ba101 engineering mathematic chapter 5 geometry coordinate and graph

COORDINATE GEOMETRY AND GRAPH BA101/CHAPTER4

1 Prepared By : Nur Haslinda Mohd Sailan

CHAPTER 5: COORDINATE GEOMETRY AND GRAPH (Graph of Linear Function) 4.0 INTRODUCTION

The relation of 2 anus can be pictured by graph drawing. What is graph? Graph is the combination

of 2 points of anus from 2 scaled-axis which is vertical axis and horizontal axis. For the graph y

against x, it means x and y is 2 anus. Vertical axis is y-axis and horizontal axis is x-axis.

The point where these two axis met is called origin. Meanwhile, the coordinate is a point which has

a pair of x value and y value. On the graph paper, these coordinates are plotted using Roman.

Combination of these coordinate will develop a line whether straight line or curve line. So, this line

will pictured the relation of 2 anus, for example x and y. Refer the Diagram 12.1 below:

absissa coordinate 3 (x,y) = (3,3) 2 ordinate 1 x -2 -1 -1 1 2 3 -2

Diagram 12.1

y



4.1 GRAPH DRAWING

Graph can be drawn using the criteria below:

a) Table of graph and/or equation of graph.

b) Graph paper to draw graph.

c) Graph title.

d) Two axis of anus, e.g. y against x.

e) Two arrows for both axis of graph, to the above and to the right of the graph.

f) 0 digit at origin of the graph

g) Scale of axis, e.g. 1 cm : 5 unit for y-axis.

h) Straight line of graph, from combination of coordinate points.

i) Balance of straight line, especially when the coordinate points at graph are not flat.

Example 1:

Plot a graph y against x from the table below:

x 10.0 20.0 30.0 40.0 50.0 60.0

y 5.0 6.3 7.5 8.7 10.0 11.0

Solution:



Example 2:

Given one straight line with the equation y = -x + 3 intersect with another one straight line with the

equation y = 2x – 4. Using graph drawing, find the intersection point.

Solution

Given the first linear equation, y = - x + 3.

At x-axis, y = 0. So 0 = - x + 3 or x = 3. Coordinate is (3, 0).

At y-axis, x = 0. So y = 0 + 3 or y = 3. Coordinate is (0, 3).

Given the second linear equation, y = 2x – 4.

At x-axis, y = 0. So 0 = 2x – 4 or x = 2. Coordinate is (2, 0).

At y-axis, x = 0. So y = 2(0) – 4 or y = -4. Coordinate is (0, -4).

From the graph drawing, we can see that the intersection point of the graph is at(2.3, 0.5).



4.2 LINEAR EQUATION

Basically, graph equation can be represented by y = mx + c where y and x are 2 anus, c is

coordinate of interception at y-axis, while m is the gradient of a straight line.

Example 1

Get the gradient of the straight line below:

a) Straight line which passing points are A (-2, 0) dan B (0, 4)

b) Straight line which passing points are P (-4, 3) dan Q (5, 3)

c) Straight line which passing points are L (4, 5) dan M (4, -6)

Then, create the straight line equation.

Solution

By using the gradient equation12

12

xx

yym

, so

a) m = )2(0

04

= 2, so y = 2x + 4 (interception at point A because y = 0)

b) m = )4(5

33

= 0, so y = 3 (constant y = 3, only x changed)

c) m = 44

56

= , so x = 3 (constant x = 3, only y changed)

Example 2

Referring to Example 12.1, draw a graph for each.

Solution

a)

y = 2x + 4

(0,4)

(-2,0) x

y



(0,3) y = b)

x y x = 3 c) (3,0) x 4.3.1 DISTANCE AND GRADIENT OF GRAPH Diagram 12.2 Referring to Diagram 12.2, by using Pythagoras theorem,

y

A(x1, y1)

x

B(x2, y2)

C(x2 ,y1)

0

y



AB2 = AC2 + CB2

= (x2 – x1)2 – (y2 – y1)2

so the distance of AB = 2

12

2

12 )()( yyxx

The formula of m gradient is given by:

12

12

xx

yym

or

21

21

xx

yym

Example 1

Find the distance and gradient of the graph which passing through coordinate point (3, -2) and

(0,8).

Solution

Let say )2,3(),( 11 yx and )8,0(),( 22 yx .

So the distance = 2

12

2

12 )()( yyxx = 22 )30()28( = 109 unit.

Gradient, 3

10

30

28

12

12

xx

yym

4.3.2 DISTANCE AND GRADIENT OF GRAPH

Diagram 10.3 Referring to diagram 10.3, by using Pythagoras theorem,

A(x1, y1)

x

y B(x2, y2)

C(x2 ,y1)

0



AB2 = AC2 + CB2

= (x2 – x1)2 – (y2 – y1)2

So the distance of AB = 2

12

2

12 )()( yyxx

The formula of m gradient is given by:

12

12

xx

yym

or

21

21

xx

yym

Example 1

Find the distance and gradient of the graph which passing through coordinate point (3, -2) and

(0,8).

Solution

Let say )2,3(),( 11 yx and )8,0(),( 22 yx .

So the distance = 2

12

2

12 )()( yyxx = 22 )30()28( = 109 unit.

Gradient, 3

10

30

28

12

12

xx

yym

4.4.1 EQUATION GRAPH OF LINEAR FUNCTION

Basic for equation graph of linear function or straight line can be recognized by the highest power

of its anu which is 1 and the lowest power which is 0.

Example 1

State whether the graph equation below is linear or not:

a) y = 3x2 + 5x – 7.

b) y = 11x + 5.

c) y + 4x = 7x3 –2.

d) y –2 = 6x.

e) y – 5x = 3 + 4x-1.

Solution

a) not linear, because the highest power of x is 2.

b) linear, because the highest power of x is 1.

c) not linear, because the highest power of x is 3.

d) linear, because the highest power of x is 1.



e) not linear, because the lowest power is –1.

12.1) Based on the data below, draw a graph y against x and get the gradient and interception at

y-axis. Then, create one complete linear equation.

x 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

y -6 -4 -3 -2 0 2 3 4 6 8

12.2) Plot a graph of linear function from table below. From the graph, create one graph equation.

x 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

y 6 5 3 1 0 -1 -3 -5 -6 -7

12.3) Given x related to y based on the data below. Draw a graph of linear function and state the

relation of x and y through the graph equation.

X -3 -2 -1 0 1 2 3 4 5

Y -1 1 3 5 7 9 11 13 15

12.4) Plot a graph for one straight line which passing point A( 2 , 7 ) and B( 4 ,10).

From the straight line u get, find the interception at y-axis, the gradient and the distance of AB.

12.5) Given one straight line which passing point J( 2 , 0 ) and K( 6 , 2 ).

Plot a linear graph and from the graph, determine the interception at y-axis, the gradient and the distance of JK.

12.6) One graph of linear function created from coordinate point P( 2 , 3 ) dan Q( 4 , 1 ). Draw a

linear graph from it and from the graph, find the coordinate of intercept point at y-axis, the gradient and the distance of PQ.

12.7) Two graph of linear function with the equation y = -x + 5 and y = 3. Draw both graph on the

same axis and from the graph, get the intersection point. 12.8) By using the suitable scale, plot two straight line with the equation y = -x + 5 and y = ½ x – 1. From the graph, get the intersection point.

12.9) Two straight line with the equation y = -x + 5 and y = ½ x – 1 are met at point A(p, q). Draw a suitable graph to find the value of p and q. 12.10) Get the intersection point for two straight line with the equation y = 2x + 3 and y = -9/2 x +

9.

ACTIVITY 12a



12.1) gradient m = 3 , intercept c (0, -6) , y = 3x – 6

12.2) y = -3x + 6 12.3) y = 2x + 5 12.4) intercept c(0, 3) , gradient m = 2 , distance of AB = 8.06 cm 12.5) intercept c(0, -1) , gradient m = 1/2 , distance of JK = 4.47 cm 12.6) intercept c(0, 5) , gradient m = -1 , distance of PQ = 7.07 cm

12.7) intersection point (2 , 3)

12.8) intersection point (4 , 1) 12.9) intersection point (1 , 5) 12.10) intersection point (4 , 1)

FEEDBACK



4.4.2 SIMULTANEOUS EQUATION OF 2 GRAPHS OF LINEAR FUNCTION

Solving two simultaneous equation can be done by using graph drawing where the intersection

point of a graph will the value for x and y.

Example 1

Given one straight line with the equation y = - x + 3 intersect with another one straight line with the

equation y = 2x – 4. By using graph drawing, find the intersection point.

Solution

Given the first linear equation, y = - x + 3.

At x-axis, y = 0. Therefore 0 = - x + 3 or x = 3. Coordinate is (3, 0).

At y-axis, x = 0. Therefore y = 0 + 3 or y = 3. Coordinate is (0, 3).

Given the second linear equation, y = 2x – 4.

At x-axis, y = 0. Therefore 0 = 2x – 4 or x = 2. Coordinate is (2, 0).

At y-axis, x = 0. Therefore y = 2(0) – 4 or y = -4. Coordinate is (0, -4).

From the graph drawn, we can see that the intersection point at the graph is (2.3, 0.5).



4.4.3 EQUATION OF GRAPH OF LINEAR FUNCTION

Basic of graph of linear function or straight line can be recognized by the highest power of its anu

which is 1 and the lowest power which is 0.

Example 1

State whether the graph equation below is linear or not:

a) y = 3x2 + 5x – 7.

b) y = 11x + 5.

c) y + 4x = 7x3 –2.

d) y –2 = 6x.

e) y – 5x = 3 + 4x-1.

Solution

a) not linear, because the highest power of x is 2.

b) linear, because the highest power of x is 1.

c) not linear, because the highest power of x is 3.

d) linear, because the highest power of x is 1.

e) not linear, because the lowest power is –1.

Example 2

Given y = x2. Prove that this equation is not linear by using graph drawing.

Solution

x = -4, y = (-4)2 = 16 x = 1, y = (1)2 = 1

x = -3, y = (-3)2 = 9 x = 2, y = (2)2 = 4

x = -2, y = (-2)2 = 4 x = 3, y = (3)2 = 9

x = -1, y = (-1)2 = 1 x = 4, y = (4)2 = 16

x = 0, y = (0)2 = 0



Table:

x -4 -3 -2 -1 0 1 2 3 4

y 16 9 4 1 0 1 4 9 16

Graph drawn is as per below.

From the graph, the equation y = x2 is not creating a straight line; therefore it is not a linear

equation.



12.1) Given parameter x related to y1 and y2 as per below. Get the intersection point by using graph drawing.

x -7 -6 -5 -4 -3 -2 -1 0 1

y1 -10 -8 -6 -4 -2 0 2 4 6

y2 6 4 2 0 -2 -4 -6 -8 -10


x -6 -5 -4 -3 -2 -1 0 1 2

y1 -13 -10 -7 -4 -1 2 5 8 11

y2 7 5 3 1 -1 -3 -5 -7 -9


x -2 -1 0 1 2 3 4 5 6

y1 -29 -22 -15 -8 -1 6 13 20 27

y2 15 11 7 3 -1 -5 -9 -13 -17


x -3 -2 -1 0 1 2 3 4 5

y1 23 18 13 8 3 -2 -7 -12 -17

y2 -25 -18 -11 -4 3 10 17 24 31


x -7 -6 -5 -4 -3 -2 -1 0 1

y1 -15 -12 -9 -6 -3 0 3 6 9

ACTIVITY 12b



y2 9 6 3 0 -3 -6 -9 -12 -15

12.6) 2 graph of linear function with the equation y = -x + 5 and y = 3. Draw both graph on the

same axis and from the graph, get the intersection point. 12.7) By using suitable scale, plot two straight line with the equation y = -x + 5 and y = ½ x –

1. From the graph, get the intersection point.

12.8) Two straight line with the equation y = -x + 5 and y = ½ x – 1 are met at point A(p, q). Draw a suitable graph to find the value of p and q.

12.9) Get the intersection point for two straight line with the equation

y = 2x + 3 and y = -9/2 x + 9.



12.1) (-3,-2) 12.2) (-2, -1) 12.3) (-2,-1) 12.4) (1, 3) 12.5) (-3, -3) 12.6) (2 , 3)

12.7) (4 , 1) 12.8) (p, q) = (1 , 5) 12.9) (4 , 1)

FEEDBACK




y-axis. Then, create a complete linear equation.

x -8 -5.5 -4.5 -3 -1.5 -0.5 0.5 2 3.5 4.5 7

y 5 3 2.5 1 -0.5 -1 -1.5 -3 -4.5 -5 -7

12.2) Draw a graph y against x from the table below and get the gradient and interception at y-

axis. Then, create a complete linear equation

x -4 -3 -2 -1 0 1 2 3 4 5

y 5 4 3 2 1 0 -1 -2 -3 -4


y-axis. Then, create a complete linear equation.

x -9 -6.5 -4 -3.5 -1.5 1.5 2 3.5 4 5.5

y -7 -3.5 -1 0 2.5 6 7 9 10 11.5

12.4) Draw a graph for one linear which passing points R( 2 , -3 ) and S( 0 , 1 ). From the linear,

get the interception at y-axis, the gradient, the distance of RS and one complete linear

equation for that line.

ASSESSMENT



12.5) One straight line passing through points X( 2 , -3 ) and Q( -3 , 1 ). Plot one graph and get

the interception at y-axis, the gradient, the distance of XQ and one complete linear

equation.

12.6) Get the intersection point for linear equation y = 15/8 x – 83 and y = 3/2 x – 15 by using

graph drawing.

12.7) Draw the graphs on the same axis and get the intersection point for linear equation y = - x

+ 1 and y = 7.5/6 x –3.5

12.8) Plot the graph from linear equation y = 4x - 7 and y = -3x +7 and get the intersection point.

12.9) Given parameter x related to y1 and y2 as per below. Get the intersection point by drawing

graph.

x -1 0 1 2 3 4 5 6 7

y1 -4 -3 -2 -1 0 1 2 3 4

y2 8 6 4 2 0 -2 -4 -6 -8

12.10) Given parameter x related to y1 and y2 as per below. Get the intersection point by drawing

graph.

x 4 5 6 7 8 9 10 11 12

y1 -2 0 2 4 6 8 10 12 14

y2 10 9 8 7 6 5 4 3 2

12.1) m = -4/5 , c(0, -7/5), y = -4/5 x – 7/5

12.2) m = -1, c(0, 1), y = -x + 1

12.3) m = 9/7, c(0, 9/2), y = 9/7 x + 9/2

12.4) c (0, 1), m = -1 , RS = 4.24 unit, y = - x + 1

12.5) c(0, -7/5), m = -4/5 , XQ = 6.40 unit, y = -4/5 x – 7/5

FEEDBACK



12.6) (20, 45)

12.7) (2, -1)

12.8) (2, -2)

12.9) (3, 0)

12.10) (8, 6)



4.5 COORDINATE GEOMETRY AND GRAPH (Quadratic Graph)

4.5.0 INTRODUCTION

If y and x is 2 anus in one quadratic graph with the situation y against x, so the equation of

quadratic graph is y = ax2 + bx + c. The a, b and c can be categorized as constant, while c is the

intercept at y-axis.

4.5.1 DRAWING QUADRATIC GRAPH FROM TABLE

Example 13.1

Given y = x2. Build a table from x = -4 until x = 4. From the table, draw a quadratic graph.

Solution

x = -4, y = (-4)2 = 16 x = 1, y = (1)2 = 1

x = -3, y = (-3)2 = 9 x = 2, y = (2)2 = 4

x = -2, y = (-2)2 = 4 x = 3, y = (3)2 = 9

x = -1, y = (-1)2 = 1 x = 4, y = (4)2 = 16

x = 0, y = (0)2 = 0

Table:

x -4 -3 -2 -1 0 1 2 3 4

y 16 9 4 1 0 1 4 9 16

The graph drawn is as below.



4.5.2 THE INTERSECTION POINT OF QUADRATIC GRAPH AND LINEAR GRAPH

Example 13.2

Draw the graph from equation y = 4x2 + 7x – 5 and y = -5x + 20, by using the value of x = -6 until

x = 4. From the graph, determine the intersection points.

Solution

Given the first equation y = 4x2 + 7x – 5. At y-axis, x = 0. Therefore y = 4(0)2 + 7(0) – 5 or y = -5.

At x-axis, y = 0. Therefore 0 = 4x2 + 7x – 5. By using the quadratic formula,

8

1297

8

80497

)4(2

)5)(4(4497

x So, x1 = -2.3 or x2 = 0.5 .

Midpoint 9.02

5.03.2

2

213

xxx and y3 = 4(-0.9)2 +7(-0.9) – 5 = -8.1

Next values can be found by using calculator as per below:

x -6 -4 -2.3 -2 -0.9 0 0.5 2 4

y 97 31 0 -3 -8.1 -5 0 25 87

Given the second equation y = -5x + 20. At y-axis, x = 0. Therefore y = -5(0) + 20 or

y= 20.

At x-axis, y = 0. Therefore 0 = -5x + 20 or x = 4. Table of linear graph is as per below:

x 0 4

y 20 0



From the graph, the intersection point is at ( -4.4, 42) and (1.4, 10). 4.5.3 THE INTERSECTION POINT OF 2 CURVES

Example 13.3

Determine the intersection points from two curves with the equation

y = x2 + x –6 and y = –x2 + x +2.

Solution

Given the first equation y = x2 + x – 6. At y-axis, x = 0. So, y = (0)2 + 0 – 6 or y = -6.

At x-axis, y = 0. Therefore 0 = x2 + x – 6. By using factorisation method,

0 = (x + 3)(x – 2), or

x1 = -3 and x2 = 2. Midpoint 5.02

233

x and y3 = (-0.5)2 + (-0.5) –6 = -6.25.

The table as per below:

x -3 -2 -1 -0.5 0 1 2 2.5

y 0 -4 -6 -6.25 -6 -4 0 2.756

Given the second equation y = –x2 + x +2. At y-axis, x = 0. Therefore y = -(0)2 + 2(0) + 2 or y = 2.

At x-axis, y = 0. Therefore 0 = -x2 + x + 2. By using factorisation method,

0 = (-x + 2)(x +1), or x1 = 2 and x2 = -1. Midpoint 5.02

123

x and

y3 = -(0.5)2 + (0.5) +2 = 3.25.

The table as per below:

x -3 -2 -1 0 0.5 1 2 3

y -10 -4 0 2 2.25 2 0 -4

A graph can be drawn by using the above table.



From the graph drawn, we can see that the intersection points of the graph is at (-2, -4) and (2, 0).

13.1 Given one curve equation y = x2 + x –1. Develop an

appropriate table and plot a quadratic graph from it. From the graph, get all the possible

values for x if 0 = x2 + x –1.

13.2 Given one quadratic graph with the equation y = 4 - x2. Draw the graph and determine the

x values if y = 0.

13.3 Given one curve with the equation y = x2 – 4x + 3. Plot a quadratic graph and get the

coordinate points of the graph when it intercept at x-axis.

13.4 Given one quadratic equation y = -x2 – 4x + 5 and one linear equation

y = -x + 5 met at two intersection points. Draw the graphs on the same axes and from the

graph, get the intersection points.

ACTIVITY 13



13.5 Two kind of graphs that are drawn on the same axes which are y = -x2 – 2x + 8 and y = -

4x + 4 have two intersection points. Find the intersection points.

13.6 Two graph equation which are y = 5x – 15 and y = x2 – 2x – 15 met for two times.

Determine the points from graph drawing.

13.3 Get the intersection coordinate for two graph equation

y = x2 – 9 and y = -x + 2 from graph drawing.

13.8 Given two quadratic equation which are y = x2 + 2x – 3 and y = -x2 + 2x + 8. Draw the

suitable graphs on the same axes and determine the intersection points of the graph.

13.9 Two quadratic graphs with the equation y = -2x2 - 7x + 6 and y = x2 + x – 2 are drawn on

one same axes. Get the intersection points of the graphs.

13.10 Two quadratic graphs with the equation y = 2x2 + 5x – 3 and

y = -2x2 - x + 6 are drawn on the same axes which met two times. Get the coordinates of

the meeting points.

13.1 x = 1, x = -2

13.2 x = 2, x = -2

13.3 x = 1, x = 3

13.4 (0, 5) and (5, 0)

13.5 (-1.3, 9.2) and (3.3, -9.2)

13.6 (0, –15) and (7, 20)

13.7 (-3.9, 5.9) and (2.9, 0.9)

13.8 (2.3, 6.9) and (-2.3, -2.3)

13.9 (0.8, -0.6) and (-3.4, 6.2)

13.10 (-2.4, -3.1) and (0.9, 3.5)

FEEDBACK 13



13.1) Based on the above data, draw the graph and get the gradient and the intercept at y-axis for

each graph. Then, create one complete linear equation and the distance for the straight

line.

x 0 1 2 3 4 5 6 7 8 9 10

y 1 3 5 7 9 11 13 15 17 19 21

13.2) Draw one linear graph which passing points A(-1 , 2) and B(-4 , -1).

From the linear graph, get the coordinate at y-intercept, the gradient, distance of AB and

the graph equation.

13.3) One linear graphs with the points C( 3 , 4 ) and D( 1 , -4 ). Plot the graph and find the

coordinate at y-intercept, the gradient, distance of CD and the graph equation.

13.4) Get the intersection points for these graph equation y = x + 3 and y = -x + 6.

13.5) Two equation of linear graph are given by y = 4x - 8 and y = 4/3 x. Plot the suitable graphs

on the same axes and from the graphs, determine the coordinate of the intersection point.

13.6) Given one quadratic graph with the equation y = 2x2 – 4x – 6. Draw the graph from the

equation given and get the coordinate points when the graph intercept at x-axis.

13.7) One curve y = x2 + 2x – 15 and one straight line y = -2x – 8 sharing two intersection points.

Create the suitable graphs on the same axes and from the graphs, get the intersection

points.

13.8) Given one linear graph with the equation y = 4x + 8 and one quadratic graph with the

equation y = 16 - x2 met twice on the same axes. Find the meeting points by using graph

drawing.

ASSESSMENT



13.9) Two graph equation y = x2 - 9 and y = -x2 - 2x + 8 have two intersection points. Plot both

graphs on the same axes and find the intersection points.

13.10) Two curves y = -x2 - x + 2 and y = x2 - x – 2 met twice on the same axes. Get both

intersection points from the graph drawing.

13.1) m = -2 , c(0, 1), y = -2x + 1 , distance = 15.65 unit

13.2) c(0, 3) , m = 1 , y = x + 3 , distance = 4.24 unit

13.3) c(0, -8) , m = 4 , y = 4x - 8 , distance = 8.25 unit

13.4) ( 1.5 , 4.5 )

13.5) ( 1.5 , -2 )

13.6) x = -1, x = 3

13.7) (2.6, -2.8) and (-2.6, -13.2)

13.8) (1.5, 14) and (-5.5, -14)

13.9) (2.5, -3.3) and (-3.5, 2.8)

13.10) (1.4, -1.4) and (-1.4, 1.4)

FEEDBACK

ba101 engineering mathematic chapter 5 geometry coordinate and graph

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