x1, y1) and B (x2, y2) be two points inthe Cartesian plane as shown below.

Triangle ABC is a right-angled triangle.

= x2 − x1

= y2 − y1

By Pythagoras’ theorem:

= +

= (x2 − x1)2 + (y2 − y1)

2

Hence =

The distance between two points A (x1, y1) and B (x2, y2) is:

x

y(x2, y2)

(x1, y1)

x1

y1

y2

A

B

C

x2

AC

BC

AB2 AC2 BC2

AB x2 x1–( )2 y2 y1–( )2+

AB x2 x1–( )2 y2 y1–( )2+=

Find the distance between the points A and B in thefigure at right.

THINK WRITECreate a right-angled triangle.

Find AC. = 3 − (−3) = 6

Find BC. = 4 − 1 = 3

Use Pythagoras’ theorem to find AB.

= = 62 + 32

= 36 + 9 = 45 = = = 6.71 (rounded to 2 decimal places)

1

x

y

B

CA

3–3

4

1

2 AC

3 BC

4 AB2 AC2 BC2+

AB 453 5

1WORKEDExample

x

y

B

A

3–3

4

1

Chapter 12 GM Page 516 Friday, May 4, 2001 4:15 PM

C h a p t e r 1 2 C o o r d i n a t e g e o m e t r y 517

Find the distance between the points P(−1, 5) and Q(3, −2).

THINK WRITE

Let P have coordinates (x1, y1). Let (x1, y1) = (−1, 5)Let Q have coordinates (x2, y2). Let (x2, y2) = (3, −2)

Find the length PQ by applying the formula for the distance between two points.

=

=

=

=

= = 8.06 (rounded to 2 decimal places)

1

2

3 PQ x2 x1–( )2 y2 y1–( )2+

3 1–( )–[ ]2 2– 5–( )2+

4( )2 7–( )2+

16 49+

65

2WORKEDExample

Prove that the points A(1, 1), B(3, −1) and C(−1, −3) are the vertices of an isosceles triangle.

THINK WRITE

Plot the points.Note: For triangle ABC to be isosceles, two sides must have the same magnitude.

Find the length AC. =

=

=

Find the length BC. =

=

=

Find the length AB. =

=

=

=

=

State your proof. Since = , triangle ABC is an isosceles triangle.

1

x

yA

B

C

3–1 1

–3

1From the diagram,

appears to have the same length as .

AC

BC

2 AC 1 1–( )–[ ]2 1 3–( )–[ ]2+

2( )2 4( )2+

20

3 BC 3 1–( )–[ ]2 −1 3–( )–[ ]2+

4( )2 2( )2+

20

4 AB 3 1( )–[ ]2 −1 1( )–[ ]2+

2( )2 2–( )2+

4 4+

8

2 2

5 AC BC

3WORKEDExample



Distance betweentwo points

1 Find the distance between each pair of points shown at right.

2 Find the distance between the following pairs of points.a (2, 5), (6, 8) b (−1, 2), (4, 14)c (−1, 3), (−7, −5) d (5, −1), (10, 4)e (4, −5), (1, 1) f (−3, 1), (5, 13)g (5, 0), (−8, 0) h (1, 7), (1, −6)i (a, b), (2a, −b) j (−a, 2b), (2a −b)

3 Prove that the points A(0, −3), B(−2, −1) and C(4, 3) are the vertices of an isosceles triangle.

4 The points P(2, −1), Q(−4, −1) and R(−1, ) are joined to form a triangle.Prove that triangle PQR is equilateral.

5 Prove that the quadrilateral with vertices A(−1, 3), B(5, 3), C(1, 0) and D(−5, 0) is aparallelogram.

6 Prove that the triangle with vertices D(5, 6), E(9, 3) and F(5, 3) is a right-angledtriangle.

7 The vertices of a quadrilateral are A(1, 4), B(−1, 8), C(1, 9) and D(3, 5).a Find the lengths of the sides.b Find the lengths of the diagonals.c What type of quadrilateral is it?

8

If the distance between the points (3, b) and (−5, 2) is 10 units, then the value of b is:

9

A rhombus has vertices A(1, 6), B(6, 6), C(−2, 2) and D(x, y). The coordinates of D are:

10 A rectangle has vertices A(1, 5), B(10.6, z), C(7.6, −6.2) and D(−2, 1). Find:a the length of CDb the length of ADc the length of the diagonal ACd the value of z.

11 Show that the triangle ABC with coordinates A(a, a), B(m, −a) and C(−a, m) is isosceles.

A −8 B −4 C 4 D 0 E 2

A (2, −3) B (2, 3) C (−2, 3) D (3, 2) E (3, −2)

rememberThe distance between two points A(x1, y1) and B(x2, y2) is:

AB x2 x1–( )2 y2 y1–( )2+=

remember

12AWWORKEDEExample

1

Mathca

d

Distancebetween2 points

EXCEL

Spreadsheet

Distancebetween2 points

654321

–1–2–3–4–5–6

–5–6 –4 –3 –2 –1 1 2 3 4 650 x

yGK

HEL

FM

A

P

BO

CN

DJIGCpro

gram

Distance between2 points

Cabri

Geometry

Distance between2 points

WORKEDExample

2

WORKEDExample

3 3 3 1–

mmultiple choiceultiple choice




Midpoint of a line segmentWe can determine the coordinates of the midpoint of a line segment either by plottingthe given coordinates and using a first-principles approach, or applying a given for-mula. The first-principles method will be examined in worked example 4; the formulamethod will be shown later, in worked example 5.

General formulaConsider the line segment connecting the points A(x1, y1) and B(x2, y2).Let P(x, y) be the midpoint of AB.AC is parallel to PD.PC is parallel to BD.AP is parallel to PB (collinear).Hence triangle APC is similar to triangle PBD.But AP = PB (since P is the midpoint of AB).Hence, triangle APC is congruent to triangle PBD.Therefore x − x1 = x2 − x

2x = x1 + x2

x =

Similarly it can be shown that .

In general, the coordinates of the midpoint of a line segment joining the points (x1, y1)and (x2, y2) can be found by averaging the x- and y-coordinates of the end points, respectively.

The coordinates of the midpoint of the line segment

joining (x1, y1) and (x2, y2) are:

First-principles methodFind the midpoint, M, of the line segment joining the points A(2, 1) and B(6, 3), shown at right.

THINK WRITE

Since M is halfway between A and B the x-coordinate of M is halfway between where x = 2 and x = 6.

x-coordinate of M is x = 4

Similarly, the y-coordinate of M is halfway between y = 1 and y = 3.

y-coordinate of M is y = 2Hence the coordinates of the midpoint M are (4, 2).

x

yB

M

A

62

3

2

1

1

2

4WORKEDExample

x

y (x2, y2)

(y2 - y)

P (x, y)

(x1, y1) (x - x1)

(y - y1)(x2 - x)

AC

D

B

x1 x2+

2----------------

yy1 y2+

2----------------=

x

y

(x1, y1)

(x2, y2)

M_____y1 + y2_____,x1 + x2

2 2( )x1 x2+

2-----------------

y1 y2+

2-----------------,



The graphics calculator program Linear may be used to find midpoints (as well as otheritems). Ensure that the programs Surd and Linear are also loaded on your TI-83 beforerunning Linear.

Note the x-coordinate of the midpoint in the last screen is .In this section, the program Linear can be loaded by clicking on the graphics calcu-

lator program icon titled ‘Midpoint of a segment’ on the Maths Quest CD-ROM.

Formula methodFind the coordinates of the midpoint of the line segment joining (−2, 5) and (7, 1).THINK WRITE

Label the given points (x1, y1) and (x2, y2). Let (x1, y1) = (−2, 5) and (x2, y2) = (7, 1)

Find the x-coordinate of the midpoint. x =

x =

x =

x = 2

Find the y-coordinate of the midpoint. y =

y =

y =

y = 3

Give the coordinates of the midpoint. Hence the coordinates of the midpoint are (2 , 3).

1

2x1 x2+

2----------------

−2 7+2

----------------

52---

12---

3y1 y2+

2----------------

5 1+2

------------

62---

412---

5WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Finding midpoints

GM SD12.7a

GCpro

gram

Midpointof asegment

52---



You can verify the result obtained for questions like that shown in worked example 6using computers or calculators. If using the graphics calculator program Linear (seepage 520), enter the coordinates of A(1, –4) and B(13, 8) and check whether you obtainthe coordinates (7, 2) for the midpoint.

Other software files can be used in a similar way.

The coordinates of the midpoint, M, of the line segment AB are (7, 2). If the coordinates of A are (1, −4), find the coordinates of B.THINK WRITE

Label the start of the line segment (x1, y1) and the midpoint (x, y).

Let (x1, y1) = (1, −4) and (x, y) = (7, 2)

Find the x-coordinate of the end point. x =

7 =

14 = 1 + x2

x2 = 13

Find the y-coordinate of the end point. y =

2 =

4 = −4 + y2

y2 = 8Give the coordinates of the end point. Hence the coordinates of the point B are (13, 8).Check that the coordinates are feasible.

1

2x1 x2+

2----------------

1 x2+

2--------------

3y1 y2+

2----------------

−4 y2+

2------------------

45

x

y

(7, 2)

(1, –4)A

M

(13, 8)B

1371

8

2

–4

6WORKEDExample

GC program

Midpointof a

segment

Mathcad

Midpointof a

segment

EXCEL Spreadsheet

Midpointof a

segment

Cabri Geometry

Midpointof a

segment

rememberThe coordinates of the midpoint of the line segment

joining (x1, y1) and (x2, y2) are:x1 x2+

2----------------

y1 y2+

2----------------,

remember

x

y

(x1, y1)

(x2, y2)

M_____y1 + y2_____,x1 + x2

2 2( )



Midpoint of a line segment

1 Use the first-principles method to find the coordinates of the midpoint of the linesegment joining the following pairs of points.

2 Use the formula method to find the coordinates of the midpoint of the line segmentjoining the following pairs of points.

3 The coordinates of the midpoint, M, of the line segment are (2, −3). If the coordi-nates of A are (7, 4), find the coordinates of B.

4 Find:a the coordinates of the centre of a square with vertices A(0, 0), B(2, 4), C(6, 2) and

D(4, −2)b the side lengthc the length of the diagonals.

5

The midpoint of the line segment joining the points (−2, 1) and (8, −3) is:

6

If the midpoint of is (−1, 5) and the coordinates of B are (3, 8), then A hascoordinates:

7 a The vertices of a triangle are A(2, 5), B(1, −3) and C(−4, 3). Find:i the coordinates of P, the midpoint of ii the coordinates of Q, the midpoint of iii the length of iv the length of .

b Hence show that .

8 a A quadrilateral has vertices A(6, 2), B(4, −3), C(−4, −3) and D(−2, 2). Find:i the midpoint of the diagonal, ii the midpoint of the diagonal .

b Comment on your findings.

9 a The points A(−5, 3.5), B(1, 0.5) and C(−6, −6) are the vertices of a triangle. Find:i the midpoint, P, of ii the length of iii the length of iv the length of .

b Describe the triangle. What could represent?

a (1, 6), (3, 4) b (2, 7), (4, 12)c (−1, 5), (3, 8) d (5, −3), (6, −11)

a (−5, 1), (−1, −8) b (4, 2), (11, −2) c (0, 4), (−2, −2)d (3, 4), (−3, −1) e (a, 2b), (3a, −b) f (a + 3b, b), (a − b, a − b)

A (6, −2) B (5, 2) C (6, 2) D (3, −1) E (5, −2)

A (1, 6.5) B (2, 13) C (−5, 2) D (4, 3) E (7, 11)

12BWORKEDExample

4

Mathca

d

Midpoint of a segment

EXCEL

Spreadsheet

Midpoint of asegment

GCpro

gram


Cabri

Geometry


WORKEDExample

5

WORKEDExample

6

AB



AB

ACAB

PQBCBC 2PQ=

ACBD

ABPCACBC

PC



Dividing a line segment internally in the ratio a:b

We can also determine the coordinates of a point dividing a line segment internally in agiven ratio either by plotting the given coordinates and using a first-principles approachor by applying a given formula.

General formulaConsider the line segment connecting the pointsA(x1, y1) and B(x2, y2).Let P(x, y) be the point on AB that divides it in the ratio a:b as shown at right.AC is parallel to PD.PC is parallel to BD.AP is parallel to PB (collinear).Hence, triangle APC is similar to triangle PBD.

First-principles methodFind the coordinates of the point, P, that divides the line segment joining the points A(2, 3) and B(6, 11) internally in the ratio 3:1.

THINK WRITE

Show the end points (x1, y1) and (x2, y2) on a sketch graph and show an estimated position of the internal point, P.

Find the x-coordinate of P.Since P divides AB in the ratio 3:1 then P is located of the length of the line segment AB from A, (P is 3 parts from A and B is 4 parts from A).

The x-coordinate of P is of the way between where x = 2 and x = 6.

x = 2 + (6 − 2)

x = 2 + × 4

x = 2 + 3x = 5

Find the y-coordinate of P. Similarly, the y-coordinate of P is of the way between y = 3 and y = 11.

y = 3 + (11 − 3)

y = 3 + × 8

y = 3 + 6y = 9

Give the coordinates of the point. Hence the coordinates of the point P are (5, 9).

1

x

y

(2, 3)A

3

1 (6, 11)B

P

62

11

3

2

34---

34---

34---

34---

334---

34---

34---

4

7WORKEDExample

a

b

x

y (x2, y2)

(y2 - y)

P (x, y)

(x1, y1) (x - x1)

(y - y1)(x2 - x)

AC

D

B



Given that = (Note: The ratio a:b may be written in fractional form as .)

then = = =

but =

so =

b(x − x1) = a(x2 − x)bx − bx1 = ax2 − axax + bx = ax2 + bx1

x(a + b) = ax2 + bx1

x =

Similarly it can be shown that

y =

The coordinates of the point that divides the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio a:b are:

Note: When a = b the formula simplifies to that for the midpoint of a line segment asdescribed earlier.

APPB-------- a

b--- a

b---

ACPD-------- PC

BD-------- AP

PB-------- a

b---

ACPD--------

x x1–

x2 x–--------------

x x1–

x2 x–-------------- a

b---

ax2 bx1+

a b+-----------------------

ay2 by1+

a b+-----------------------

ax2 bx1+

a b+------------------------

a y2 b y1+

a b+------------------------,

Formula methodFind the coordinates of the point, P, that divides the line segment joining A(2, 3) andB(6, 11) internally in the ratio 3:1.

THINK WRITE

Label the end points (x1, y1) and (x2, y2). Let (x1, y1) = (2, 3) and (x2, y2) = (6, 11)Find a and b. a:b = 3:1

Hence a = 3, b = 1

Use the formula to find thex-coordinate and the y-coordinate of P.

x =

x =

x =

x =

x = 5Give the coordinates of the point. Hence, the coordinates of the point dividing the

line segment in the ratio 3:1 are (5, 9).

12

3ax2 bx1+

a b+-----------------------

3 6( ) 1 2( )+3 1+

----------------------------

18 2+4

---------------

204

------

4

8WORKEDExample

y =

=

=

=

= 9

ay2 by1+

a b+-----------------------

3 11( ) 1 3( )+3 1+

-------------------------------

33 3+4

---------------

364

------



If P (3, −4) is the point that divides the line segment AB internally in the ratio 1:2, find the coordinates of point A if the coordinates of point B are (5, 8).

THINK WRITE

Label the end point (x2, y2) and the point P(x, y).

Let (x2, y2) = (5, 8) and (x, y) = (3, −4)

Find a and b. a:b = 1:2Hence a = 1, b = 2

Find the x1-coordinate. Let A be (x1, y1).Hence, if:

x =

then

3 =

=

9 = 5 + 2x1

2x1 = 4x1 = 2

Find the y1-coordinate. y =

then

−4 =

=

−12 = 8 + 2y1

2y1 = −20y1 = −10

Give the coordinates of A. Hence, the coordinates of the point A are (2, −10).Check that the coordinates are feasible.

1

2

3

ax2 bx1+

a b+-----------------------

1 5( ) 2x1+

1 2+--------------------------

5 2x1+

3-----------------

4ay2 by1+

a b+-----------------------

1 8( ) 2y1+

1 2+--------------------------

8 2y1+

3-----------------

56 y

B

P

2

1

253

8

–10

–4x

A

9WORKEDExample

rememberThe coordinates of the point P that divides the linesegment joining the points (x1, y1) and (x2, y2) internallyin the ratio a:b are:

x

y

A (x1, y1)

P (x, y)

(x2, y2)B

a

b

ax2 bx1+

a b+-----------------------

ay2 by1+

a b+-----------------------,

remember



Dividing a line segmentinternally in the ratio a:b

1 Use the first-principles method to find the coordinates of the point that divides the linesegment joining the following pairs of points internally in the given ratio.

2 Use the formula method to find the coordinates of the point that divides the linesegment joining the following pairs of points internally in the given ratio.

3 If P(6, 1) is the point that divides the line segment AB internally in the ratio 2:5, findthe coordinates of A if the coordinates of B are (1, 7).

4 The point P(5, 4) divides the line segment joining A(2, −5) and B(c, d) in the ratio 3:2.Find c and d.

5a The point, P, divides the line segment AB internally in the ratio 1:4. If A is (−3, 2)

and B is (7, −3), then the coordinates of P are:

b Point Q(3, 10) divides the line segment joining the points A(7, 4) and B(1, 13)internally in the ratio:

c Points P and Q are the points of trisection of AB in thediagram at right. The coordinates P and Q respectively are:

a (1, 7), (4, 1) 1:2 b (1, 7), (4, 1) 2:1c (5, −1), (3, 3) 1:1 d (2, 13), (6, 9) 1:1

a (−3, 5), (1, −5) 3:1 b (−3, 5), (1, −5) 1:3 c (2, 8), (7, 3) 2:3d (2, 8), (7, 3) 3:2 e (2, −9), (−2, −5) 5:3 f (2, −9), (−2, −5) 3:5

A (5, −2) B (−1, 1) C (−4, −1) D (10, 5) E (2, 0.5)

A 1:2 B 1:1 C 2:1 D 3:1 E 1:3

A (0, −1), (2, 3) B (1, 0), (3, 4) C (0, 1 ), (6 , 3 )

D (0, 1), (4, 3) E (0, 2), (5, 3 )

Finding the centroid of a triangle1 a Triangle ABC has vertices A(3, 15), B(6, 9) and C(−3, −6). Find:

i the coordinates of L, the midpoint of BCii the coordinates of M, the midpoint of ACiii the coordinates of N, the midpoint of AB.

b AL, BM and CN are the medians of triangle ABC. A median is a line drawnfrom the vertex of a triangle to the midpoint of the opposite side. Find:

i the coordinates of the point on AL that divides it in the ratio 2:1ii the coordinates of the point on BM that divides it in the ratio 2:1iii the coordinates of the point on CN that divides it in the ratio 2:1.

c Comment on your findings from b i, ii and iii.d The three medians are concurrent. Their common point, usually labelled G,

is called the centroid of the triangle.Graph the triangle ABC and draw the medians AL, BM and CN. Mark the centroid.

2 Triangle PQR has vertices P(−6, 3), Q(−3, 9) and R(3, 12). Find:a the midpoints U, V and W of QR, PR and PQ respectivelyb the coordinates of the centroid, Gc the ratio PG:GU and PG:PU.

12C

SkillSH

EET 12.1

WWORKEDEExample

7

EXCEL

Spreadsheet

Dividinga segmentinternally

WORKEDExample

8

WORKEDExample

9


x

y

A

BQ

8

–4

5

P

–1

WorkS

HEET 12.113--- 1

3--- 2

3---

12---



Dividing a line segment externally in the ratio a:b

We can also determine the coordinates of a point dividing a line segment externally ina given ratio either by plotting the given coordinates and using a first principlesapproach, or by applying a given formula.

First-principles methodFind the coordinates of the point, P, that divides the line segment joining A(2, 3) and B(6, 11) externally in the ratio 3:1.

THINK WRITE

Show the end points A(x1, y1) and B(x2, y2) on a sketch graph and an estimated position of the external point P (x, y).

Find the x-coordinate of P.Since P divides AB externally in the ratio 3:1 then P is located of the length of the line segment AB from A. (P is 3 parts from A and B is 2 parts from A.)

The x-coordinate of P is of the way between where x = 2 and where x = 6.

x = 2 + (6 − 2)

x = 2 + × 4

x = 2 + 6x = 8

Find the y-coordinate of P. Similarly, the y-coordinate of P is of the way between y = 3 and y = 11.

y = 3 + (11 − 3)

y = 3 + × 8

y = 3 + 12y = 15

Give the coordinates of thepoint.

Hence the coordinates of the point Pare (8, 15).

Check that the coordinates are feasible.

1

x

yP (x, y)

B (6, 11)

A (2, 3)

2 6

11

3

13

2

32---

32---

32---

32---

332---

32---

32---

4

5

x

yP (8, 15)

B (6, 11)

A (2, 3)

2 6 8

11

15

3

31

10WORKEDExample



General formulaConsider the line segment connecting the points A(x1, y1) and B(x2, y2).Let P(x, y) be an external point on the extension of AB that divides it in the ratio a:b as shown at right.AD is parallel to BC.PD is parallel to PC (collinear).AP is parallel to BP (collinear).Hence, triangle APD is similar to triangle BPC.

Given that =

then = = =

but =

so =

a(x − x2) = b(x − x1)ax − ax2 = bx − bx1

ax − bx = ax2 − bx1

x(a − b) = ax2 − bx1

x =

Similarly it can be shown that:

y =

The coordinates of the point P that divides the line segment joining the points (x1, y1) and (x2, y2) externally in the ratio a:b are:

y

x - x1

y - y2

y - y1x - x2

(x1, y1)A D

CB

(x, y)P

a b

(x2, y2)

x

APBP-------- a

b---

ADBC--------- PD

PC-------- AP

BP-------- a

b---

ADBC---------

x x1–

x x2–--------------

ab---

x x1–

x x2–--------------

ax2 bx1–

a b–-----------------------

ay2 by1–

a b–-----------------------

y (x, y)

A (x1, y1)

P

a b

B(x2, y2)

x

ax2 bx1–

a b–------------------------

a y2 b y1–

a b–------------------------,

Formula methodFind the coordinates of the point that divides the line segment joining the points (2, 3) and (6, 11) externally in the ratio 3:1.

THINK WRITE

Label the end points A(x1, y1) and B(x2, y2).

Let (x1, y1) = (2, 3) and (x2, y2) = (6, 11)

Find a and b. a:b = 3:1Hence a = 3, b = 1

1

2

11WORKEDExample



THINK WRITE

Use the formula to find the x-coordinate and the y-coordinate of the required point.

x =

x =

x =

x =

x = 8Give the coordinates of the point. Hence the coordinates of the point dividing

the line externally in the ratio 3:1 are (8, 15).

3ax2 bx1–

a b–-----------------------

3 6( ) 1 2( )–3 1–

----------------------------

18 2–2

---------------

162

------

4

y =

=

=

=

= 15

ay2 − by1

a b–------------------------

3 11( ) 1 3( )–3 1–

-------------------------------

33 3–2

---------------

302

------

If P(3, −4) is the point that divides the line segment AB externally in the ratio 1:2, find the coordinates of A if the coordinates of B are (5, 8).

THINK WRITELabel the end point (x2, y2) and the point P(x, y). Let (x2, y2) = (5, 8) and (x, y) = (3, −4)Find a and b. a:b = 1:2

Hence a = 1, b = 2Find the x1-coordinate. Let A be (x1, y1).

Hence if:

x =

then 3 =

3 =

−3 = 5 − 2x1

2x1 = 5 + 3 = 8

x1 = 4

Find the y1-coordinate. y =

then −4 =

−4 =

4 = 8 − 2y1

2y1 = 8 − 4 = 4

y1 = 2Give the coordinates of A. Hence, the coordinates of the point A are (4, 2).

12

3

ax2 bx1–

a b–-----------------------

1 5( ) 2x1–

1 2–-------------------------

5 2x1–

−1-----------------

4ay2 by1–

a b–-----------------------

1 8( ) 2y1–

1 2–-------------------------

8 2y1–

1–-----------------

5

12WORKEDExample



Dividing a line segmentexternally in the ratio a :b

1 Use the first-principles method to find the coordinates of the point that divides the linesegment joining the following pairs of points externally in the given ratio.

2 Use the formula method to find the coordinates of the point that divides the linesegment joining the following pairs of points externally in the given ratio.

3 The point P(6, 1) is the point that divides the line segment AB externally in the ratio2:5. Find the coordinates of A if the coordinates of B are (1, 7).

4 The point P(5, 4) divides the line segment joining A(2, −5) and B(c, d) externally in theratio 3:2. Find c and d.

5P is the point that divides the line segment AB externally in the ratio 4:1. If A is (−2, 3)and B is (7, −3) then the coordinates of P are:

6Point Q(−5, 22) divides the line segment joining the points A(7, 4) and B(1, 13) exter-nally in the ratio:

a (1, 7), (4, 1) 1: 2 b (1, 7), (4, 1) 2:1c (−3, 5), (1, −5) 3 :1 d (−3, 5), (1, −5) 1:3

a (2, 8), (7, 3) 2:3 b (2, 8), (7, 3) 3:2 c (2, −9), (−2, −5) 5:3d (2, −9), (−2, −5) 3:5 e (5, −1), (3, 3) 4:1 f (5, −1), (3, 3) 1:4

A (10, −5) B (−5, 5) C (8 , −3) D (− , 3) E (6, 3)

A 1:2 B 2:1 C 3:1 D 1:1 E 1:3

Give wayA give way sign has the shape of an equilateral triangle. The

sign is attached in two places to a metal pole.1 How far from the top of the sign should the holes be drilled if

the top hole divides the vertical height of the sign in the ratio 1:9 andthe bottom hole in the ratio 8:9?

2 How high is the top of the sign from the ground if the distance to thebase of the pole from the top and bottom of the sign is in the ratio 7:6?

rememberThe coordinates of the point, P, that divides the line segmentjoining the points (x1, y1) and (x2, y2) externally in the ratioa:b are:

y (x, y)

A (x1, y1)

P

a b

B(x2, y2)

x

ax2 bx1–

a b–-----------------------

ay2 by1–

a b–-----------------------,

remember

12DWORKEDEExample

10EXCEL

Spreadsheet

Dividinga segmentexternally

WORKEDExample

11

WORKEDExample

12


23--- 1

3---


87 cm



Parallel linesIn a previous chapter, ‘Linear graphs and modelling (Chapter 6)’, we investigated lineargraphs and equations. We are now going to investigate further properties of straight lines.The equation of a straight line may be expressed in the form:

y = mx + cwhere m is the gradient of the line, and c is the y-intercept.

The gradient can be calculated if two points, (x1, y1) and (x2, y2) are given.

An alternative form for the equation of a straight line is:Ax + By + C = 0

where A, B and C are constants.

Investigating linear equations1 Using a graphics calculator, investigate the following linear equations.

a y = 3x b y = x c y = −2xd y = 3x + 4 e y = −2x + 2 f y = 3x − 5g y = x − 5 h y = − x + 4

2 On the same set of axes, sketch each of the graphs.3 Which graphs have a positive gradient? Why?4 Which graphs have the same y-intercept? Why?5 Which graphs are parallel? Why?6 Copy and complete the following.

For straight lines to be parallel they must have gradient.

my2 y1–

x2 x1–----------------=

15---

34--- 1

5---

Show that AB is parallel to CD given that A has coordinates (−1, −5), B has coordinates (5, 7), C has coordinates (−3, 1), and D has coordinates (4, 15).

THINK WRITE

Find the gradient of AB. Let A(−1, −5) = (x1, y1) and B(5, 7) = (x2, y2)

Since m =

mAB =

=

= 2Find the gradient of CD. Let C(−3, 1) = (x1, y1) and D(4, 15) = (x2, y2)

mCD =

=

= 2Compare the gradients to determine if they are parallel. (Note: || means ‘is parallel to’.)

Since parallel lines have the same gradient and mAB = mCD = 2 then AB||CD

1

y2 y1–

x2 x1–----------------

7 5–( )–5 1–( )–--------------------

126------

2

15 1–4 3–( )–--------------------

147------

3

13WORKEDExample



Collinear points lie on the same straight line.

The Maths Quest spreadsheet ‘Plotting points’ may be used to visualise the previousworked example.

Show that the points A (2, 0), B (4, 1) and C (10, 4) are collinear.

THINK WRITE

Find the gradient of AB. Let A(2, 0) = (x1, y1)and B(4, 1) = (x2, y2)

Since m =

mAB =

=

Find the gradient of BC. Let B(4, 1) = (x1, y1)and C(10, 4) = (x2, y2)

mBC =

=

=

Show that A, B and C are collinear. Since mAB = mBC = then AB||BCSince B is common to both line segments A, B and C must lie on the same straight line. That is A, B and C are collinear.

1

y2 y1–

x2 x1–----------------

1 0–4 2–------------

12---

2

4 1–10 4–---------------

36---

12---

312---

14WORKEDExample

EXCEL

SpreadsheetPlottinggraphs

remember1. The equation of a straight line may be expressed in the form:

y = mx + cwhere m is the gradient of the line and c is the y-intercept.

2. The gradient can be calculated if two points, (x1, y1) and (x2, y2) are given.

3. Parallel lines have the same gradient.

4. Collinear points lie on the same straight line.

my2 y1–

x2 x1–----------------=

remember



Parallel lines

1 Find if AB is parallel to CD given the following coordinates.a A(4, 13), B(2, 9), C(0, −10), D(15, 0).b A(2, 4), B(8, 1), C(−6, −2), D(2, −6).c A(−3, −10), B(1, 2), C(1, 10), D(8, 16).d A(1, −1), B(4, 11), C(2, 10), D(−1, −5).e A(1, 0), B(2, 5), C(3, 15), D(7, 35).f A(1, −6), B(−5, 0), C(0, 0), D(5, −4).

2 Which pairs of the following straight lines are parallel?a 2x + y + 1 = 0 b y = 3x − 1c 2y − x = 3 d y = 4x + 3e f 6x − 2y = 0g 3y = x + 4 h 2y = 5 − x

3 Show that the points A(0, −2), B(5, 1) and C(−5, −5) are collinear.

4 Show that the line that passes through the points (−4, 9) and (0, 3) also passes throughthe point (6, −6).

5 In each of the following, show that ABCD is a parallelogram.a A(2, 0), B(4, −3), C(2, −4), D(0, −1)b A(2, 2), B(0, −2), C(−2, −3), D(0, 1)c A(2.5, 3.5), B(10, −4), C(2.5, −2.5), D(−5, 5)

6 In each of the following, show that ABCD is a trapezium.a A(0, 6), B(2, 2), C(0, −4), D(−5, −9)b A(26, 32), B(18, 16), C(1, −1), D(−3, 3)c A(2, 7), B(1, −1), C(−0.6, −2.6), D(−2, 3)

7

The line that passes through the points (0, −6) and (7, 8) also passes through:

8

The point (−1, 5) lies on a line parallel to 4x + y + 5 = 0. Another point on the same lineas (−1, 5) is:

9 Prove that the quadrilateral ABCD is a rhombus, given A(2, 3), B(3, 5), C(5, 6) and D(4, 4).

A (4, 3)B (5, 4)C (−2, 10)D (1, −8)E (1, 4)

A (2, 9)B (4, 2)C (4, 0)D (−2, 3)E (3, −11)

12E

EXCEL

Spreadsheet

Mathca

d

Gradient

GCpro

gram

Cabri

Geometry

Gradient

WORKEDExample

13

SkillSH

EET12.2 EXCEL Spreadsheet

Parallelcheckery

x2--- 1–=

WORKEDExample

14



WorkS

HEET 12.2



Perpendicular linesIn this section, we shall examine some of the properties of perpendicular lines.Graphics calculators and graphing software can be very useful in investigating theseproperties.

Consider (the diagram below) where the line segment AB is perpendicular to the linesegment BC. Line AC is parallel to the x-axis. Line BD is the height of the resultingtriangle ABC.

Let mAB = m1

=

= tan θLet mBC = m2

= −

= − tan α

= −

= −

Pairs of perpendicular lines1 Using a graphics calculator or graphing software, investigate the

following pairs of straight lines and copy and complete the table below.Note: It is important that both axes have the same scale, hence use and 5:ZSquare for graphics calculators and make appropriate adjustments for graphing software.

2 Copy and complete the following.Two lines that are at right angles or degrees to one another are said to be perpendicular. If two lines are perpendicular to one another then the product of their gradients is .

ZOOM

Equation 1 Equation 2 m1 m2 m1 × m2

Angle between the lines

a y = 2x y = − x

b y = 3x − 1 y = − x + 5

c y = − x + 2 y = x − 4

d y = x + 1 y = −x

e y = 0.4x + 3 y = −2.5x − 10

12---

13---

32--- 2

3---

θ

θ

α

αb

a

c

x

y

A C

B

D

ab---

ac---

ba---

1m1------


C h a p t e r 1 2 C o o r d i n a t e g e o m e t r y 535Hence m2 = −

or m1m2 = −1Hence, if two lines are perpendicular to each other then the product of their gradients is −1.

Two lines are perpendicular if and only if:m1m2 = −1

or m2 =

To check visually if two lines are perpendicular using a graphics calculator, ensure equationsare entered in ‘y = form’ in the menu, and press and select 5:ZSquare.

The perpendicular lines from the previous worked example are used below — notehow they only appear perpendicular after performing a 5:ZSquare and not a

6:ZStandard.

1m1------

1m1-------–

Show that the lines y = −5x + 2 and 5y − x + 15 = 0 are perpendicular to one another.

THINK WRITE

Find the gradient of equation 1. y = −5x + 2Hence m1 = −5

Find the gradient of equation 2. 5y − x + 15 = 0Rewrite in the form y = mx + c

5y = x − 15

y = − 3

Hence m2 =

Test for perpendicularity. m1m2 = −5 ×

= −1Hence the two lines are perpendicular to each other.

1

2

x5---

15---

3 15---

15WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Viewing perpendicular lines

Y= ZOOM

ZOOMZOOM

GM SD12.26a2 6:ZStandardZOOM 5:ZSquareZOOM



Perpendicular lines

1 Show that the lines y = 6x − 3 and x + 6y − 6 = 0 are perpendicular to one another.

2 Determine if AB is perpendicular to CD, given the following coordinates.a A(1, 6), B(3, 8), C(4, −6), D(−3, 1)b A(2, 12), B(−1, −9), C(0, 2), D(7, 1)c A(1, 3), B(4, 18), C(−5, 4), D(5, 0)d A(1, −5), B(0, 0), C(5, 11), D(−10, 8)e A(−4, 9), B(2, −6), C(−5, 8), D(10, 14)f A(4, 4), B(−8, 5), C(−6, 2), D(3, 11)

3 Determine which pairs of the following straight lines are perpendicular.a x + 3y − 5 = 0 b y = 4x − 7c y = x d 2y = x + 1e y = 3x + 2 f x + 4y − 9 = 0g 2x + y = 6 h x + y = 0

4 Show that the following sets of points form the vertices of a right-angled triangle.a A(1, −4), B(2, −3), C(4, −7) b A(3, 13), B(1, 3), C(−4, 4)c A(0, 5), B(9, 12), C(3, 14)

5 Prove that the quadrilateral ABCD is a rectangle when A is (2, 5), B(6, 1), C(3, −2) andD(−1, 2).

6 In each of the following, prove that ABCD is a rectangle.a A(2, 8), B(6, 6), C(2, −2), D(−2, 0)b A(1, 3), B(2, 0), C(−1, −1), D(−2, 2)

7 Prove that the quadrilateral ABCD is a rhombus, given A(2, 3), B(3, 5), C(5, 6) andD(4, 4). Hint: The diagonals of a rhombus intersect at right angles.

8

The gradient of the line perpendicular to the line with equation 3x − 6y = 2 is:

9

Triangle ABC has a right angle at B. The vertices are A(−2, 9), B(2, 8) and C(1, z). Thevalue of z is:

A 3 B −6 C 2 D E −2

A 8 B 4 C 12 D 7 E −4

rememberTwo lines are perpendicular if and only if:

m1m2 = −1

or m2 = − .1

m1------

remember

12FWORKEDExample

15

EXCEL

Spreadsheet

Perpendicularchecker


12---


14--- 3

4---



ApplicationsThe equation of a straight line can be determined by two methods.

The y = mx + c method requires the gradient, m, and a given point to be known, inorder to establish the value of c.Note : Since the value of c represents the y-intercept, it can be substituted directly if known.

The alternative method comes from the gradient definition.

m =

Hence m(x2 − x1) = y2 − y1

Using the general point (x, y) instead of the specific point (x2, y2) gives the generalequation:

y − y1 = m(x − x1)This requires the gradient, m, and a given point (x1, y1) to be known.

Find the equation of the straight line that passes through the point (3, −1) and is parallel to the straight line with equation y = 2x + 1.

THINK WRITE

Write the general equation. y = mx + cFind the gradient of the given line. y = 2x + 1 has a gradient of 2

Hence m = 2Substitute for m in the general equation. so y = 2x + cSubstitute the given point to find c. (x, y) = (3, −1)

∴ −1 = 2(3) + c−1 = 6 + c

c = −7Substitute for c in the general equation. y = 2x − 7

or2x − y − 7 = 0

12

34

5

16WORKEDExample

y2 y1–

x2 x1–----------------

Find the equation of the line that passes through the point (0, 3) and is perpendicular to a straight line with a gradient of 5.

Continued over page

THINK WRITE

Find the gradient of the perpendicular line.

Given m = 5

m⊥ = −1

15---

17WORKEDExample



THINK WRITE

Substitute for m and (x1, y1) in the general equation.

Since y − y1 = m(x − x1)and (x1, y1) = (0, 3)

then y − 3 = − (x − 0)

y − 3 = −

5(y − 3) = −x5y − 15 = −x

x + 5y − 15 = 0

2

15---

x5---

Find the equation of the perpendicular bisector of the line joining the points (0, −4) and (6, 5).

THINK WRITE

Find the gradient of the line joining the given points using the general equation.

Let (0, −4) = (x1, y1)Let (6, 5) = (x2, y2)

m =

m =

=

=

Find the gradient of the perpendicular line.

For lines to be perpendicular, m2 = −m⊥ = −

Find the midpoint of the line joining the given points.

x =

=

= 3

y =

=

=

Hence (3, ) are the coordinates of the

midpoint.

1

y2 y1–

x2 x1–----------------

5 4–( )–6 0–

--------------------

96---

32---

21

m1------

23---

3x1 x2+

2----------------

0 6+2

------------

y1 y2+

2----------------

4– 5+2

----------------

12---

12---

18WORKEDExample



THINK WRITE

Substitute for m and (x1, y1) in the general equation.

Simplify (remove fractions).(a) Multiply both sides by 3.

(b) Multiply both sides by 2.

Since y − y1 = m(x − x1)

and (x1, y1) = (3, ) and m⊥ = −

then y − = − (x − 3)

3(y − ) = −2(x − 3)

3y − = −2x + 6

6y − 3 = −4x + 124x + 6y − 15 = 0

Note: The diagram at right shows the geometric situation.

4

5

12--- 2

3---

12--- 2

3---

12---

32---

x

y(6, 5)

3–4 6

5

2

–4

1–21–2

ABCD is a parallelogram. The coordinates of A, B and C are (1, 5), (4, 2) and (2, −2) respectively. Find:a the equation of AD. b the equation of DC. c the coordinates of D.THINK WRITE

a Draw the parallelogram ABCD.Note: The order of the lettering of the geometric shape determines the links in the diagram. For example: ABCD means join A to B to C to D to A. This avoids any ambiguity.

a

Find the gradient of BC. mBC =

=

= 2State the gradient of AD. Since mBC = 2

and AD||BCthen mAD = 2

Using the given coordinates of A and the gradient of AD find the equation of AD.

y = 2x + cLet (x, y) = (1, 5)

5 = 2(1) + cc = 3

Hence the equation of AD is y = 2x + 3.

1

x

yA

B

C

D

21–1 4

5

2

–2

22– 2–

2 4–----------------

4–2–

------

3

4

19WORKEDExample

Continued over page



The Maths Quest CD-ROM files‘Equation of a straight line’ maybe used to find the equation of astraight line given either two pointsor a point and a gradient.

The screen shows an exampleusing the Excel file.

THINK WRITE

b Find the gradient of AB. b mAB =

=

= −1State the gradient of DC. Since mAB = −1

and DC||ABthen mDC = −1

Using the given coordinates of C and the gradient of DC find the equation of DC.

y = −x + cLet (x, y) = (2, −2)

−2 = −(2) + cc = 0

Hence the equation of DC is y = −x.c Solve simultaneously to find D, the point of

intersection of the equations AD and DC.c Equation of AD: y = 2x + 3 [1]

Equation of DC: y = −x [2][1]–[2]: 0 = 3x + 3

3x = −3x = −1

Substituting x = −1 in [2]:y = −(−1)y = 1

Hence the coordinates of D are (−1, 1).Note: Alternatively, a graphics calculator could be used to determine the point of intersection of AD.

12 5–4 1–------------

3–3

------

2

3

EXCEL

Spreadsheet

Equationof astraightline

Mathca

d

Equationof astraightline

rememberThe equation of a straight line can be determined by two methods:1. The y = mx + c method.

This requires the gradient, m, and a given point to be known, in order to establish the value of c.If the y-intercept is known, then this can be directly substituted for c.

2. Alternative method: y − y1 = m(x − x1)This requires the gradient, m, and a given point (x1, y1) to be known.

remember



Applications

1 Find the equation of the straight line that passes through the point (4, −1) and isparallel to the straight line with equation y = 2x − 5.

2 Find the equation of the line that passes through the point (−2, 7) and is perpendicularto a line with a gradient of .

3 Find the equations of the following straight lines.a Gradient 3 and passing through the point (1, 5).b Gradient −4 and passing through the point (2, 1).c Passing through the points (2, −1) and (4, 2).d Passing through the points (1, −3) and (6, −5).e Passing through the point (5, −2) and parallel to x + 5y + 5 = 0.f Passing through the point (1, 6) and parallel to x − 3y − 2 = 0.g Passing through the point (−1, −5) and perpendicular to 3x + y + 2 = 0.

4 Find the equation of the line which passes through the point (−2, 1) and is:a parallel tob perpendicular tothe straight line with equation 2x − y − 3 = 0.

5 Find the equation of the line that contains the point (1, 1) and is:a parallel tob perpendicular tothe straight line with equation 3x − 5y = 0.

6 The triangle ABC has vertices A(9, −2), B(3, 6) and C(1, 4).a Find the midpoint, M, of BC.b Find the gradient of BC.c Show that AM is the perpendicular bisector of BC.d Describe triangle ABC.

7 Find the equation of the perpendicular bisector of the line joining the points (1, 2) and(−5, −4).

8 Find the equation of the perpendicular bisector of the line joining the points (−2, 9)and (4, 0).

9 ABCD is a parallelogram. The coordinates of A, B and C are (4, 1), (1, −2) and(−2, 1) respectively. Find:a the equation of ADb the equation of DCc the coordinates of D.

10 ABCD is a quadrilateral with vertices A(4, 9), B(7, 4), C(1, 2) and D(a, 10).Given that the diagonals are perpendicular to each other, find:a the equation of the diagonal ACb the equation of the diagonal BDc the value of a.

12GWORKEDExample

16Mathcad

Equationof a

straightline

EXCEL Spreadsheet

Equationof a

straightline

WORKEDExample

1723---

WORKEDExample

18

WORKEDExample

19



11

a The equation of the line passing through the point (4, 3) and parallel to the line 2x − 4y + 1 = 0 is:

b The equation of the perpendicular bisector of the line segment AB where A is (−3, 5) and B is (1, 7) is:

c The coordinates of the centroid of triangle ABC with vertices A(1, 8), B(9, 6) andC(−1, 4) are:

A x − 2y + 2 = 0 B 2x − y − 5 = 0C 2x − y − 10 = 0 D 2x − y − 11 = 0

A 2y = x + 13 B y = 2x − 8 C 2y = x + 11D y = −2x + 4 E y = 2x − 4

A (4, 5) B (0, 6) C (3, 6) D (5, 7) E (2, 7)

A Roman aqueduct


To supply cities with water when the source is a long distance away, artificialchannels, called aqueducts, may be built. More than 2000 years after it was built,a Roman aqueduct still stands in southern France. It brought water from a sourcein Uzès to the city of Nîmes. The aqueduct does not follow a direct routebetween these two locations as there is a mountain range between them. The tableshows the approximate distance from Uzès along the aqueduct to each town (or inthe case of Pont du Gard, a bridge) and the aqueduct’s height above sea level ateach location.

1 Show the information in the table as a graph with the distance from Uzès alongthe horizontal axis. Join the plotted points with straight lines.

2 Calculate the gradient of the steepest part of the aqueduct (in m/km).

3 Suppose the aqueduct started at Uzès and ended at Nîmes, but had a constantgradient. Write a linear equation to describe its course.

4 Using the equation found in part 3, calculate the height of the aqueduct at thePont du Gard. This calculated height is higher than the actual height. How muchhigher?

5 Why do you think the Romans made the first part of the aqueduct steeper thanthe rest?

LocationDistance from

Uzès (km)Height of aqueduct above sea level (m)

Uzès 0 76

Pont du Gard (bridge) 16 65

St. Bonnet 25 64

St. Gervasy 40 61.5

Nîmes 50 59



Distance between two points• The distance between two points A (x1, y1) and B (x2, y2) is:

Midpoint of a line segment• The coordinates of the midpoint of the line segment

joining (x1, y1) and (x2, y2) are:

Dividing a line segment internally in the ratio a:b• The coordinates of the point that divides the line segment

joining the points (x1, y1) and (x2, y2) internally in the ratio a:b are:

Dividing a line segment externally in the ratio a:b• The coordinates of the point that divides the line segment

joining the points (x1, y1) and (x2, y2) externally in the ratio a:b are:

Parallel lines• Parallel lines have the same gradient.• Collinear points lie on the same straight line.

Perpendicular lines• Two lines are perpendicular if and only if:

m1m2 = −1 or

Applications• Equations of a straight line:

1. Gradient and y-intercept form: y = mx + c

where

2. General form: Ax + By + C = 0• To find the equation of a straight line:

1. Given gradient and y-intercept y = mx + c2. Given gradient and a point y − y1 = m(x − x1) or

y = mx + c method3. Given two points Find m, then use: y − y1 = m(x − x1) or

y = mx + c method

summaryAB x2 x1–( )2 y2 y1–( )2+=

x

y

(x1, y1)

(x2, y2)

M_____y1 + y2_____,x1 + x2

2 2( )x1 x2+

2----------------

y1 y2+

2----------------,

x

y

A (x1, y1)

P (x, y)

(x2, y2)B

a

b

ax2 bx1+

a b+-----------------------

ay2 by1+

a b+-----------------------,

y (x, y)

A (x1, y1)

P

a b

B(x2, y2)

x

ax2 bx1–

a b–-----------------------

ay2 by1–

a b–-----------------------,

m21

m1------–=

my2 y1–

x2 x1–----------------=



Multiple choice

1 The distance between the points (1, 5) and (6, −7) is:

2 The midpoint of the line segment joining the points (−4, 3) and (2, 7) is:

3 If the midpoint of the line segment joining the points A (3, 7) and B (x, y) has coordinates (6, 2), then the coordinates of B are:

4 C is a point that divides the line segment AB internally in the ratio 1:2. If A is the point(−4, 1) and B is the point (2, −5), then the coordinates of C are:

The following information refers to questions 5, 6, 7 and 8.Triangle ABC has vertices A (1, 5), B (4, −1) and C (−6, −3).

5 The median from A meets the line segment, BC, at M. The coordinates of M are:

6 The centroid divides a median in the ratio 2:1. The coordinates of the centroid are:

7 The gradient of the median, AM, is:

8 The equation of the median, AM, is:

9 D is a point that divides the line segment AB externally in the ratio 2:3. If A is the point (2, 3) and B is the point (7, 6), then the coordinates of D are:

10 The gradient of the line joining the points (2, 7) and (5, −8) is:

11 If the points (−6, −11), (2, 1) and (x, 4) are collinear then the value of x is:

12 The gradient of the line perpendicular to 3x − 4y + 7 = 0 is:

13 The equation of the line perpendicular to 2x + y − 1 = 0 and passing through the point (1, 4) is:

A B C 13 D E 12

A (−1, 5) B (−2, 10) C (−6, 4) D (−2, 4) E (−1, 2)

A (15, 3) B (0, −6) C (9, −3) D (4.5, 4.5) E (−9, 3)

A (0, −3) B (−2, −1) C (−3 , 2 ) D (2 , −3 ) E (−1, −2)

A (2 , 2) B (−3 , 1) C (−1, −2) D (2, −4) E (10, 2)

A (0, 1 ) B ( , 2 ) C (0, 3) D ( , ) E ( , 1)

A 3 B C undefined D 2 E −2

A 2x − 7y − 12 = 0 B 7x − 2y + 3 = 0 C 7x − 2y − 3 = 0D x + 1 = 0 E 2x − 7y + 12 = 0

A (−8, −3) B (8, −3) C (4, 4 ) D (17, 12) E (8, 3)

A −5 B −1 C 5 D − E 1

A 4 B 3.2 C D E 3

A B C − D 3 E −4

A 2x + y − 6 = 0 B 2x + y − 2 = 0 C x − 2y + 7 = 0D x + 2y + 9 = 0 E x − 2y = 0

CHAPTERreview

12A53 29 193

12B

12B

12C13--- 1

3--- 2

3--- 2

3---

12C 12--- 1

2---

12C 12--- 1

3--- 2

3--- 1

3---–

13--- 1

2---

12C 12--- 2

7---

12C

12D15---

12E 37---

12E 14--- 5

16------

12F 34--- 4

3--- 4

3---

12G


C h a p t e r 1 2 C o o r d i n a t e g e o m e t r y

545

The following information refers to questions

14

,

15

,

16

and

17

. The diagram at right shows a square inscribed in a circle. The square has coordinates A(1, 4), B(2,

−

3), C(

−

5, 4) and D(

a

,

b

).

14

The circle has a radius of:

15

The coordinates of the centre are:

16

The gradient of the diagonal, BD, is:

17

The coordinates of the point D are:

Short answer

1

Find the distance between the points (1, 3) and (7,

−

2).

2

Prove that triangle ABC is isosceles given A(3, 1), B(

−

3, 7) and C(

−

1, 3).

3

Show that the points A(1, 1), B(2, 3) and C(8, 0) are the vertices of a right-angled triangle.

4

The midpoint of the line segment AB is (6,

−

4). If B has coordinates (12, 10), find the coordinates of A.

5

Find the coordinates of the point which divides the line joining the point A(

−

2, 6) and the point B(2,

−

4) internally in the ratio 3:1.

6

Find the coordinates of the point which divides the line joining the points (2, 8) and (5, 2) externally in the ratio 1:2.

7

Show that the points A(3, 1), B(5, 2) and C(11, 5) are collinear.

8

Show that the lines

y

=

2

x

−

4 and

x

+ 2y − 10 = 0 are perpendicular to one another.

9 Find the equation of the straight line passing through the point (6, −2) and parallel to the line x + 2y − 1 = 0.

10 Find the equation of the line perpendicular to 3x − 2y + 6 = 0 and having the same y-intercept.

11 Find the equation of the perpendicular bisector of the line joining the points (−2, 7) and (4, 11).

12 Find the equation of the straight line joining the point (−2, 5) and the point of intersection of the straight lines with equations y = 3x − 1 and y = 2x + 5.

A 10 units B 7.07 units C 6 units D 5 units E 12 units

A (−4, 0) B (−2, 0) C (0, −2) D (−1, 1) E (0, −4)

A −1 B − C 1 D − E

A (−3, 6) B (3, −6) C (−6, 2) D (−2, 4) E (−6, 3)

x

y

(–5, –4)

(a, b)

(2, –3)

(1, 4)A

BC

D

12G

12G

12G34--- 3

5--- 4

3---

12G

12A12A12A12B

12C

12D

12E12F12G

12G

12G

12G


546

G e n e r a l M a t h e m a t i c s

13

Using the information given in the diagram.

a

Find:

i

the gradient of AD

ii

the gradient of AB

iii

the equation of BC

iv

the equation of DC

v

the coordinates of C.

b

Describe quadrilateral ABCD.

14

In triangle ABC, A is (1, 5), B is (

−

2,

−

3) and C is (8,

−

2).

a

Find:

i

the gradient of BC

ii

the midpoint, P, of AB

iii

the midpoint, Q, of AC.

b

Hence show that:

i

PQ is parallel to BC

ii

PQ is half the length of BC.

15

Triangle ABC has vertices A(

a

,

b

), B(

−

3, 6) and C(5,

−

2). The centroid, G, of the triangle has coordinates (

−

2,

−

1).

a

Find:

i

the midpoint, M, of BC

ii

the coordinates of A

iii

the gradient of BC

iv

the gradient of AM

v

the length of AB

vi

the length of AC.

b

Describe triangle ABC.

Analysis

1

The map shows the proposed course for a yacht race.Buoys have been positioned at A(1, 5), B(8, 8) and C(12, 6), but the last buoy’s placement, D(10,

w

), is yet to be finalised.

a

How far is the first stage of the race, that is, from the start, O, to buoy A?

b

The race marshall boat, M, is situated halfway between buoys A and C. What are the coordinates of the boat?

c

Stage 4 of the race (from C to D) is perpendicular to stage 3 (from B to C). What is the gradient of CD?

d

Find the linear equation that describes stage 4.

e

Hence determine the exact position of buoy D.

f

An emergency boat is to be placed at point E, of the way from buoy A to buoy D. Into what internal ratio does point E divide the distance from A to D?

g

Determine the coordinates of the emergency boat.

h

How far is the emergency boat from the hospital, located at H, 2 km North of the start?

2

The centroid, G, of a triangle ABC divides the medians internally in the ratio 2:1. For example:AG:GD

=

2:1 where D is the midpoint of BC.A

′

B

′

C

′

is a triangle with coordinates A

′

(5, 4), B

′

(

−

2, 5) and C

′

(6, 9). Find the coordinates of the centroid, G

′

.

12G

x

y

(4, 9)

D

C

B

A

94 5

9

4

12G

12G

1110987654321

1 2 3 4 6 7 8 9 10 11 125

O(start) x

y

Buoy B

Buoy C

Buoy DE

M

NScale: 1 unit = 1 km

H

BuoyA

23---

testtest

CHAPTERyyourselfourself

testyyourselfourself

12

A C

B

DFG

E

2

1


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