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]
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.
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.
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:
$m=-\frac{4}{6}=-\frac{2}{3}$ (remember to simplify the fraction).
This value is now substituted for m in the equation:
$y=-\frac{2}{3}x-1$ . 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.
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:
$m=\frac{\text{vertical}\text{distance}}{\text{horizontal}\text{distance}}=\frac{-1-2}{3-4}=\frac{-3}{-1}=3$
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