Applying graphs
By Clement Radcliffe, Contributor

The solution of last week's homework is an appropriate opening to this lesson.

Homework
(1) Solve 2x²-3x-4=2-4x

Solution
By transposing then
2x²-3x-4-2+4x=0
... 2x² + x - 6 =0 Factorizing
(2x-3)(x+2)=0
...2x-3=0, that is, x=3/2
OR x+2=0, that is, x =-2
Answer x=3/2 or -2

Using the appropriate scale, draw the curve y=2x²-5x-3. Hence solve the equation:
(a) 2x²-5x-3=0
(b) 2x²-5x=4
By drawing the appropriate line, solve the equation 2x-4x-5=0

Solution

(a) From the graph of y=2x²-5x-3, the solution of the equation 2x²-5x-3=0 is the x coordinates of the points of intersection of the curve and the line y=0 or the x axis.
...x=-1/2 or 3

(b) Given the plot of y=2x²-5x-3, then the equation of 2x²-5x-3=0 may be reorganized to 2x-5x-3=1.

The solution is therefore the x coordinates of the points of intersection of the curve y=2x²-5x-3 and the line y=1. By drawing the line y = 1 on the graph, the solution of the equation is seen as:
x=-0.63 or 3.13

Similarly, the equation 2x²-4x-5=0 may be expressed as 2x²-5x-3=2-x
NB: I am sure you are able to see that both equations are identical.
ie. By drawing the line y=2-x on the graph reading off the points of intersection, then the solution is
x = -0.87 or 2.87

Please attempt the following:
Given the following graphs, please determine the line which is required to solve the equation.
a) Solve 2x²+x-3=0, given the graph y = 2x² + x .3
b) 2x² + x = 3, given the graph 2x² + x -3
c) x² = 1, given the graph y =x²
d) 2x² + 2x - 1 = 0, given the graph y=2x²+y-3

Now to COORDINATE GEOMETRY
This topic will highlight aspects of straight line graph. They include Gradient, Intercept, Mid-Point, Length of line and equation of lines.

Reminders
A graph must satisfy the following:
* The axes must be correctly labelled
* Appropriate scales should be accurately used.
* The coordinates of a point is always expressed in the form: (x,y)
* At least three points are required to draw a straight line. A ruler must always be used to join the points.

GRADIENT
The gradient of a line is a measure of its slope. The value is denoted by m and is defined as:

All of the above are illustrated in the following example.

LENGTH OF LINE
The length of AB is found using Pythagoras 'Theorem as triangle ABC is right angled.
... AB² = BC² + AC²
...AB² = (x2-x1)² + (y2 - y1)²
See the diagram above.

EXAMPLE
A straignt line is drawn through the points A(2,5) B (1,2) Find the length of AB

We will try another example.

A straight line is drawn through the points A(1,2) and B(-5,3)
Find: (i) the gradient of AB
(ii) the mid-point of AB
(iii) the length of AB

SOLUTION
i) The gradient of AB = m = y2 -y1/x2-x1

ii) The mid-point of AB is

iii) In order to find the length of AB, we use th formula AE² = (y2 - y1)² + (x2 - x1)²
ie. AB² = (3-2)² + (-5-1)²
1² + (-6)² = 37
AB =

The following points should be noted as you attempt questions:
*Always begin by presenting the required formula
* To calculate the gradient, you may use one of the following:

I am sure you can prove that both are equivalent.
* In evaluating the values be careful to ensure the accuracy of the substitution and please watch the negative signs. (Directed numbers).

On your own, find the gradient, midpoint and length of the line joining the points.
a) A(-3, -4), B (3,7)
b) C(8,8), D(-4,2)

Clement Radcliffee is Principal of Glenmuir High School in Clarendon. Send your questions and comments to the CXC Study Guide, the Gleaner Company Ltd., 7 North Street, Kingston; or email us at jcampbell@gleanerjm.com.