Spatial Reasoning

Learning about straight lines is important because it helps us understand direction and route-planning to ensure efficient travel and transportation. We can also apply the concept of straight lines in architecture and different kinds of art forms. This section introduces you to the concept of straight lines, how to determine whether they are parallel or perpendicular and the properties associated with such lines.

You also learn how to determine the midpoint of a straight line, divide such lines in a given ratio either internally or by extending the line segment and find the equation of straight lines. You will explore the ways of determining the sizes of acute angles formed from the intersection of two straight lines.

At the end of this section, you will be able to:

• Describe the properties of lines including parallel, perpendicular and midpoints.

• Work out the midpoint of a line segment given two points and find the generalization of the midpoint of a line segment.

• Apply the knowledge of ratio to divide a line segment in a given ratio either internally or externally.

• Recall the formula for finding the gradient of a line and apply it to find the equation of a straight line in various forms

• Use standard algebraic manipulations to find the equation of parallel and perpendicular lines including the equation of perpendicular bisector of a line

• Deduce the shortest distance between a point and a line and use the knowledge of intercepts and right-angled triangles to find the perpendicular distance from an external point to a line.

• Determine the acute angles between two intersecting lines with the aid of technological tools e.g. GeoGebra.

Key Ideas:

• When a line is straight there is no curve on the line.

• Parallel lines are straight lines that do not meet.

• Perpendicular lines are straight lines that meet at an angle of 90⁰.

• A midpoint is a point that divides a straight line into two equal parts.

• A straight line is the shortest distance between two points.

• A straight line can be divided internally using a given ratio.

• Acute angles are less than 90⁰, right angles are exactly 90⁰and obtuse angles are greater than 90⁰but less than 180⁰.

• Equations of lines can be written in the slope-intercept form and point- slope form.

Like every artifact, some features are specific to it. Dealing with straight lines, certain characteristics determine whether a line is straight or not and these traits are referred to as properties of straight lines. The straight lines can be horizontal, vertical, slanted, parallel and perpendicular.

Activity 5.1

Figure 1: Vertical Line Figure 2: Perpendicular Lines Figure 3: Parallel lines

Figure 4: Horizontal line Figure 5: Slanted line

i. Choose a destination (Dining hall, School field, Assembly Hall, etc.).

ii. Walk along a straight path to your chosen destination.

iii. Record the time and distance from your starting point to your destination.

iv. Now, walk along a path with bends and curves to your chosen destination.

v. Record the time and distance from your starting point to your destination.

vi. Compare the distances covered in both scenarios and time taken and share your findings with a colleague.

Some of the properties include:

i. A straight line is formed when two points are joined with the shortest distance, it can be extended forever in both directions.

ii. A straight line has no curves within.

iii. A straight line is one-dimensional and has no width.

Parallel and Perpendicular Lines

Parallel lines are two or more straight lines that are always the same distance (equidistant) apart. They never intersect. Examples include opposite ends of a goal post, railway tracks, edges of a ruler, zebra crossing (parallel white lines).

Perpendicular lines are two lines that intersect, but all the angles at that intersection are the same, that is 90°. For example, “T” junctions on roads, corners of a football pitch etc.

Note:

1. Distances are always positive.

2. Distance can only be zero if the points coincide.

3. The distance from P to Q is the same as the distance from Q to P

Activity 5.2

i. Pick a coordinate grid paper/graph sheet

Figure 6: Coordinate grid paper/graph sheet

ii. Plot points B(3, 2) and E(4, 6) on the grid and draw a straight line from one point to the other.

Figure 7: Straight Line

iii. Is line BE slanted, horizonal or vertical?

iv. Measure the length of the line starting from point B and ending at point E and record.

v. Repeat steps 2 and 3 using different self - selected points

vi. Now, use d = √_______________ (x₂− x₁)²+ (y₂− y₁)²to calculate the distance of line BE (|BE|) and the other lines created from exploring with other points.

vii. Record your observations and discuss with a classmate.

Generalisation: Distance of a line (d) = √_______________ (x₂− x₁)²+ (y₂− y₁)²where x₁, x₂ and y₁, y₂ are x and y coordinates of the first and second points respectively.

Ready to consolidate your knowledge? Let’s try out some examples!

Example 1

Given that U (-7, 23) and V (6, 18) are the endpoints of a line UV, calculate |UV|.

Solution

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁ in the distance formula |ᵁⱽ| = √__________________ (6− ( − 7))²+ (18− 23)²Step 2: Simplify the terms under the square root |ᵁⱽ| = √169 + 25

Step 3: Add the terms under the square root |UV| = √194

Step 4: Write out final answer in decimal unless instructed otherwise |UV| = 13.93 (to two decimal places)

Example 2

Calculate the distance between the points N (12, 5) and R (-3, -6).

Solution

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁ in the distance formula |ᴺᴿ| = √________________ (− 3− 12)²+ (− 6− 5)²

Step 2: Simplify the terms under the square root |ᴺᴿ| = √225 + 121

Step 3: Add the terms under the square root |NR| = √346

Step 4: Write out final answer in decimal unless instructed otherwise |NR| = 18.60 (to two decimal places) Midpoint of a Line Segment A midpoint is a point that divides a line segment into two equal parts. The midpoint is equally distant from both ends of the line segment.

Activity 5.3

i. Choose a coordinate grid paper

Figure 8: Coordinate grid paper/graph sheet

ii. Plot points A(1, 2) and Y(5, 6) on the grid and draw a straight line from one point to the other.

iii. Fold the grid paper such that the line is divided into two equal parts.

Figure 8: Straight line divided into two equal parts

iv. Indicate the coordinates of the point where the straight line gets divided into two equal parts.

v. Repeat the steps ii to iv using different sets of points to draw different straight lines

vi. Record your observations on the values that make up the coordinates of the points that divide the lines into two equal parts in relation to the end points.

Generalisation: Midpoint (M) = (1/2 (x₁ + x₂) , 1/2 (y₁ + y₂)) where x₁, y₁ and x₂, y₂ are x and y coordinates of the first and second points respectively.

Are you Ready to consolidate your knowledge? Let’s try out some examples!

Example 3

Find the midpoint of the line KG with coordinates K (16, 24) and G (-8, 10).

Solution

Step 1: Let M be the midpoint of line KG Substitute the corresponding values for x₂, x₁, y₂, y₁ in the midpoint formula M (1/2 (16 + ( − 8)) , 1/2 (24 + 10))

Step 2: Simplify the terms in the brackets multiplied by 1/2 M(1/2(8), 1/2(34))

Step 3: Divide where necessary M(4 , 17)

Example 4

Given the coordinates of line RP as R (1, 17) and P (0, 4), determine the midpoint of line RP.

Solution

Step 1: Let M be the midpoint of line RP Substitute the corresponding values for x₂, x₁, y₂, y₁ in the midpoint formula M (1/2 (1 + 0) , 1/2 (17+ 4))

Step 2: Simplify the terms in the brackets multiplied by 1/2 M (1/2 (1) , 1/2 (21))

Step 3: Divide where necessary M(0.5 , 10.5)

Example 5

W is the midpoint of DV. If the coordinates of D are (− 5, 4) and W is (− 2, 1), find the co-ordinates of V.

Solution

Step 1: Let point V be (xᵥ, yᵥ) Substitute the corresponding values for x₂, x₁, y₂, y₁ in the midpoint formula W = (1/2 (− 5+ xᵥ) , 1/2 (4 + yᵥ))

Step 2: Remember W is (− 2, 1), now substitute W to get (− 2, 1) = (1/2 (− 5+ xᵥ) , 1/2 (4 + yᵥ))

Step 3: Treat the inputs in step 2 as points, pair corresponding x and y coordinates.

For x coordinate;

− 2 = 1/2 (− 5+ xᵥ) For y coordinate;

1= 1/2 (4 + yᵥ)

Step 4: Simplify for each of the coordinates and make xᵥ and yᵥ the subjects to get;

xᵥ = 1 , yᵥ = − 2

Step 5: Write out your conclusion Therefore, the coordinates of V are (1, − 2 ).

Lines can be divided by cutting them into specific parts either equally or unequally.

This division can be done with the help of ratios. Take a broomstick for example, you can divide this stick into parts internally (within) or externally (outwards).

Let’s talk about internal division of straight lines!

To divide a line segment into two parts internally, place a point somewhere between the two endpoints. This new point creates a ratio between the two resulting segments. For example, dividing the line in half creates a 1:1 ratio which is the same as finding the midpoint discussed earlier. Suppose you are asked to divide a line segment KE in the ratio 2: 3, it means that you find a point, say C such that KC: CE = 2: 3 which is the same as KC___ CE = 2/3.

Now, let’s move on to external divisions of straight lines!

To divide a line segment into two parts externally, place a point somewhere outside the original line segment. This point also creates a ratio but it describes the lengths relative to the original segment not parts within it.

Come along as we do this activity for a better understanding

Activity 5.3

Internal Division

i. Draw a line segment KE on a coordinate grid paper.

ii. Choose a ratio m:n (e.g. 2:1).

iii. Divide the line segment into m + n equal parts.

iv. Label the point dividing the line segment according to the ratio as N

Figure 9: Line Segments

v. Measure the length of KN and NE on the coordinate grid paper.

vi. Verify the length by calculating the length of KN and NE using the ratio External Division

i. Draw a line segment MQ on a coordinate grid paper. Q is an extension of the line MN, N will lie somewhere along the line MQ.

ii. Choose a ratio m:q (e.g. 5:1), where m is the ratio of the length of MQ and q is the ratio of the length NQ

iii. Divide the extended line segment into m equal parts (where m>q).

iv. Label the point dividing the line segment as N

Figure 10: Line Segment

v. Measure the length of MQ and QN on the coordinate grid paper.

vi. Verify the length by calculating MQ and QN using the ratio.

Let’s tap into your creativity, critical thinking and collaborative skills!

1. Pick out your own sets of points for different line segments and choose different ratios.

2. Now, apply the steps in the activities on internal and external division of lines and document the different results you get.

3. How about using the generalizations below on the points and ratios you picked?

4. Did you get the same results as that generated from the activity?

5. Discuss your observations with a friend or classmate.

Generalisations:

1. The coordinates P, which divides the line segment K (x₁, y₁) and E (x₂, y₂) internally in the ratio m:n is given by:

P = (mx₂ + n x₁________ m + n , my₂ + n y₁________ m + n ).

2. The coordinates Q, which divides the line segment A(x₁, y₁) and B(x₂, y₂) externally in the ratio m:n is given by:

Q = (ᵐˣ²− n x₁________ m − n , my₂ − n y₁________ m − n ).

Note: You do not always negate ‘n’ but you negate the smaller number in the ratio for external division of lines.

Understanding both internal and external division of lines is important in various applications like scaling distances on maps or solving geometric problems.

Are you up for a challenge? Let’s try out some examples!

Example 6

G and K divide the line FH, F(13, 9) and H(6, -17) internally and externally respectively in the ratio 11:3. Find the coordinates of G and K.

Solution

Focusing the internal division, Let m = 11 and n = 3.

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁, m and n in the internal division formula.

G(11(6)+ 3(13)__________ 11 + 3 , 11( − 17)+ 3(9)____________ 11+ 3 ),

Step 2: Simplify the terms substituted G(7.5, − 11.43).

Focusing the external division, Let m = 11 and n = 3.

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁, m and n in the external division formula.

K(11(6) − 3(13)___________ 11 − 3 , 11( − 17) − 3(9)____________ 11 − 3 ),

Step 2: Simplify the terms substituted K(3.375, − 26.75).

Example 7

Find the coordinates of the point that divides the lines segment (−4, 3) and (6, -12) in the ratio 3: 2, internally and externally.

Solution

Focusing the internal division, Let m = 3 and n = 2.

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁, m and n in the internal division formula.

(3(6)+ 2( − 4)___________ 3 + 2 , 3( − 12)+ 2(3)___________ 3+ 2 ),

Step 2: Simplify the terms substituted (2, − 6).

Focusing the external division, Let m = 3 and n = 2.

Step 1: Substitute the corresponding values for x₂, x₁, y₂, y₁, m and n in the external division formula.

(3(6)− 2( − 4)___________ 3 − 2 , 3( − 12)− 2(3)___________ 3 − 2 ),

Step 2: Simplify the terms substituted (26, − 42).

If A (x₁, y₁) and B (x₂, y₂) then given any arbitrary point P(x, y) on the line AB we can generate the equation of a line.

Let’s derive the formula for the equation of a line using standard algebraic manipulations.

Recall that the formula for finding the gradient of a line is given by:

m = y₂ −y₁_____ x₂ − x₁ or y₁ − y₂_____ x₁ − x₂.

Well, if there is another point P(x, y) on the line segment with endpoints A(x₁, y₁) and B(x₂ , y₂), then the gradient of line AP becomes m = y −y₁_____ x − x₁ which is equal to m = y₂ −y₁_____ x₂ − x₁.

Activity 5.4

i. Equate the gradient for lines AP and AB y − y₁_ x – x₁ = y₂ − y₁_ x₂ – x₁

ii. Multiply both sides by (x – x₁) y − y₁ = ( y₂ −y₁_____ x₂ – x₁) (x – x₁) ….......(equation 1)

iii. Remember m = y₂ −y₁_____ x₂ − x₁ , now substitute m = y₂ −y₁_____ x₂ − x₁ into equation 1 y − y₁ = m (x − x₁)

iv. Expand terms on the RHS of the equation y −y₁ = mx − m x₁

v. Add y₁ to both sides:

y = mx − m x₁ + y₁

vi. Represent − m x₁ + y₁ by a variable (c) Now, y = mx + c Congratulations!! You have used algebra to generate the equation of a straight line in two forms. First the point-slope form (y −y₁ = m ( x − x₁) ) and the slope- intercept form (y = mx + c).

Note, horizontal lines have zero gradient, so these are written in the form y = c .

Vertical lines have infinite gradient, so these are written in the form x = c.

Two lines are perpendicular when they meet (intersect) at a right angle (90⁰) and two lines are said to be parallel when they never intersect no matter how far you extend the lines.

Whether perpendicular or parallel, they are lines and thus their equations can be determined.

Let us explore the relationship between parallel and perpendicular lines!

Activity 5.5

i. Pick a coordinate grid paper

ii. Choose two points and draw a line segment

iii. Choose another set of points and draw a different line segment (Note:

this line should not intersect the first no matter how far it is extended).

iv. Find the gradients of the two lines

v. Record your observation

vi. Pick a different set of points and draw a line (Select your points in such a way that the line intersects the first line in step iii at 90⁰).

vii. Find the gradient of the third line.

viii. What conclusion can you draw on the gradients of the first and the third lines; second and third lines?

Generalisations:

1. Parallel lines have the same gradient

2. When the slope of one line is m, then the slope of the line perpendicular to it will be − 1/m .

3. If two lines are perpendicular then the product of their gradients, m × − 1/m is − 1 Let’s try out some examples!

Example 8

a) Find the equation of the line that passes through the points S (4, −2) and T (8, 4), leaving your final answer in the form ax + by + c = 0 where a, b and c are integer values.

b) If line JK is perpendicular to line ST and passes through the point W (-7, 5) find the equation of line that passes through JK.

c) Suppose line ED is parallel to line JK, and passes through L (2, 9), find the equation of line that passes through ED.

Solution

a)

Step 1: Find the gradient of line ST.

m₁ = 4 − ( − 2)________ 8− 4 m₁ = 3/2

Step 2: Use the point-slope form for equation of line and substitute values accordingly.

y − y₁ = m₁ (x − x₁) y − ( − 2) = 3/2 (x − 4)

Step 3: Expand the brackets and multiply both sides of the equation by 2.

2y + 4 = 3 x – 12

Step 4: Group all terms on one side of the equation and simplify.

2y + 4 – 3x + 12 = 0

– 3x + 2y + 16 = 0 The equation of the line that passes through ST is –3x + 2y + 16 = 0 or 3x – 2y–16 = 0.

b)

Step 1: Determine gradient of JK based on relations between perpendicular lines.

m₂ = –1/m₁ m₂ = –2/3

Step 2: Use the point-slope form for equation of line and substitute values accordingly.

y − y₁ = m₂ (x − x₁) y − 5 = –2/3 (x − ( − 7))

Step 3: Expand the bracket and multiply both sides of the equation by 3.

3y − 15 = − 2 x − 14

Step 4: Group all terms on one side of the equation and simplify.

3y − 15 + 2x + 14 = 0 2x + 3y − 1 = 0 The equation of the line that passes through JK is 2x + 3y − 1 = 0 .

c)

Step 1: Determine gradient of ED based on relations between parallel lines.

m₃ = m₂ m₃ = –2/3

Step 2: Use the point-slope form for equation of line and substitute values accordingly.

y − y₁ = m₃ (x − x₁) y − 9 = –2/3 (x − 2)

Step 3: Expand the bracket and multiply both sides of the equation by 3.

3y − 27 = − 2 x + 4

Step 4: Group all terms on one side of the equation and simplify.

3y − 27 + 2x − 4 = 0 2x + 3y − 31 = 0 The equation of the line that passes through ED is 2x + 3y − 31 = 0 .

Investigate the shortest distance between a point and a line, as well as the shortest distance between two lines. Establish that the shortest distance between a line and a point is the perpendicular distance, D.

Activity 5.6

Perpendicular Distance

i. Graph a line on the coordinate grid, y = mx + c.

ii. Plot a point not on the line: (x₁, y₁).

iii. Draw a perpendicular line from the point (x₁, y₁) to the original line

iv. Label the point of intersection: (x₂, y₂)

v. Calculate the distance between (x₁, y₁)and (x₂, y₂)) using the distance formula:

d = √_______________ (x₂− x₁)²+ (y₂− y₁)²

vi. Use the formula to find the distance from a point to a line D = | ᵃˣ1 + by₁ + c |___________ √a²+ b²vii. Plug in values for a, x₁, y₁, b and c.

viii. Record your observations and discuss with a classmate.

Example 9

Find the shortest distance between the perpendicular line drawn from the point J (1, 5) to the straight line 5x + 12y + 7 = 0.

Solution

Step 1: Substitute the corresponding values for a, b, c, x₁, y₁ in the shortest distance formula D = | ᵃˣ1 + by₁ + c |___________ √a²+ b²D = |5(1) + 12(5) + 7 |_____________ √5²+ 12²Step 2: Simplify the terms in absolute and those under the square root D = |72 |____ √169

Step 3: Simplify the result D = 5.54 units to two decimal places

Example 10

Telecel is setting up a new cell tower at a location C (6, −2) on the map of Accra.

The main road, which serves as a reference line, is represented by the equation 3x − 4y = 12. Determine the perpendicular distance from point C (6, −2) to the main road.

Solution

Step 1: Substitute the corresponding values for a, b, c, x₁, y₁ in the shortest distance formula D = | ᵃˣ1 + by₁ + c |___________ √a²+ b² D = |3(6) − 4( − 2) − 12 |_______________ √3²+ 4²Step 2: Simplify the terms in absolute and those under the square root D = |14 |___ √25

Step 3: Simplify the result D = 2.8

In geometry, a triangle is a closed, three-sided polygon. Each of the three sides meet at a point called a vertex and the angles formed between these sides within the triangle are called interior angles. Every triangle has exactly three interior angles. The measures of the three interior angles of any triangle sums up to 180⁰.

Triangles can be classified into categories based on the measure of their interior angles, namely:

• Acute Triangle: All three interior angles are less than 90 degrees (acute).

• Right Triangle: One interior angle is exactly 90 degrees (right angle).

• Obtuse Triangle: One interior angle is greater than 90 degrees (obtuse).

Let’s focus on measuring acute angles through this activity!

Activity 5.7

i. Plot the points (1, -1) and (3, 3) and join them with a straight line on a coordinate grid paper.

ii. Plot the different sets of point (1, -1) and (-1, 5) and connect with a straight line on the same coordinate grid paper.

iii. Measure the interior angle formed by the lines at the point of intersection using a protractor.

iv. Label the angle as θ (theta).

v. Calculate the difference between the slopes (gradient) of the two lines.

vi. Multiply the gradients of the two lines.

vii. Compute m₁ − m₂___________ ( 1 + m₁ × m₂) where m₁, m₂ are the gradients of the two lines

viii. Calculate tan⁻¹( m₁ − m₂___________ ( 1 + m₁ × m₂) ) on your calculator.

ix. Compare the values recorded in steps 3 and 8.

x. Write your observations Do you have the GeoGebra Software? If yes, let’s explore how to measure angles!

Activity 5.8

i. Open GeoGebra and create a new worksheet

ii. Construct two lines using the “Line” tool, y = 2x – 3 and y = -3x + 2

iii. Use the “Slider” tool to adjust the slopes (m₁ and m₂) and y-intercepts ( c₁ and c₂)

iv. Now, use the “Angle” tool to measure the angle between the two lines

v. Label the angle as θ (theta)

vi. Use the “Slider” tool to adjust the slopes and y-intercepts and observe how the angle changes.

vii. Use the “Calculate” tool to calculate the difference between the slopes , m₁ and m₂.

viii. Calculate the product of the slopes, m₁ and m₂.

ix. Explore the relationship between the angle θ and the slope calculations

x. Now find the value of θ using tan(θ) = ( m₁ − m₂___________ (1 + m₁ × m₂) ), with the help of a calculator.

xi. What are your observations? Record and discuss with a classmate.

Note: For acute angles tanθ = ( m₁ – m₂_______ 1 + m₁m₂) Intersection of Straight Lines The point of intersection of straight lines refer to the exact points where straight lines meet.

When dealing with two straight lines, given by equations y = m₁x +c₁and y = m₂ x + c₂, to find their intersection point, we equate the two equations: m₁x +c₁ = m₂ x + c₂.

Then, solve for x to determine the x-coordinate of the intersection point. Once x is found, substitute it into either of the original equations to obtain the corresponding y-coordinate.

Example 11

Given two lines y = 7x − 13 and y = − 2x + 6 find their point of intersection.

Solution

Step 1: Equate the two equations 7x – 13 = − 2x + 6

Step 2: Make x the subject 7x + 2x = 6 + 13 9x = 19 x = 19/9

Step 3: Substitute x in any of the original equations to find y y = 7(19_ 9 )– 13

Step 4: simplify the equation y = 16/9 The lines intersect at (19/9 , 16/9 )

Example 12

A transportation planner is designing two new bus routes in Kasoa. The routes will be represented by straight lines on the map of Kasoa. The goal is to identify the most favourable point of intersection between the two routes that can serve as a major bus stop. Route A is designed to pass through the points (2, 5) and (8, 17). Route B is designed to pass through the points (3, 20) and (9, 2). Calculate the point of intersection between Route A and Route B.

Solution

Step 1: Find the equation of the line passing through Route A m_(A) = 17 − 5/8 − 2 m_(A) = 12/6 m_(A) = 2 y − 5 = 2 (x − 2) y − 5 = 2x − 4 y = 2x + 1

Step 2: Find the equation of the line passing through Route B m_(B) = 2− 20/9 − 3 m_(B) = –18/6 m_(B) = − 3 y − 20 = − 3 (x − 3) y − 20 = − 3x + 9 y = − 3x + 29

Step 3: Equate the two equations 2x + 1 = − 3x + 29

Step 4: Make x the subject 5x = 28 x = 28/5

Step 5: Substitute x in any of the original equations to find y y = 2(28_ 5 ) + 1

Step 6: Simplify the equation y = 61/5 The lines intersect at (28/5 , 61/5 ).

Example 13

Determine the acute angle between two straight lines having slopes of 4 and 2/7.

Give your answer to two decimal places.

Solution

Step 1: Substitute the gradients in the formula for finding acute angles tan( θ ) = | 4 − 2__ 7/1 + 4 (2/7)|,

Step 2: Simplify terms on RHS of the equation tan (θ) = 26/15

Step 3: Take tan⁻¹of the terms θ = tan⁻¹(26_ 15) θ = 60.02⁰The acute angle between the two straight lines is 60.02⁰.

Example 14

Find the acute angle formed between the lines y = 19x − 5 and y = − 6x + 4

Solution

Step 1: Determine the gradients of the two lines.

m₁ = 19 , m₂ = − 6

Step 2: Substitute the gradients into the acute angle formwula tan( θ ) = | |19 − (− 6)|__________ (1 + 19(− 6))| ,

Step 3: Simplify terms on RHS of the equation tan (θ) = 25/113

Step 4: Take tan⁻¹of the terms θ = tan⁻¹( 25_ 113) θ = 12.48⁰The acute angle between the two straight lines is 12.48⁰.

Now, pick up some challenging tasks in the Review Questions. You’ve got this!

1. The coordinates of two points are F (3, 7) and Y (-2, 8).

Determine the distance between F and Y, |FY|.

2. An engineer is designing a suspension bridge that spans two river banks. The coordinates of the feet of the bridge on the river banks are R (3, 5) and G (11, 7).

Find the coordinates of the midpoint of the river where the main support pillar will be anchored.

3. Given three points A (3, 2), B (8, 11) and C (14, 5) on a coordinate plane:

i. Determine a point P on segment AB such that the line segment AB is divided internally in the ratio 3:2.

ii. Determine a point Q on segment BC such that the line segment BC is divided internally in the ratio 4:1.

iii. Find the equation of the line passing through P and Q.

iv. Find a point R on segment AC such that AR is divided internally in the ratio 5:3.

v. Analyse the position of this line relative to the original triangle △ABC.

Does this line bisect any sides of the triangle?

Provide a reasoned argument based on the coordinates and the properties of line division.

4. Given three points A (-6, 4), B (2, 0) and M (12, -5) on a coordinate plane.

If M divides line AB externally in the ratio 9:n, find the value of ‘n’.

5. Find the equation of the line that passes through the points D (25, 6) and C (29, -14).

6. The equation of a line passing through the points M (4, 2) and T (− 8, − 2) is:

3y = ax + b, where a and b are constants. Find the values of a and b.

7. Find the equation of the line perpendicular to 5x – 3y -18 = 0 that passes through the points (-5, 6).

8. Determine the equation of the line parallel to 3x + 5y = 108 and passes through the point (5, 10).

9. Find the measure of acute angles between the lines y = 2x +10 and y = – 5x + 4.

10. Find the length of the perpendicular line drawn from the point B(−1, − 7) to the straight line passing through the points E(6, − 4) and Y(9, − 5).

11. Determine the equation of the line passing through points (3, 6) and (1, 2) using the point-slope form.

12. The Agona West Municipal Assembly wants to ensure that specific streets are parallel or perpendicular to each other to create an organised layout in its residential area. Main Street runs through the city and follows the equation y = 2x + 3. Nana Botwe Street runs parallel to the Main Street and pass through the point (4, 1). Bebianiha Street will be perpendicular to the Main Street and intersect it at the point (1, 5).

i. Determine the equation of Nana Botwe Street.

ii. Determine the equation of Bebianiha Street.

iii. The Assembly also wants to build a new playground (Children’s Park), at a point that intersects with Bebianiha Street and a new road called Park Road. Park Road should be parallel to the Nana Botwe Street and pass through the point (2, 3).

iv. Determine the coordinates of the playground.

13. Discuss the geometric relationship that exists between the four Roads and their equations.

i. Road A, y = −12x + 10

ii. Road B, −x + 12y = 56

iii. Road C, 12x = 76 − y

iv. Road D, 12y = x −52 GLOSSARY

• Straight line - this line is the shortest distance between any two points. It extends infinitely in both directions without curving.

• Perpendicular lines – these are lines that intersect at a right angle (90⁰). If two lines are perpendicular, the product of their slopes (gradients) is −1.

• Parallel lines – these are lines in the same plane that never intersect, no matter how far they are extended. These lines have the same slope (gradient) but different y-intercepts.

• Gradient of a line - The gradient (or slope) of a line measures its steepness and direction. It is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line. Mathematically, it is expressed as:

Gradient = change in y_________ change in x = ∆ y___ ∆ x .

• Acute angle - this is an angle that measures greater than 0⁰but less than 90⁰. It is smaller than a right angle.

• Obtuse angle - it is an angle that measures greater than 90⁰but less than 180⁰. It is larger than a right angle but smaller than a straight angle.

• Right angle – this is an angle that measures exactly 90⁰. It is the angle formed when two perpendicular lines intersect.

Did you know that the Global Positioning System (GPS) used by Uber, Bolt and Yango drivers for pick-up and delivery is made possible by knowledge in Vectors?

How about the paths in your favourite video games, did you know vectors are used for calculating those paths? Well, let’s come to your football games. Vectors help to determine the direction of a ball and how far the ball moves when kicked by a player. As we go through this section, it is expected that you will learn about the types and forms of vectors as well as the algebraic and geometric operations of vectors. The concepts of vectors are applied in many fields such as physics, engineering and computer science.

At the end of this section, you will be able to:

• Recognise and explain various forms of vectors and apply the knowledge to find unit vectors.

• Perform algebraic and graphical operations (addition, subtraction, scalar multiplication) and their geometrical interpretation.

• Determine the resultant of vectors using triangle and parallelogram laws of addition.

Key Ideas:

• Vectors are mathematical quantities that represent both magnitude and direction. The magnitude of a vector is the length or distance of the vector.

• Types of vectors to be discussed are position, collinear, unit, free, negative, parallel, equal and co-initial.

• Vectors can be represented in column/component and magnitude and direction forms.

On a daily basis, specific quantities can be defined mathematically with a single number, which represents their magnitude or size. Mass, volume, distance and temperature are some examples of such quantities. Also, there are many other quantities that require both magnitude and direction to be fully described. These quantities are represented mathematically by vectors.

For, example, you can decide to throw a ball forward or backwards (direction) at a particular distance (magnitude). If you push a car forward, backwards or sideways, you will get different results. Thus, force, velocity and acceleration are examples of vector quantities. Vectors can come in various types and represented in different ways. The vector that represents the movement from point B to point A can be represented graphically with ⟶ BA or i, while vector that represents the movement from point F to point E can be represented graphically with ⟶ FE or h as shown in Figure 1.

Figure 1: Vector Representation

Vectors are mathematical quantities that represent both magnitude and direction.

The magnitude of a vector is the length or distance of the vector.

The Pythagorean theorem x²+ y²= z ²is used to calculate the magnitude of a vector.

Activity 6.1

i. On a graph paper, plot points O (0, 0), B (4, 5), Z (4, 0) and D (12, 10).

ii. Draw vectors OB and ZD.

Figure 2: Collinear Vectors

iii. On graph paper, plot points A (2, 1), B (5, 5), C (6, 1), and D (9, 5).

iv. Draw vectors AB and CD.

Figure 3:Parallel Vectors

v. Plot the point A (2, 2), B (5, 5), C (6, 4) and D (3, 6) on the graph.

vi. Draw vectors AB, AC and AD originating from A.

Figure 4: Co-initial vectors

vii. Plot points A (3, 1), B (7, 4), C (10, 4) and D (6, 1).

viii. Draw vectors AB and CD.

ix. Plot the points A (3, 1), B (7, 4), C (10, 4), and D (6, 1) on the graph.

x. Draw vectors AB and DC.

Figure 6: Equal vectors

xi. Plot points O (0, 0) and P (5, 4).

xii. Draw vector OP

Figure 7: Position Vector

xiii. Plot points A₁(4, 2), A₂(3, 1), B₁(9, 4), B₂(8, 3), A(7, 2) and B(12, 4).

xiv. Draw vectors AB, A₁ B₁ and A₂ B₂.

Figure 8: Free Vector

xv. Record your observations from the vectors drawn and discuss with a friend Generalisations

i. Vectors that lie on the same straight line (with a common slope) are collinear.

ii. Vectors starting from the same point are co-initial.

iii. Vectors are parallel because they have the same direction and proportional components and are scalar multiples of each other.

• Position vectors represent the positions of points A, B, and C relative to the origin.

• Free vectors are positioned differently, they are identical in magnitude and direction, making them free vectors.

• Negative Vectors have the same magnitude but opposite direction.

• Equal Vectors have the same magnitude and direction even though they are placed at different locations.

• Unit vector has a magnitude of 1 unit (√x²+y²= 1).

Forms of Vectors

The varied ways in which vectors can be written or represented are what we term forms of vectors. The forms we are going to focus on are:

• Column/component form (ˣy) where x represents the x (horizontal) direction and y the y (vertical) direction.

• Magnitude and direction form (r, θ ) where r represents the magnitude (distance or length) and θ the direction.

Note: Algebraically, we can rewrite a vector as xi + yj.

Example 1

Given that Benyiwa and Ebo move from the same point (0, 0) towards a church building. If Benyiwa walks 5 units to the right and 2 units upward while Ebo moves 3 units to the left and 8 units upward, express their position as column vectors.

Solution

Step 1: Let O represent point of origin O (0, 0)

Step 2: Let B represent point of Benyiwa’s movement B (5, 2)

Step 3: Let E represent point of Ebo’s movement E (-3, 8)

Step 4: Write out column vector for Benyiwa ⟶ OB = (5/2)

Step 5: Write out column vector for Ebo ⟶ OE = (− 3/8 ) A person positioned at O moving 5 units to the right denotes a positive displacement and 2 units upwards (positive displacement) to get to B. The vector that depicts this movement is ⟶ OB = (5/2). Similarly, a movement of 3 units to the left and 8 units upward will move a point from O to E giving, ⟶ OE = (− 3/8 ).

Example 2

If the motion represented by ⟶ MN = 6i − 3j is the translation of a particle on the i − j plane from M to N. Represent ⟶ MN as a component vector.

Solution

Step 1: Let the coefficient of i represent the x-coordinate for the column vector

Step 2: Let the coefficient of j represent the y-coordinate for the column vector Thus, ⟶ MN = ( 6___ − 3).

Example 3

Juliet’s house is 15 metres away from her school and the bearing is 35°. Represent the location of Juliet’s house in magnitude-direction vector form.

Solution

(15m, 35⁰)

Example 4

Indicate the parallel vectors from the following given vectors;

u = (− 2/3 ), v = (1/3), w = (− 6/9 ), a = (− 6___ − 9).

Solution

Step 1: Find the vectors that are scalar multiples of others.

Figure 9: Parallel vector identification (− 2/3 ), (− 6/9 ) the scalar multiple is 3.

Therefore, vectors u and w are parallel.

Example 5

Given that the vector ⟶ AB = ( 3___ − 4), find the magnitude of ⟶ AB .

Solution

|⟶ AB| = √3²+ ( − 4)²= √9 + 16 = √25 = 5 units

Example 6

Find the unit vector for g = (− 5 − 6).

Solution

The unit vector can be determined by ˆg = g__ |g |.

Step 1: Find the magnitude of the vector |g| = √___________ ( − 5)²+ ( − 6)²|g| = √25 + 36 |g| = 7.810 to three decimal places

Step 2: Substitute the g and |g| in the unit vector formula.

ˆg = (− 5 − 6)_____ √__ 61 ˆg = ( − 5___ √61 − 6___ √61 )

The basic mathematic operations used are +, − , ÷, ×. For vectors, the operations that are applicable are addition, subtraction and scalar multiplication. We are going to explore how to operate on vectors both algebraically and geometrically.

Starting with the algebraical operations, let’s discuss vector addition, subtraction and scalar multiplication!

In adding of vectors, sum the corresponding x− coordinates and corresponding y− coordinates. In the case of subtraction, subtract the corresponding x− coordinates and corresponding y− coordinates. In the case of scalar multiplication, the x and y coordinates are both multiplied by the scalar k .

Given that a = ( x₁ y₁), b = ( x₂ y₂), a + b = ( x₁+ x₂ y₁+ y₂ ) Also given that a = ( x₁ y₁), b = ( x₂ y₂), a – b = ( x₁− x₂ y₁− y₂) Given that a scalar multiplier k and a = ( x₁ y₁), ka = (kx₁ y₁ ) = ( k x₁ k y₁)

Example 7

Given that u = (12 5 ) and v = (2 3), find u + v, u – v and 5v.

Solution

Step 1: Add the corresponding coordinates for vector u and v u + v = (12 + 2 5 + 3 ) = (14 8 )

Step 2: Subtract the corresponding coordinates for vector u and v u – v = ( 12 − 2 5 − 3 ) = (10 2 )

Step 3: Multiply the coordinates of vector v by the scalar multiplier 5.

5v = 5(2 3) = (10 15 ) Triangular Law of Vector Addition Imagine that you walk from your classroom (point A) to the School Assembly Hall (point B), then to School Administration block (point C). The vectors, ⟶ AB and ⟶ BC can be used to represent the movement. It would be much simpler and straightforward to move from point A to point C. This movement can be represented by ⟶ AC geometrically in Figure 10.

(Assuming a straight path connects the classroom to the Assembly Hall, Assembly Hall to Administration, and Classroom to Administration block).

Figure 10: Triangular law of vector addition Now ⟶ AC is what we refer to as the resultant vector of ⟶ AB and ⟶ BC .

⟶ AC = ⟶ AB + ⟶ BC = (6 5 ) + (− 1 − 6 ) = ( 5 − 1)

Note: The triangular law of vector addition can be applied to establish the relationship between a free vector, say ⟶ BC , and the position vectors, ⟶ OB also written b, ⟶ OC and written as c.

Recall that ⟶ OB + ⟶ BC = ⟶ OC from triangular law of vector addition.

Figure 11: Triangular law of vector addition Applying change of subject, ⟶ BC = ⟶ OC - ⟶ OB Now, ⟶ BC = ⟶ OC - ⟶ OB ⟶ BC = c - b Parallelogram Law of Vector Addition Recall that a parallelogram is a four-sided figure with specific properties? Great!

We are going to explore how vectors relate when they come together to form a parallelogram.

Analyse Figure 12 and share your observations with a classmate.

Figure 12: Parallelogram law of vector addition Generalisation

• Given two vectors, ⟶ OB and ⟶ OD , being co-initial vectors and representing adjacent sides of a parallelogram, OBCD, as in Figure 11, the resultant vector ⟶ OC can be represented by the diagonal of the parallelogram passing through O.

• ⟶ OC = ⟶ OB + ⟶ BC and ⟶ OC = ⟶ OD + ⟶ DC by triangular law of vector addition

• ⟶ BC = ⟶ OD and ⟶ DC = ⟶ OA , hence ⟶ OC = ⟶ OB + ⟶ OD

Example 8

The vertices of a parallelogram reservoir are at M (6, − 1), N (5, 1), S (9, 3) and T ( x, y ).

a) Find the coordinates of T

b) Find |⟶ MN |

Solution

a) ⟶ MT = ⟶ NS ( ˣy ) − ( 6 − 1) = ( 9 3 ) − ( 5 1 ) (x − 6 y + 1) = (9 − 5 3 − 1) x – 6 = 4 x = 10 y + 1 = 2 y = 1 Hence T (10, 1)

b) |⟶ MN | = √_________________ (5 − 6 )²+ (1 − ( − 1))²= √1 + 4 = √5 units

1. Given that a = (3

1) and b = (− 2 4 ), find:

i. 5a + 2b

ii. 3a − 1/2 b

2. Given that a = ( 4 − 8) and b = (− 2 3 ) find r such that 1/4 a − b + r = (1 2)

3. A hiker travels 4 km north, then turns and walks 3 km west. The hiker then walks directly back to the starting point.

Represent the hiker’s journey using vectors and apply the triangular law of vector addition to determine the resultant vector. How far is the hiker from the starting point after the first two legs of the journey?

4. Ama and Kojo are preparing to go to the market. Ama walks 3 km due east to reach the main road. From there, she walks 4 km due north to reach the market. Kojo, on the other hand, starts from a different point and walks 5 km due west and then 2 km due south to reach the same market.

i. Represent Ama’s and Kojo’s displacements as vectors.

ii. Determine the resultant vector of Ama’s displacement.

iii. If Ama and Kojo were to return home directly, calculate the vector representing their journey back.

5. A delivery van needs to transport goods from Kumasi to Nsawam. The driver first travels 120 km southeast to Konongo and then 80 km due east to Nsawam.

i. Represent the two parts of the journey as vectors.

ii. Determine the total displacement of the van from Kumasi to Nsawam.

iii. If the driver takes an alternative route that goes directly from Kumasi to Nsawam, calculate the vector of this direct route and compare it with the total displacement vector.

6. A cyclist rides 6 km due east from Madina to Legon, then 8 km due north to Achimota.

i. Represent the cyclist’s journey as vectors.

ii. Calculate the resultant vector representing the cyclist’s total displacement.

iii. Determine the vector representing the cyclist’s journey if they return directly to Madina from Achimota.

7. Two fishing boats are on Lake Volta. Boat A moves 4 km due east from its initial position, while Boat B moves 8 km due east from a point 2 km north of Boat A’s starting position.

i. Are the displacement vectors of the two boats parallel? Explain your reasoning.

ii. If the boats were to return to their starting points, would the return vectors be parallel? Justify your answer.

8. A delivery van travels 10 km due north from Adabraka to Achimota.

Another delivery van starts from Kaneshie and travels 20 km due north to Kwabenya.

i. Represent the displacement vectors of the two vans.

ii. Determine if the vectors are parallel.

iii. If the second van’s route was due south instead of north, would the vectors still be parallel?

9. Two Minibuses operate in Kejetia. The first one travels 5 km due west from Kejetia to Santasi, while the second one travels 15 km due west from Kejetia to Abuakwa.

i. Are the routes of the two Minibuses represented by parallel vectors?

ii. If a third Minibuses traveled 5 km due east from Kejetia, would its vector be parallel to the first two?

10. A construction worker in Ghana is tasked with marking a 20-meter segment of a new road, which is oriented in the direction of the vector v = ( 5

12) meters.

i. Find the unit vector in the direction of v.

ii. If the worker needs to mark a 20-meter segment in the same direction, determine the vector that represents this segment.

11. Kofi and Kwame are paddling a canoe on Lake Bosomtwe. Kofi paddles 4 km directly north, while Kwame paddles 3 km directly east.

i. Represent Kofi and Kwame’s movements as vectors.

ii. Use the triangular law of vector addition to determine Kwame’s position relative to Kofi’s position.

iii. Calculate the magnitude and direction of this vector.

12. Two tugboats are towing a cargo ship off the coast of Tema. The first tugboat exerts a force of 10 kN in a direction due east, and the second tugboat exerts a force of 6 kN in a direction 60° north of east.

i. Represent the forces exerted by the tugboats as vectors.

ii. Use the parallelogram law of vector addition to determine the resultant force acting on the cargo ship.

iii. Calculate the magnitude and direction of the resultant force.

GLOSSARY

• Collinear vectors are vectors that lie along the same line or along parallel lines. These vectors can have the same or opposite directions.

Mathematically, if two vectors a and b are collinear, then there exists a scalar k such that a = kb.

• Co-initial vectors are vectors that have the same starting point (initial point). Even though they may point in different directions or have different magnitudes, they all originate from the same location.

• Parallel vectors are vectors that have the same or exactly opposite direction.

They may differ in magnitude but lie along lines that are parallel to each other.

• The triangular law of vector addition states that if two vectors are represented as two sides of a triangle in sequence, then the third side of the triangle (taken in the reverse order) represents the resultant vector.

• The parallelogram law of vector addition states that if two vectors are represented by adjacent sides of a parallelogram, then the resultant vector is represented by the diagonal of the parallelogram that starts from the same point.