(B) Gradient of a Straight Line
If the points $A(x_1,y_1)$  and $B(x_2,y_2)$   lie on the straight line $y=mx+c$ , then gradient of ,the straight line
$$m=\frac{y_2-y_1}{x_2-x_1}or\frac{y_1-y_2}{x_1-x_2}$$

Question 3
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation $\frac{a}{y}=\frac{b}{x}+1$ , where k and p are constants.


(a) Based on the table above, construct a table for the values of $\frac{1}{x}$ and $\frac{1}{y}$ . Plot $\frac{1}{y}$ against $\frac{1}{x}$ , using a scale of  2 cm to 0.1 unit on the $\frac{1}{x}$ - axis and  2 cm to 0.2 unit on the $\frac{1}{y}$ - axis. Hence, draw the line of best fit.
(b) Use the graph from  (b)  to find the value of
(i)  a,
(ii)  b.

Solution
Step 1 : Construct a table consisting X and Y.




Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here




Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph




Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

Steps to draw a line of Best Fit
(i) Select suitable scales for the x-axis and the y-axis, make sure the points plotted accurately and the graph produced is large enough on the graph paper,
(ii) Mark the points correctly,
(iii) Use a long and transparent ruler to draw the line of best fit.

Linear Function (Straight Line)

Examples :
(a) y = 3x + 5
(b) 3y + 2x - 7 = 0

Non-Linear Function

Examples :
(a) $y=a{x}^3+b{x}^2$

(b) $y=ax+\frac{b}{x}$

(c) $y=a{b}^x$

(d) $y=\frac{x}{p+qx}$

A set of two variables are related non linearly can be converted to a linear equation.  The line of best fit can be written in the form

Y = mX + c

where
X and Y are in terms of x and/or y
m is the gradient,
c is the Y-intercept

Examples:
Reduce each of the following equations to the linear form. Hence, state the gradient and the Y-intercept of the linear equations in terms of  a and b.
(a)  $y=a{x}^3+b{x}^2$
(b)  $y=ax+\frac{b}{x}$
(c)  $y=ax-b{x}^2$
(d)  $xy=\frac{p}{x}+qx$
(e)  $y=a\sqrt{x}+\frac{b}{\sqrt{x}}$
(f)  $\frac{a}{y}=\frac{b}{x}+1$

[Note :
X and Y cannot have constant, but can have the variables (for example x and y)
 m and c can only have the constant(for example a and b), cannot have the variables x and y]

Solution:

(C) Mid Point 



Mid Point AB is given by $$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

2.2 The Line of Best Fit

Line of best fit has 2 characteristics:
(i) it passes through as many points as possible,
(ii) the number of points which are not on the line of best fit are equally distributed on the both sides of the line.

Question 2 (10 marks):
Use a graph to answer this question.
Table shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of $\frac{y^2}{x}\text{ against }\frac{1}{x}$ is plotted.


(a) Based on Table, construct a table for the values of $\frac{1}{x}\text{ and }\frac{y^2}{x}.$  
$\begin{array}{l}\left(b\right)\text{ Plot }\frac{y^2}{x}\text{ against }\frac{1}{x}\text{, using a scale of 2 cm to 0}\text{.1 unit on the }\frac{1}{x}\text{-axis}\\ \text{ and 2cm to 2 units on the }\frac{y^2}{x}\text{-axis}\text{.}\\ \text{ Hence, draw the line of best fit}\text{.}\end{array}$

(c) Using the graph in 11(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.

Solution:
(a)


(b)


(c)(i)
$\begin{array}{l}\text{When }x=2.7,\frac{1}{x}=0.37\\ \text{From graph,}\\ \frac{y^2}{x}=5.2\\ \frac{y^2}{2.7}=5.2\\ y=3.75\end{array}$


(c)(ii)
$\begin{array}{l}\text{Form graph, }y\text{-intercept, }c\text{ = –4}\\ \text{gradient, }m=\frac{16-\left(-4\right)}{0.8-0}=25\\ Y=mX+c\\ \frac{y^2}{x}=25\left(\frac{1}{x}\right)-4\\ y=\sqrt{25-4x}\end{array}$

(A) Equation of a straight Line 

 An equation of straight line is given by y = mx + c.
• The variables x and y are linearly related
• The term c is known as y-intercept. It represents the y value where the line cuts the y-axis
• The term m is the gradient of the straight line and its value is constant

(B) Gradient of a Straight Line
If the points $A(x_1,y_1)$  and $B(x_2,y_2)$   lie on the straight line $y=mx+c$ , then gradient of ,the straight line
$$m=\frac{y_2-y_1}{x_2-x_1}or\frac{y_1-y_2}{x_1-x_2}$$

(C) Mid Point 



Mid Point AB is given by $$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

(D) Distance Between Two Points
Distance between point $A(x_1,y_1)$ and point is $B(x_2,y_2)$ given by $$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$$