3.3 Finding the equation of a straight line graph from a diagram  (Page 2/2)

• If you do the subtraction the other way round, then you must do it for both coordinates, like this:

$m=\frac{\text{vertical}\text{distance}}{\text{horizontal}\text{distance}}=\frac{2-\left(-1\right)}{4-3}=\frac{+3}{+1}=3$ , 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 = ½ x + 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 = ½ x – 2 and y = 0

2 Look at the diagrams in the previous e x ercise 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 $y=-\frac{1}{8}x+\frac{1}{2}$ . (This equation can also be written as x + 8 y = 4. Confirm that this is so by writing x + 8 y = 4 in the standard form.)
• When we solve these two equations simultaneously, we substitute from y = 0 into x + 8 y = 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

Memorandum

2.1 m = –1; c = 1

y = – x + 1

2.2 m = –1,5; c = –1,5

y = –1½ x – 1½

2.3 m = $\frac{5}{6}$ ; c = –0,4

y = $\frac{5}{6}$ x – 0,4

2.4 m = 2; c = –1

y = 2 x – 1

2.5 m = –1; c = 0

y = – x

2.6 m = $-\frac{2}{3}$ ; c = 0

y = $-\frac{2}{3}$ x

2.7 m = $\frac{1}{3}$ ; c = 0

y = $\frac{1}{3}$ x

2.8 m = $\frac{2}{3}$ ; c = 0

y = $\frac{2}{3}$ x

3. A: y = 3

B: y = –½ x

C: y = ½ x + 2

D: x = –1

E: y = –3

F: x = 2

G: y = x

H: y = x – 2

I: y = –¼ x + ½

J: y = 0

K: y = ½ x – 2

L: y = –½ x + 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, m ₁ = m ₂ , and for perpendicular lines, m ₁ × m ₂ = –1.

Gradients between two points

2.1 $m=\frac{6-4}{2-4}=\frac{2}{-2}=-1$

2.2 $m=\frac{2-\left(-1\right)}{1-\left(-2\right)}=\frac{2+1}{1+2}=\frac{3}{3}=1$

2.3 $m=\frac{5-0}{1-0}=\frac{5}{1}=5$

2.4 $m=\frac{4-4}{-1-5}=\frac{0}{-6}=0$

2.5 $m=\frac{0-\left(-3\right)}{7-7}=\frac{3}{0}$ 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)