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Finding the equation of a straight line graph from a diagram

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MATHEMATICS

Grade 9

NUMBER PATTERNS, GRAPHS, EQUATIONS,

STATISTICS AND PROBABILITY

Module 15

FINDING THE EQUATION OF A STRAIGHT LINE GRAPH FROM A DIAGRAM

ACTIVITY 1

To find the equation of a straight line graph from a diagram

[LO 2.5]

  1. If we can find out the values of m and c, then we simply substitute them in the general equation y = mc + c to give us the defining equation of the line. Let’s do an example from the given diagram.
Figure 1
Figure 1 (Picture 37.png)

To find c is easy as it is the value (positive or negative or zero) where the line cuts the y–axis. Substitute this value (it is –1) for c.

The equation now becomes y = mx – 1. To find the gradient (the value of m) we construct the right-angled triangle between two suitable points where the graph goes exactly through corners on the graph paper.

Table 1
  • Remembering that m is a fraction:
change in vertical distance
change in horizontal distance
  • We read off the number of units of the height and the length of the triangle to give us the numerator and denominator respectively
  • We also have to decide whether the sign is negative or positive by looking at which way the line slopes.
  • This gives us: m=−46=−23m=−46=−23 size 12{ size 11{m```=``` - { { size 8{4} } over { size 8{6} } } ```=``` - { { size 8{2} } over { size 8{3} } } }} {} (remember to simplify the fraction).
  • This value is now substituted for m in the equation: y=−23x−1y=−23x−1 size 12{ size 11{y```=``` - ``` { { size 8{2} } over { size 8{3} } } x``` - ```1}} {}. This gives us the defining equation of the line in the diagram.
  • Going back to the previous section, use this method to find the defining equations of the eight graphs in the first two diagrams.

2 How do we deal with horizontal and vertical graphs? They are the easiest.

  • If the line is horizontal, then the equation is y = c. We have to replace the c by a value. We read this value off the graph – it is the y–intercept! Substitute this into y = c, and you have the defining equation.
  • If the line is vertical, the equation is x = k. Find k by reading from the graph where the line cuts the x–axis and substitute this number for k. This gives the defining equation.
  • From the previous section, find the equations for the four graphs in the last diagram.

Here are the answers:y = 1 and y = –1,5 are the two horizontal lines, and x = –1 and x = –2,5 are the two vertical lines.

3 The following diagrams have a mixture of lines for you to test your skills on.

Figure 2
Figure 2 (Picture 38.png)

4 Did you notice that the gradients (m) of lines G and H are the same? Why is this?

ACTIVITY 2

To calculate the gradient of a straight line from two points on the line

[LO 2.5]

  • If you know the coordinates of two points on a certain straight line, then you can draw that line, as you have seen. And from the sketch you can find the gradient as you have already learnt. But it is not necessary to have a graph to find the gradient.
  • Here is an example: The points (3 ; –1) and (4 ; 2) are on a certain straight line.
  • First we calculate the vertical distance between the two points by subtracting the second point’s y-coordinate from the first point’s y–coordinate. This is the numerator of the gradient.
  • Then we calculate the horizontal distance between the two points by subtracting the second point’s x-coordinate from the first point’s x-coordinate. This is the denominator of the gradient.
  • So, the gradient is: m=verticaldistancehorizontaldistance=−1−23−4=−3−1=3m=verticaldistancehorizontaldistance=−1−23−4=−3−1=3 size 12{ size 11{m``=`` { { size 11{"vertical"``"distance"}} over { size 11{"horizontal"```"distance"}} } ``=`` { { size 11{ - 1` - `2}} over { size 11{3 - 4}} } ``=`` { { size 11{ - 3}} over { size 11{ - 1}} } ``=`} size 13{`}3} {}
  • If you do the subtraction the other way round, then you must do it for both coordinates, like this:

m=verticaldistancehorizontaldistance=2−−14−3=+3+1=3m=verticaldistancehorizontaldistance=2−−14−3=+3+1=3 size 12{ size 11{m``=`` { { size 11{"vertical"```"distance"}} over { size 11{"horizontal"```"distance"}} } ``=`` { { size 11{2` - ` left ( size 11{ - 1} right )}} over { size 11{4 - 3}} } ``=`` { { size 11{+3}} over { size 11{+1}} } ``=``3}} {}, the same answer!

1 On squared paper, mark the two points (3 ; –1) and (4 ; 2) and draw the line. Then use the graphical method you used before to calculate the gradient, to confirm that it agrees with the answer from the calculation above.

2 Below you are given five pairs of coordinates. Calculate the five gradients between the points.

2.1 (2 ; 6) and (4 ; 4)

2.2 (1 ; 2) and (–2 ; –1)

2.3 (0 ; 0) and (1 ; 5)

2.4 (–1 ; 4) and (5 ; 4)

2.5 (7 ; 0) and (7 ; –3)

ACTIVITY 3

To graphically solve two linear equations simultaneously

[LO 2.5]

1 Solve the following five sets of equations simultaneously (you can refer to the chapter where you learnt to do this).

1.1 y = ½x + 2 and y = 3

1.2 y = x and y = –3

1.3 y = x – 2 and y = –3

1.4 y = –x + 4 and y = 0

1.5 y = ½x– 2 and y= 0

2 Look at the diagrams in the previous exercise and write down the coordinates of the points where the following lines cross:

2.1 A and C

2.2 E and G

2.3 E and H

2.4 J and L

2.5 K and J

3 Study these answers together with the equations for lines A to L that you found in problem three of the previous section.

  • An example:
  • Line J above has the equation y = 0, and for line I you should have found the equation y=−18x+12y=−18x+12 size 12{y= - { {1} over {8} } x+ { {1} over {2} } } {}. (This equation can also be written as x + 8y = 4. Confirm that this is so by writing x + 8y = 4 in the standard form.)
  • When we solve these two equations simultaneously, we substitute from y = 0 into x + 8y= 4.

So, x + 8(0) = 4

x + 0 = 4

x = 4

The solution is ( 4 ; 0). Checking this with the graph, we see that the lines I and J do indeed intersect at the point ( 4 ; 0 ).

  • Confirm that your answers are correct by comparing the answers you found when solving the equations algebraically, and those found by solving them graphically.

Source:

New Scientist, 27 April 2002 for Graphs A and B.

Assessment

Table 2
LO 2
Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
We know this when the learner:
2.1 investigates, in different ways, a variety of numeric and geometric patterns and relation­ships by representing and generalising them, and by explaining and justifying the rules that generate them (including patterns found in nature and cultural forms and patterns of the learner’s own creation;
2.2 represents and uses relationships between variables in order to determine input and/or output values in a variety of ways using:
2.2.1 verbal descriptions;
2.2.2 flow diagrams;
2.2.3 tables;
2.2.4 formulae and equations;
2.3 constructs mathematical models that repre­sent, describe and provide solutions to pro­blem situations, showing responsibility to­ward the environment and health of others (including problems within human rights, social, economic, cultural and environmental contexts);
2.4 solves equations by inspection, trial-and-improvement or algebraic processes (additive and multiplicative inverses, and factorisa­tion), checking the solution by substitution;
2.5 draws graphs on the Cartesian plane for given equations (in two variables), or deter­mines equations or formulae from given graphs using tables where necessary;
2.6 determines, analyses and interprets the equivalence of different descriptions of the same relationship or rule presented:
2.6.1 verbally;
2.6.2 in flow diagrams;
2.6.3 in tables;
2.6.4 by equations or expressions;
2.6.5 by graphs on the Cartesian plane in order to select the most useful represen­ta­tion for a given situation;
2.8 uses the laws of exponents to simplify expressions and solve equations;
2.9 uses factorisation to simplify algebraic expressions and solve equations.

Memorandum

2.1 m = –1; c = 1

y = –x + 1

2.2 m = –1,5; c = –1,5

y = –1½x – 1½

2.3 m = 5656 size 12{ { {5} over {6} } } {}; c = –0,4

y = 5656 size 12{ { {5} over {6} } } {}x – 0,4

2.4 m = 2; c = –1

y = 2x – 1

2.5 m = –1; c = 0

y = –x

2.6 m = −23−23 size 12{ - { {2} over {3} } } {}; c = 0

y = −23−23 size 12{ - { {2} over {3} } } {}x

2.7 m = 1313 size 12{ { {1} over {3} } } {}; c = 0

y = 1313 size 12{ { {1} over {3} } } {}x

2.8 m = 2323 size 12{ { {2} over {3} } } {}; c = 0

y = 2323 size 12{ { {2} over {3} } } {}x

3. A: y = 3

B: y = –½ x

C: y = ½x + 2

D: x = –1

E: y = –3

F: x = 2

G: y = x

H: y = x – 2

I: y = –¼x + ½

J: y = 0

K: y = ½x – 2

L: y = –½x + 4

4. The lines are parallel. At this point, depending on the class, the educator may want to introduce the facts that for parallel lines, m1 = m2, and for perpendicular lines, m1 × m2 = –1.

Gradients between two points

2.1 m=6−42−4=2−2=−1m=6−42−4=2−2=−1 size 12{m= { {6 - 4} over {2 - 4} } = { {2} over { - 2} } = - 1} {}

2.2 m=2−−11−−2=2+11+2=33=1m=2−−11−−2=2+11+2=33=1 size 12{m= { {2 - left ( - 1 right )} over {1 - left ( - 2 right )} } = { {2+1} over {1+2} } = { {3} over {3} } =1} {}

2.3 m=5−01−0=51=5m=5−01−0=51=5 size 12{m= { {5 - 0} over {1 - 0} } = { {5} over {1} } =5} {}

2.4 m=4−4−1−5=0−6=0m=4−4−1−5=0−6=0 size 12{m= { {4 - 4} over { - 1 - 5} } = { {0} over { - 6} } =0} {}

2.5 m=0−−37−7=30m=0−−37−7=30 size 12{m= { {0 - left ( - 3 right )} over {7 - 7} } = { {3} over {0} } } {} which is undefined.

  • Learners often confuse the meanings of the zero numerator and the zero denominator. It is wise to emphasize that a 0 denominator must be dealt with first.

If time allows, ask the learners to sketch the lines above by connecting the two given points and to confirm that their answers are reasonable.

1.1 (2 ; 3)

1.2 (–3 ; –3)

1.3 (–1 ; –3)

1.4 (4 ; 0)

1.5 (4 ; 0)

2.1 (2 ; 3)

2.2 (–3 ; –3)

2.3 (–1 ; –3)

2.4 (4 ; 0)

2.5 (4 ; 0)

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