WorkSHEET 3.2 Midpoint of a line segment Name:

© John Wiley & Sons Australia, Ltd Page 1

WorkSHEET 3.2 Midpoint of a line segment Name: ___________________________ 1 Find the equation of the graph shown.

2 Find the midpoint of the axis intercepts of the function 𝑦 = − !

"𝑥 − 8.

Refer to graph of 𝑦 = − !"𝑥 − 8 in question 1.

Axis intercepts are (-12,0) and (0,-8) Midpoint Rule: 𝑀 = '#!$#"

! , %!$%"

!*

= +−12 + 0

2 ,0 − 82 0

= (−6,−4)

3 Find the distance between the axis intercepts of the function 𝑦 = − !

"𝑥 − 8.

Refer to graph of 𝑦 = − !"𝑥 − 8 in question 1.

Axis intercepts are (-12,0) and (0,-8) Distance rule: 𝐷 = 6(𝑦! − 𝑦&)! + (𝑥! − 𝑥&)!

= 6(−8 − 0)! + (0 − −12)! = √208 = 14.42

( )

8 212 3

23

Using 0, 8 :28 038

2 83

m

y x c

c

c

y x

-= = -

= - +

-

- = - ´ +

= -

\ = - -


4 Clearly that last question HAS to be in the “calculator” exam? NO! Try this one … Find the distance between the axis intercepts of the function 𝑦 = "

'𝑥 + 3.

Find the axis intercepts; y-axis intercept, set 𝑥 = 0

𝑦 =34 × 0 + 3 𝑦 = 3

x-axis intercept, set 𝑦 = 0

0 =34𝑥 + 3

𝑥 = −4 Points are (0,3) and (-4,0) Distance rule: 𝐷 = 6(𝑦! − 𝑦&)! + (𝑥! − 𝑥&)!

= 6(0 − 3)! + (−4 − 0)!

= √9 + 16

= √25

= 5

5 How about; Find the distance between the points (1,9) and (−2,5).

Distance rule: 𝐷 = 6(𝑦! − 𝑦&)! + (𝑥! − 𝑥&)!

= 6(5 − 9)! + (−2 − 1)!

= √16 + 9

= √25

= 5

6 Sketch the graph of y = x - 1 using the

gradient–intercept method.

32 2

31

m

c

=

= -


7 If the x-intercept for a linear graph is 4 and the y-intercept is -2 then. (a) Sketch the graph.

(b) Find the equation of the linear graph.

(a)

(b)

8 Find the equation of a line with the same gradient as the line y = -2x + 1 and passing through the point (2, 4).

9 The length of a line segment is 10 units. If the endpoints of the line segment are (3,6) and (9, 𝑘), determine the value of 𝑘.

Distance rule: 𝐷 = 6(𝑦! − 𝑦&)! + (𝑥! − 𝑥&)!

10 = 6(𝑘 − 6)! + (9 − 3)! Square both sides

100 = (𝑘 − 6)! + 36 Recognise this is a Quadratic equation, so need to get it in the right form … expand brackets and rearrange;

100 = 𝑘! − 12𝑘 + 36 + 36

0 = 𝑘! − 12𝑘 − 28

0 = (𝑘 + 2)(𝑘 − 14) NFL:

𝑘 = −2𝑜𝑟14 Plot the two solutions so you can see how it makes sense there are 2 answers!

221

-= xy

22

Using (2,4) :4 2 24 4

4 48

2 8

my x c

cc

cc

y x

= -= - +

= - ´ += - += +=\ = - +


10 Consider the function defined by the 2 points 𝐴(−1,1) and 𝐵(−2,−1). Determine the value of 𝛼, given that the point 𝐶(1, 𝛼) lies on the line.

First find the line defined by AB. Takes the form … 𝑦 = 𝑚𝑥 + 𝑐

𝑚 = 𝑠𝑙𝑜𝑝𝑒 =𝑅𝑖𝑠𝑒𝑅𝑢𝑛 =

𝑦! − 𝑦&𝑥! − 𝑥&

=1 − −1−1 − −2 = 2

So have; 𝑦 = 2𝑥 + 𝑐

Because the point 𝐴(−1,1) lies on the line, it must satisfy the equation, by sub;

1 = 2 × −1 + 𝑐 𝑐 = 3

So have;

𝑦 = 2𝑥 + 3 Now, because the point 𝐶(1, 𝛼) lies on the line, it must also satisfy the equation, by sub;

𝛼 = 2 × 1 + 3

∴ 𝛼 = 5


11 Determine of the following points are colinear: A(-5, 20), B(2, -1) and C(4, -7). Colinear means the points all lie on the same line. These points all lie on the same straight line if it can be shown that the gradient of AB is equal to the gradient of BC. Show that A, B and C lie on the same straight line.

Let (x1, y1) = (-5, 20) and (x2, y2) = (2, -1).

Thus, the gradient of the line AB = -3. Similarly, let (x1, y1) = (2, -1) and (x2, y2) = (4, -7).

Thus, the gradient of the line BC = -3. Since , the points A, B and C are on the same straight line.

2 1AB

2 1

AB

AB

1 202 52173AB

y ymx x

m

m

m

-=

-- -

=- -

-=

= -

2 1BC

2 1

BC

BC

BC

7 14 2623

y ymx x

m

m

m

-=

-- - -

=-

-=

= -

AB BCm m=


12 Determine of the following points are colinear: A(5, 7), B(9, 11) and C(1, 3). Colinear means the points all lie on the same line. These points all lie on the same straight line if it can be shown that the gradient of AB is equal to the gradient of BC. Show that A, B and C lie on the same straight line.

This is a different strategy to the last question. I will find the equation of the line through the point A and B and then I will see if C lies on that line. Find the line defined by point A and B; In the form:

𝑦 = 𝑚𝑥 + 𝑐 Find 𝑚

𝑚 =𝑟𝑖𝑠𝑒𝑟𝑢𝑛 =

𝑦! − 𝑦&𝑥! − 𝑥&

=11 − 79 − 5

=44

= 1 So we have;

𝑦 = 𝑥 + 𝑐 Because the point A(5,7) lies on the line it must satisfy the equation, so by substitution;

7 = 5 + 𝑐 𝑐 = 2

And the line defined by A(5,7) and B(9,11) is

𝑦 = 𝑥 + 2 Check to see if C(1,3) lies on this line; by substitution we have,

3 = 1 + 2 3 = 3

Therefore all three points A, B and C all lie on the same line J


13 Determine the rule for the straight line whose x-intercept = 10 and y-intercept = –5.

Use the two intercepts (10, 0) and (0, –5) to determine the gradient.

Therefore the rule is of the form:

y = x + c.

But c = y-intercept = –5, so y = x – 5.

14 Find the distance between the pairs of points (-6, 5) and (-2, -7) correct to 1 decimal place.

15 Find the midpoint of the line segment joining the points (-3, -5) and (-2, 11).

Let = (-3, -5) and = (-2, 11).

21

105

10005 Gradient

=

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=

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çèæ ++

=

3,212

26,

25

2115,

223

2

,2

Midpoint 2121 yyxx


16 Find the midpoint between the axis intercepts of the function;

𝑦 = 2𝑥 − 8


𝑦 = 2 × 0 − 8 𝑦 = −8

x-axis intercept, set 𝑦 = 0 0 = 2𝑥 − 8 𝑥 = 4

Points are (0,-8) and (4,0) Midpoint 𝑀 = '#!$#"

!, %!$%"

!*

= +0 + 42 ,

−8 + 02 0

= (2,−4)

17 Find the midpoint between the axis intercepts of the function;

6𝑥 − 2𝑦 + 12 = 0


0 − 2𝑦 + 12 = 0 𝑦 = 6

x-axis intercept, set 𝑦 = 0 6𝑥 − 0 + 12 = 0

𝑥 = −2 Points are (0,6) and (-2,0) Midpoint 𝑀 = '#!$#"

!, %!$%"

!*

= +0 + −22 ,

6 + 02 0

= (−1,3)


18 The midpoint M of a line segment PQ has the coordinates (5, 4). If the coordinates of P are (1, -6), find the coordinates of Q.

Coordinates of Q are (9, 14).

19 The line AB passes through the points A(4, 8) and B(10, 17). The line PQ passes through the points P(-2, -9) and Q(6, 3). Show that AB is parallel to PQ.

Let (x1, y1) = (4, 8) and (x2, y2) = (10, 17).

Similarly, let (x1, y1) = (-2, -9) and (x2, y2) = (6, 3).

Since the gradients of lines AB and PQ are equal (m=1.5), the lines are parallel.

9110

101

521

1

1

1

=-=

=+

=+

xx

x

x

x

1468

86

426

1

1

1

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=+=

=-

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y

y

5.169

410817

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12

=

=

--

=

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=

m

m

m

xxyy

m

2 1

2 1

Using

3 96 2128

1.5

y ymx x

m

m

m

-=

-- -

=- -

=

=


20 Blackbeard the pirate is searching for buried treasure using a map. He is located at a point represented on his map by the point (−7 − 1). There is a shack on the map at point (6, −1). There is also a tree on the map at point (1,1). The map states the treasure is buried in a straight line with the tree and the shack. Also, the tree is located in the middle of the treasure and the shack. By finding the location of the treasure, determine how far Blackbeard has to walk to the treasure. Can you do this withOUT a calculator?

First, determine where the Treasure is. The tree is MIDPOINT between the Treasure and the shack, so use the Midpoint rule;

𝑀 = +𝑥& + 𝑥!2 ,

𝑦& + 𝑦!2 0

So,

(1,1) = +6 + 𝑥2 ,

−1 + 𝑦2 0

Clearly 6 + 𝑥2 = 1

𝑥 = −4

And further −1 + 𝑦2 = 1

𝑦 = 3

So the treasure is located at a point (−4,3). Now the distance from Blackbeard, who is standing at (−7,−1), to the treasure. Use the distance formula;

𝐷 = 6(𝑦! − 𝑦&)! + (𝑥! − 𝑥&)!

= 6(−1 − 3)! + (−7 − −4)!

= 6(−4)! + (−3)!

= √16 + 9

= √25

= 5 So, Blackbeard is 5 km from being RICH!

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WorkSHEET 3.2 Midpoint of a line segment Name: