On standard x-y coordinate axes, the equation of a straight line has the general form
y = mx + c , where m and c are particular numbers (positive or negative or zero).

For example,
y = 3x + 2     y = 0.1x - 12.5      y = -5x     y = 2.8 - 4x
are all equations of straight lines.

For example, take the line y = 2x + 6.5 .

If we are told x = 4 , then substituting 4 for x gives
y = 2(4) + 6.5
and so y = 14.5
Point is (4 , 14.5)

If we are told y = 2 , then substituting 2 for y gives
2 = 2x + 6.5
and rearranging this algebraically gives x = -2.25
Point is (-2.25 , 2)

Alternatively, we could plot the graph y = 2x + 6.5 and read off the required values.

Fill in the blanks in the coordinates for the given graphs:

1) y = 2x + 7       (4 ,    ) and (    , 21)

2) y = 0.5x - 1       (1 ,    ) and (    , 3)

3) y = 4x       (7 ,    ) and (    , -10)

4) y = 100 - 5x       (-2 ,    ) and (    , 300)

5) y = x - 0.3       (-1.5 ,    ) and (   , 0)

6) y = 0.2x + 2.4       (-13 ,    ) and (   , -1)

7) y = -0.1x - 1.7       (-8 ,    ) and (    , -2)

8) y = 2.5 - 2.5x       (-2 ,    ) and (    , 2)

Fill in the blanks in the coordinates for the given graphs:

1)                        Ans: (1 , 2) and (3 , 6)

2)                        (3 , 0) and (0 , -3)

3)                        (-6 , -3) and (0 , 0)

4)                        (1 , 1) and (1.5 , 2.5)

5)                        (4 , -1.5) and (-2 , 1.5)

6)                        (-8 , -1) and (-5 , 0.5)

7)                        (1 , 7) and (6 , 4)

8)                        (-3 , 8) and (1.5 , -5.5)

How could you calculate directly the equation of the line if you know the coordinates of two points?

This activity is best carried out with the screen divided into an algebra window and a plot window. Do this by:

<W>indow <S>plit <V>ertical 40 <enter>

<W>indow <G>oto 2 <enter>

<W>indow <D>esignate <2>D-plot

The keystroke <A>lgebra takes you from the plot window to the algebra window;

The keystroke <P>lot takes you from the algebra window to the plot window.

You should do this pen-and-paper first (or even in your head!)

To check using DERIVE, you could first <A>uthor the equation of the line. Then use <M>anage <S>ubstitute to replace x by its given value. Then so<L>ve for y. Similarly, <M>anage <S>ubstitute to replace y by its given value. Then so<L>ve for x.

Alternatively, <A>uthor and <P>lot the equation of the line, and use the arrow keys to position the cross at the point on the line which has the given x coordinate, and check the y coordinate shown at the bottom of the screen. (And vice-versa.)

Th aim of the exercise is for you to get a feeling for the effect of different values of m and c on the graph of the line y = mx + c . Then, by intelligent trial and error, you should be able to choose the values of m and c which make your graph pass through the two coordinates given.

First, try the following:

<A>uthor y = x + c

Use <M>anage <S>ubstitute to replace c by 0, 1, 2, 3, -1, -2, -3, etc

and then <P>lot

What happens to the graph as c varies?

<D>elete all the graphs and now try:

<A>uthor y = mx + 1

Use <M>anage <S>ubstitute to replace m by 0.5, 1, 1.5, 2, 0, -0.5, -1, -1.5 etc

and then <P>lot

What happens to the graph as m varies?

You should now be ready to have a go at the questions.

Before you try to fit the line, it is a good idea to highlight the points whose coordinates you are given, to see what you are aiming at. For the first question, you would do this by:

<A>uthor [[1,2],[3,6]] <P>lot <P>lot

David Bowers,
Mathematics Workshop,
Suffolk College.