coordinate geometry - wiley · distance between two points coordinate geometry is a branch of...

7Coordinate

geometry7.1 Kick off with CAS

7.2 Distance between two points

7.3 Midpoint of a line segment

7.4 Parallel lines and perpendicular lines

7.5 Applications

7.6 Review

c07CoordinateGeometry.indd 252 6/21/15 8:20 PM

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Please refer to the Resources tab in the Prelims section of your ebookPlUs for a comprehensive step-by-step guide on how to use your CAS technology.

7.1 Kick off with CASTo come


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Distance between two pointsCoordinate geometry is a branch of mathematics with many practical applications. The distance between two points can be calculated easily using Pythagoras’ theorem. It is particularly useful when trying to fi nd a distance that is diffi cult to measure directly; for example, fi nding the distance from a point on one side of a lake to a point on the other side.

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

Triangle ABC is a right-angled triangle.AC = x2 − x1

BC = y2 − y1

By Pythagoras’ theorem:

AB2 = AC2 + BC2

= (x2 − x1)2 + (y2 − y1)2

Hence AB = "(x2 − x1)2 + (y2 − y1)2

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

AB = "(x2 − x1)2 + (y2 − y1)2

7.2A

B

x

y

(x1, y1)

x1

y1

y2

A

B(x2, y2)

C

x2

a Find the distance between the points A and B in the fi gure at right.

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

tHinK WritE

a 1 From the graph fi nd points A and B. A (−3, 1) and B (3, 4)

2 Let A have coordinates (x1, y1). Let (x1, y1) = (−3, 1).

3 Let B have coordinates (x2, y2). Let (x2, y2) = (3, 4).

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

AB = "(x2 − x1)2 + (y2 − y1)2

= "[3 − (−3)]2 + (4 − 1)2

= "(6)2 + (3)2

= !36 + 9

= !45

= 3!5

= 6.71 1correct to 2 decimal places 2

A 1

4

3–3 x

yWorKEd EXAMPLE 111

254 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2


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b 1 Let P have coordinates (x1, y1). Let (x1, y1) = (−1, 5).

2 Let Q have coordinates (x2, y2). Let (x2, y2) = (3, −2).

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

PQ = "(x2 − x1)2 + (y2 − y1)2

= "[3 − (−1)]2 + (−2 − 5)2

= "(4)2 + (−7)2

= !16 + 49

= !65

= 8.06 1correct to 2 decimal places 2

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

tHinK WritE/drAW

1 Plot the points.

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

A1

–1

–3

1 3–1

y

C

x

From the diagram, AC appears to have the same length as BC.

2 Find the length AC. AC = "[1 − (−1)]2 + [1 − (−3)]2

= "(2)2 + (4)2

= !20

= 2!5

3 Find the length BC. BC = "[3 − (−1)]2 + [−1− (−3)]2

= "(4)2 + (2)2

= !20

= 2!54 Find the length AB. AB = "[3 − (1)]2 + [−1 − (1)]2

= "(2)2 + (−2)2

= !4 + 4

= !8

= 2!2

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

WorKEd EXAMPLE 222

Tutorial eles‐1520 Worked example 2

Topic 7 CoordInATE gEoMETry 255


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Distance between two points1 WE1 a Find the distance between the points A and B

shown at right.

b Find the distance between the points. (2, 5), (6, 8).

2 a Find the distance between the points C and D shown at right.

b Find the distance between the points (−1, 2) and (4, 14).

3 WE2 Prove that the points A (0, −3), B (−2, −1) and C (4, 3) are the vertices of an isosceles triangle.

4 Prove that the points A (3, 1), B (−3, 7) and C (−1, 3) are the vertices of an isosceles triangle.

5 The points P (2, −1), Q (−4, −1) and R (−1, 3!3 − 1) are joined to form a triangle. Prove that triangle PQR is equilateral.

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

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

8 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?

9 Calculate the distance between each of the pairs of points below, accurate to 3 decimal places.

a (–14, 10) and (–8, 14) b (6, –7) and (13, 6) c (–11, 1) and (2, 2)

10 Find the distance between each of the following pairs of points in terms of the given variables.

a (a, 1), (2, 3) b (5, 6), (0, b)c (c, 2), (4, c) d (d, 2d), (1, 5)

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

A −8 b −4 c 4 d 0 E 2

12 A rhombus has vertices A (1, 6), B (6, 6), C (−2, 2) and D (x, y). The coordinates of D are:A (2, −3) b (2, 3) c (−2, 3)d (3, 2) E (3, −2)

13 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 CD b the length of ADc the length of the diagonal AC d the value of z.

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

ExErcisE 7.2

PrActisE56

4321

–1–2

–5–6

1 2 3 4 5 6–2 –1 0 x

y

C

D

A

B

–3–4

consolidAtE

Digital docdoc-9928Spreadsheetdistance between two points



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15 Two hikers are about to hike from A to B (shown on the map below). How far is it from A to B ‘as the crow flies’, that is, in a straight line?

16 Using the coordinates shown on the aerial photo of the golf course, calculate (to the nearest metre):

a the horizontal distance travelled by the golf ball for the shot down the fairway

b the horizontal distance that needs to be covered in the next shot to reach the point labelled A in the bunker.

Midpoint of a line segmentWe can determine the coordinates of the midpoint of a line segment by applying the midpoint formula shown below.

Midpoint 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 =x1 + x2

2

Similarly it can be shown that y =y1 + y2

2.

MAstEr

N

Grid spacing : 1 km S

EW

B (E7, N4)

Lake Phillios

A (W12, S5)

300 m200 m100 m

N

Grid spacing : 1 km S

EW

B (E7, N4)

Lake Phillios

100 m200 m

A (W12, S5)

300 m200 m100 m

50 m

y (in metres)

x (in metres)

A (320, 148)

(225, 96)

(80, –64)

7.3

x

y

(y2 ‒ y)

P (x, y)

A

B

C

D(x2 ‒ x)(y ‒ y1)

(x ‒ x1)(x1, y1)

(x2, y2)



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Midpoint of a line segmentExErcisE 7.3

PrActisE

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.

x

y

M

(x1, y1)

(x2, y2)

x1 + x2–2

y1 + y2–2( ),

The coordinates of the midpoint of the line segment joining

(x1, y1) and (x2, y2) are: ax1 + x2

2, y1 + y2

2b.

Find the coordinates of the midpoint of the line segment joining (−2, 5) and (7, 1).

tHinK WritE

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

2 Find the x-coordinate of the midpoint. x =x1 + x2

2

= −2 + 72

= 52

= 212

3 Find the y-coordinate of the midpoint. y =y1 + y2

2

= 5 + 12

= 62

= 3

4 Give the coordinates of the midpoint. Hence, the coordinates of the midpoint are (212, 3).

WorKEd EXAMPLE 333

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

tHinK WritE/drAW

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

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

2 Find the x-coordinate of the end point. x =x1 + x2

2

7 =1 + x2

2

14 = 1 + x2

x2 = 13

WorKEd EXAMPLE 444



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Midpoint of a line segment1 WE3 Find the coordinates of the midpoint of the line segment joining

(−5, 1) and (−1, −8).

2 Find the coordinates of the midpoint of the line segment joining

(4, 2), (11, −2).

3 WE4 The coordinates of the midpoint M of the line segment AB are (2, −3). If the coordinates of A are (7, 4), find the coordinates of B.

4 The coordinates of the midpoint M of the line segment AB are (−2, 4). If the coordinates of A are (1, 8), find the coordinates of B.

5 The vertices of a square are A (0, 0), B (2, 4), C (6, 2) and D (4, −2). Find:

a the coordinates of the centre of the squareb the side lengthc the length of the diagonals.

6 The midpoint of the line segment joining the points (−2, 1) and (8, −3) is:A (6, −2) b (5, 2) c (6, 2) d (3, −1) E (5, −2)

7 If the midpoint of AB is (−1, 5) and the coordinates of B are (3, 8), then A has coordinates:A (1, 6.5) b (2, 13) c (−5, 2) d (4, 3) E (7, 11)

8 Find the coordinates of the midpoint of each of the following pairs of points, in terms of a variable or variables where appropriate.a (2a, a), (6a, 5a) b (5, 3c), (11, 3c) c (3f, 5), (g, –1)

9 Find the value of a below so M is the midpoint of the segment joining points A and B.a A (–2, a), B (–6, 5), M (–4, 5) b A (a, 0), B (7, 3), M (8, 3

2)

10 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 ACii the coordinates of Q, the midpoint of ABiii the length of PQiv the length of BC.

b Hence show that BC = 2PQ.

ExErcisE 7.3

PrActisE

Digital docdoc-9929SpreadsheetMidpoint of a segment

consolidAtE

3 Find the y-coordinate of the end point. y =y1 + y2

2

2 =−4 + y2

2

4 = −4 + y2

y2 = 8

4 Give the coordinates of the end point. Hence, the coordinates of the point B are (13, 8).

5 Check that the coordinates are feasible.

(1, ‒4)

(13, 8)

(7, 2)

A

M

B8

2

–4

1 13 x

y

7



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11 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 ACii the midpoint of the diagonal BD.

b Comment on your finding.

12 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 ABii the length of PCiii the length of ACiv the length of BC.

b Describe the triangle ABC. What could PC represent?

13 Find the equation of the straight line that passes through the midpoint of A (−2, 5) and B (−2, 3) and has a gradient of −3.

14 Find the equation of the straight line that passes through the midpoint of

A (−1, −3) and B (3, −5) and has a gradient of 23.

15 A fun-run course is drawn (not to scale) at right. If drink stations D1, D2 and D3 are to be placed at the middle of each straight section, give the map coordinates of each drink station.

16 Find the equation of a line that has a gradient of 5 and passes through the midpoint of the segment joining (−1, −7) and (3, 3).

Parallel lines and perpendicular linesParallel LinesThe equation of a straight line may be expressed in the form:

y = mx + c

where 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.

m =y2 − y1

x2 − x1

An alternative form for the equation of a straight line is:

ax + by + c = 0

where a, b and c are constants.

Another alternative form is:

y − y1 = m(x − x1)

where m is the gradient and (x1, y1) is a point on the line.

START/FINISH

Of�cial tentD2

D3Coordinatesare in kilometres.

y

x

D1(‒4.5, 5)

(1.5, ‒2)

(3, ‒7) (13, ‒8)

(1.5, 3.5)

MAstEr

7.4



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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

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

m =y2 − y1

x2 − x1

mAB = 7 − (−5)5 − (−1)

= 126

= 2

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

mCD = 15 − 14 − (−3)

= 147

= 23 Compare the gradients to determine if they

are parallel. (Note: || means ‘is parallel to’.)Parallel lines have the same gradient. mAB = mCD = 2, hence AB||CD.

WorKEd EXAMPLE 555

Collinear points lie on the same straight line.

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

tHinK WritE

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

Since m =y2 − y1

x2 − x1

mAB = 1 − 04 − 2

= 12

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

mBC =4 − 1

10 − 4

= 36

= 12

WorKEd EXAMPLE 666

Tutorialeles‐1523Worked example 6



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Parallel lines and perpendicular lines1 WE5 Show that AB is parallel to CD given that A has coordinates

(2, 4), B has coordinates (8, 1), C has coordinates (−6, −2), and D has coordinates (2, −6).

2 Show that AB is parallel to CD given that A has coordinates (1, 0), B has coordinates (2, 5), C has coordinates (3, 15), and D has coordinates (7, 35).

3 WE6 Show that the points A (0, −2), B (5, 1) and C (−5, −5) are collinear.

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

5 WE7 Find the equation of a straight line given the following conditions. The line passes through the point (−1, 3) and is parallel to y = −2x + 5.

6 Find the equation of a straight line given that the line passes through the point (4, −3) and is parallel to 3y + 2x = −3.

7 WE8 Show that the lines y = 6x − 3 and x + 6y − 6 = 0 are perpendicular to one another.

ExErcisE 7.4

PrActisE

DIGITAL DOCdoc-9933SpreadsheetGradient

DIGITAL DOCdoc-9935SpreadsheetPerpendicular checker

Perpendicular linesIn this section, we examine some of the properties of perpendicular lines. Observing the graphs can be very useful in investigating these properties.

Consider the diagram below, where the line segment AB is perpendicular to the line segment BC. Line AC is parallel to the x-axis. Line BD is the height of the resulting triangle ABC.Let mAB = m1

= ab

= tan (θ )Let mBC = m2

= −

ab

= −tan (θ )

= −

ba

= −

1m1

Hence m2 = −

1m1

or m1m2 = −1

Hence, if two lines are perpendicular to each other, then the product of their gradients is –1.

θ

θ

α

αb

a

c

x

y

A C

B

D

Find the equation of the straight line that passes through the point (2, 5) and is parallel to the line y = 3x + 1.

tHinK WritE

1 In order to fi nd the equation of a straight line, we need to know the gradient and a point on the line. One point is given, and because the line is parallel to y = 3x + 1, the gradients will be the same.

Point on the line: (2, 5)Gradient: m = 3.

2 Use the formula y − y1 = m(x − x1) and substitute the coordinates of the point and the gradient to fi nd the equation of the line.

y − y1 = m(x + x1) y − 5 = 3(x − 2) y − 5 = 3x − 6 y = 3x − 1

WorKEd EXAMPLE 777

3 Show that A, B and C are collinear. Since mAB = mBC = 12

then AB||BC.Since 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.



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Parallel lines and perpendicular lines1 WE5 Show that AB is parallel to CD given that A has coordinates

(2, 4), B has coordinates (8, 1), C has coordinates (−6, −2), and D has coordinates (2, −6).

2 Show that AB is parallel to CD given that A has coordinates (1, 0), B has coordinates (2, 5), C has coordinates (3, 15), and D has coordinates (7, 35).

3 WE6 Show that the points A (0, −2), B (5, 1) and C (−5, −5) are collinear.

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

5 WE7 Find the equation of a straight line given the following conditions. The line passes through the point (−1, 3) and is parallel to y = −2x + 5.

6 Find the equation of a straight line given that the line passes through the point (4, −3) and is parallel to 3y + 2x = −3.

7 WE8 Show that the lines y = 6x − 3 and x + 6y − 6 = 0 are perpendicular to one another.

ExErcisE 7.4

PrActisE

DIGITAL DOCdoc-9933SpreadsheetGradient

DIGITAL DOCdoc-9935SpreadsheetPerpendicular checker

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

tHinK WritE

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

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

5y = x − 15

y = x5

− 3

Hence m2 = 15

3 Test for perpendicularity. (The two lines are perpendicular if the product of their gradients is –1.)

m1m2 = −5 × 15

= −1

Hence, the two lines are perpendicular to each other.

WorKEd EXAMPLE 888

Tutorial eles‐1524 Worked example 8

Two lines are perpendicular if and only if:

m1m2 = −1

or m2 = − 1m1



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8 Show that the lines y = 2x − 4 and x + 2y − 10 = 0 are perpendicular to one another.

9 Which pairs of the following straight lines are parallel?a 2x + y + 1 = 0 b y = 3x − 1c 2y − x = 3 d y = 4x + 3

e y = x2

− 1 f 6x − 2y = 0

g 3y = x + 4 h 2y = 5 − x

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

11 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)

12 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)

13 The line that passes through the points (0, −6) and (7, 8) also passes through:A (4, 3) b (5, 4) c (−2, 10) d (1, −8) E (1, 4)

14 The point (−1, 5) lies on a line parallel to 4x + y + 5 = 0. Another point on the same line as (−1, 5) is:A (2, 9) b (4, 2) c (4, 0) d (−2, 3) E (3, −11)

15 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

16 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)

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

18 Find the equation of the straight line that cuts the x-axis at 3 and is perpendicular to the line with equation 3y − 6x = 12.

19 Calculate the value of m for which the following pairs of equations are (i) parallel (ii) perpendicular.

a 2y − 5x = 7 and 4y + 12 = mxb 5x − 6y = −27 and 15 + mx = −3y

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

Hint: The diagonals of a rhombus intersect at right angles.

consolidAtE

DIGITAL DOCdoc-9935SpreadsheetParallel checker

MAstEr



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ApplicationsIn this section we look at two important applications: the equation of a straight line, and equations of horizontal and vertical lines.

The equation of a straight lineThe 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, in order to establish the value of c.

Note: Because the value of c represents the y-intercept, it can be substituted directly if known.

The alternative method comes from the gradient defi nition:

m = y2 − y1

x2 − x1

m(x2 − x1) = y2 − y1Hence

Using the general point (x, y) instead of the specifi c point (x2, y2) gives the general equation:

7.5

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 WritE1 Write the general equation. y = mx + c2 Find the gradient of the given line. y = 2x + 1 has a gradient of 2

Hence m = 2.3 Substitute for m in the general equation. so y = 2x + c4 Substitute the given point to fi nd c. (x, y) = (3, −1)

∴ −1 = 2(3) + c= 6 + c

c = −7

5 Substitute for c in the general equation. y = 2x − 7or2x − y − 7 = 0

WorKEd EXAMPLE 999

y − y1 = m(x = x1)

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

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.

tHinK WritE

1 Find the gradient of the perpendicular line. Given m = 5

m1 = −

15

WorKEd EXAMPLE 101010



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Horizontal and vertical linesFor horizontal lines the gradient is equal to zero, so the equation y = mx + c becomes y = c. Notice that x does not appear in the equation because there is no x-intercept. Horizontal lines are parallel to the x-axis. In the case of vertical lines, the gradient is infi nite or undefi ned. The general equation for a vertical line is given by x = a. In this case, just as the equation suggests, a represents the x-intercept. Notice that y does not appear in the equation because there is no y-intercept. Vertical lines are parallel to the y-axis. The graphs of y = 4 and x = −3 are shown at right to highlight this information.

x

y

0

4y = 4

x

y

0

x = ‒3

–3

Find the equation of:

a the vertical line that passes through the point (2, −3)

b the horizontal line that passes through the point (−2, 6).

tHinK WritEa For a vertical line, there is no y-intercept, so y does not appear in

the equation. The x-coordinate of the point is 2.a x = 2

b For a horizontal line, there is no x-intercept, so x does not appear in the equation. The y-coordinate of the point is 6.

b y = 6


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

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

then y − 3 = −

15(x − 0)

= −

x5

5(y − 3) = −x5y − 15 = −x

x + 5y − 15 = 0

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

tHinK WritE1 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 =y2 − y1

x2 − x1

m = 5 − (−4)6 − 0

= 96

= 32




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2 Find the gradient of the perpendicular line. For lines to be perpendicular, m2 = −

1m1

.

m1 = −

23

3 Find the midpoint of the line joining the given points.

x =x1 + x2

2

= 0 + 62

= 3

y = y1 + y2

2

= −4 + 52

= 12

Hence the coordinates of the midpoint are (3, 12).

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

Since y − y1 = m(x − x1)

and (x1, y1) = (3, 12) and m1 = −

23

then y − 12

= − 23(x − 3)

5 Simplify by removing the fractions.

(a) Multiply both sides by 3.

(b) Multiply both sides by 2.

3(y − 12) = −2(x − 3)

3y − 32 = −2x + 6

6y − 3 = −4x + 124x + 6y − 15 = 0

Note: The diagram at right shows the geometric situation.

(6, 5)5

–4

–4 6 x

y

30

2 1–2

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/drAW

a 1 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 A

B

C

D

5

2

–2–1

y

31 42


Tutorialeles‐1525Worked example 13



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Applications1 WE9 Find the equation of the straight line that passes through the point (4, −1)

and is parallel to the straight line with equation y = 2x − 5.

2 Find the equation of the line that passes through the point (3, −4) and is parallel to the straight line with equation y = −x − 5.

3 WE10 Find the equation of the line that passes through the point (−2, 7) and is perpendicular to a straight line with a gradient of 2

3.

4 Find the equation of the line that passes through the point (2, 0) and is perpendicular to a straight line with a gradient of −2.

5 WE11 Find the equation of:a the vertical line that passes through the point (1, −8)b the horizontal line that passes through the point (−5, −7).

PrActisE

ExErcisE 7.5

Digital docdoc-9936SpreadsheetEquation of a straight line

2 Find the gradient of BC. mBC = −2 − 22 − 4

= −4−2

= 2

3 State the gradient of AD. Since mBC = 2and AD||BCthen mAD = 2.

4 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 = 3Hence, the equation of AD is y = 2x + 3.

b 1 Find the gradient of AB. b mAB = 2 − 54 − 1

= −33

= −1

2 State the gradient of DC. Since mAB = −1and DC||ABthen mDC = −1.

3 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) + c c = 0Hence, the equation of DC is y = −x.

c To find D, solve simultaneously 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) = 1

Note: Alternatively, a calculator could be used to determine the point of intersection of AD.

Hence, the coordinates of D are (−1, 1).



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6 Find the equation of:a the vertical line that passes through the point (−1, 4)b the horizontal line that passes through the point (5, −2).

7 WE12 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 (−4, 0) and (2, −6).

9 WE13 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 AD b the equation of DC c the coordinates of D.

10 ABCD is a parallelogram. The coordinates of A, B and C are Q−13

, −53R, (1, 1) and

Q32, −1R respectively. Find:

a the equation of AD b the equation of DC c the coordinates of D.

11 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 = 0f Passing through the point (1, 6) and parallel to x − 3y − 2 = 0g Passing through the point (−1, −5) and perpendicular to 3x + y + 2 = 0

12 Find the equation of the line that passes through the point (−2, 1) and is:a parallel to the straight line with equation 2x − y − 3 = 0b perpendicular to the straight line with equation 2x − y − 3 = 0.

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

14 a The vertical line passing through the point (3, −4) is given by:A y = −4 b x = 3 c y = 3x − 4d y = −4x + 3 E x = −4

15 Which of the following points does the horizontal line given by the equation y = −5 pass through?A (−5, 4) b (4, 5) c (3, −5) d (5, −4) E (5, 5)

16 Which of the following statements is true?A Vertical lines have a gradient of zero.b The y-coordinates of all points on a vertical line are the same.c Horizontal lines have an undefined gradient.d The x-coordinates of all points on a vertical line are the same.E A horizontal line has the general equation x = a.

17 Which of the following statements is false?A Horizontal lines have a gradient of zero.b The straight line joining the points (1, −1) and (−7, −1) is vertical.c Vertical lines have an undefined gradient.d The straight line joining the points (1, 1) and (−7, 1) is horizontal.E A horizontal line has the general equation y = c.

consolidAtE



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18 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.

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

20 a The equation of the line passing through the point (4, 3) and parallel to the line 2x − 4y + 1 = 0 is:A x − 2y + 2 = 0 b 2x − y − 5 = 0 c 2x − y − 10 = 0d 2x − y − 11 = 0 E 2y + x + 2 = 0

b The equation of the perpendicular bisector of the line segment AB where A is (−3, 5) and B is (1, 7) is:A 2y = x + 13 b y = 2x − 8 c 2y = x + 11d y = −2x + 4 E y = 2x − 4

c The coordinates of the centroid of triangle ABC with vertices A (1, 8), B (9, 6) and C (−1, 4) are:A (4, 5) b (0, 6) c (3, 6) d (5, 7) E (2, 7)

21 The map at right 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, 2

3 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?

22 To supply cities with water when the source is a long distance away, artificial channels 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 source in Uzès to the city of Nîmes. The aqueduct does not follow a direct route between these two locations as there is a mountain range between them. The table shows the approximate distance from Uzès along the aqueduct to each town (or in the case of Pont du Gard, a bridge) and the aqueduct's height above sea level at each location.

MAstEr

56789

1011

4321

1 2 3 4 5 6 7 8 9 10 11 12 x

y

BBuoy

Scale: 1 unit ⇔ 1 km

A

H

M

E

O

Buoy

(Start)

CBuoy

DBuoy

N



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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

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

b Calculate the gradient of the steepest part of the aqueduct (in m/km).c Suppose the aqueduct started at Uzès

and ended at Nîmes but had a constant gradient. Write a linear equation to describe its course.

d Using the equation found in part c, calculate the height of the aqueduct at the Pont du Gard. This calculated height is higher than the actual height. How much higher?

e Why do you think the Romans made the fi rst part of the aqueduct steeper than the rest?



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studyON is an interactive and highly visual online tool that helps you to clearly identify strengths and weaknesses prior to your exams. You can then con� dently target areas of greatest need, enabling you to achieve your best results.


Units 1 & 2 Coordinate geometry

AOS #

Sit Topic test

Topic 7

Concept #

ONLINE ONLY 7.6 Review www.jacplus.com.au

the Maths Quest review is available in a customisable format for you to demonstrate your knowledge of this topic.

the review contains:• short-answer questions — providing you with the

opportunity to demonstrate the skills you have developed to ef� ciently answer questions using the most appropriate methods

• Multiple-choice questions — providing you with the opportunity to practise answering questions using CAS technology

• Extended-response questions — providing you with the opportunity to practise exam-style questions.

A summary of the key points covered in this topic is also available as a digital document.

REVIEW QUESTIONSDownload the Review questions document from the links found in the Resources section of your eBookPLUS.


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7 AnswersExErcisE 7.2 1 a AB = 5

b 5

2 a CD = 2!10 or 6.32

b 13

3, 4, 5, 6 and 7 Answers will vary.

8 a AB = 4.47, BC = 2.24, CD = 4.47, DA = 2.24

b AC = 5, BD = 5

c Rectangle

9 a 7.211

b 14.765

c 13.038

10 a "a2 − 4a + 8

b "b2 − 12b + 61

c "2 1c2 − 6c + 10 2d "5d2 − 22d + 26

11 B

12 D

13 a 12

b 5

c 13

d −2.2

14 Answers will vary.

15 21.024 km

16 a 216

b 108

ExErcisE 7.3 1 (−3, −3 1

2)

2 (7 12, 0)

3 (−3, −10)

4 (−5, 0)

5 a (3, 1)

b 4.47

c 6.32

6 D

7 C

8 a (4a, 3a)

b (8, 3c)

c a3f + g

2, 2b

9 a 5

b 9

10 a i (−1, 4)

ii (1 12, 1)

iii 3.9

iv 7.8

b Answers will vary.

11 a i (1, −0.5)

ii (1, −0.5)

b Answers will vary.

12 a i (−2, 2)

ii 8.94

iii 9.55

iv 9.55

b Isosceles triangle, height

13 y = −3x −2

14 3y − 2x + 14 = 0

15 D1 (−1.5, 4.25), D2 (−1.5, 1.5), D3 (8, −7.5)

16 y = 5x − 7

ExErcisE 7.4 1, 2, 3 and 4 Answers will vary.

5 5y = −2x + 1

6 3y + 2x + 1 = 0

7 and 8 Answers will vary.

9 b, f; c, e

10, 11, 12 Answers will vary.

13 B

14 E

15 a, e; b, f; c, h; d, g

16, 17 Answers will vary.

18 y = −

12x + 3

2

19 a m = −

85

b m = 185

20 Answers will vary.

ExErcisE 7.5 1 y = 2x − 9

2 y = −x − 1

3 3x + 2y − 8 = 0

4 x − 2y − 2 = 0

5 a x = 1

b y = −7

6 a x = −1

b y = −2



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7 y = −x − 3

8 y = x − 2

9 a y = −x + 5

b y = x + 3

c (1, 4)

10 a y = –4x − 3

b y = 2x − 4

c 116, −

1132

11 a y = 3x + 2

b y = −4x + 9

c 3x − 2y − 8 = 0

d 2x + 5y + 13 = 0

e x + 5y + 5 = 0

f x − 3y + 17 = 0

g x − 3y − 14 = 0

12 a 2x − y + 5 = 0

b x + 2y = 0

13 a 3x − 5y + 2 = 0

b 5x + 3y − 8 = 0

14 B

15 C

16 D

17 B

18 a (2, 5)

b 1

c Answers will vary.

d Isosceles triangle

19 4x − 6y + 23 = 0

20 a A

b D

c C

21 a 5.10 km

b (6.5, 5.5)

c 2

d y = 2x − 18

e (10, 2)

f 2:1

g (7, 3)

h 7.071 km

22 a

Distance from Uzès (km)

70

80

60

10 20 30 40 500

Height of aqueduct above sea level (m)

50

b −0.69 m/km

c y = −0.34x + 76, where 76 is the height in metres above sea level and x is the distance in km from Uzès

d 5.56 m

e Check with the teacher.

274 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE UnITS 1 And 2


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