Reduce Non-Linear Function To Linear Function – Examples (G) To (L)

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.
(g)  kx2+ty2=x  
(h)  y=xp+qx  
(i)  hy=x+kx  
(j)  y=abx  
(k)  y=axb  
(l)  y=abx+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:












Gradient Of A Straight Line


(B) Gradient of a Straight Line
If the points A(x1,y1)  and B(x2,y2)   lie on the straight line y=mx+c , then gradient of ,the straight line
m=y2y1x2x1ory1y2x1x2


SPM Practice 3 (Linear Law) – Question 3

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 ay=bx+1 , where k and p are constants.


(a) Based on the table above, construct a table for the values of 1x and 1y . Plot 1y against 1x , using a scale of  2 cm to 0.1 unit on the 1x - axis and  2 cm to 0.2 unit on the 1y - 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 The Line Of Best Fit

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.

Step 1 : Select the suitable scale on x and y axis 

(the graph produced must be more than 50% of the graph paper)
 

 
Step 2 : Mark the points correctly
 

 
Step 3 : Draw the Line of Best Fit
 
* Note
 -the line passes through four points 
-one point is above the line
-one point is below the line


 

SPM Practice 2 (Question 1 – 3)

Question 1:
Reduce non-linear relation, y=pxn1, where k and n are constants, to linear equation.  State the gradient and vertical intercept for the linear equation obtained.
[Note : Reduce No-linear function to linear function]

Solution:


 

Question 2:
The diagram shows a line of best fit by plotting a graph of  y2against x.

  1. Find the equation of the line of best fit.
  2. Determine the value of
    1. x when y = 4,
    2. y when x = 25.
Solution:



 

Question 3:
The diagram shows part of the straight line graph obtained by plottingy against x2.

Express y in terms of x.

Solution:



 

SPM Practice 3 (Linear Law) – Question 1


Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment.
The variables x and y are related by the equation yh=hkx , where h and k are constants.


(a) Plot xy against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the xy-axis.
Hence, draw the line of best fit.

(b) Using the graph in 9(a), find
(i) the value of h and of k,
(ii) the correct value of y if one of the values of y has been wrongly recorded during the experiment.

Solution: 
(a)




(b)
yh=hkxxyhx=hkxy=hx+hkY=mX+CY=xy, m=h, C=hk


(b)(i)
m=36.55.1h=36.55.1h=7.157h=51.22C=4hk=4k=4hk=451.22k=0.0781


(b)(ii)
xy=213.5y=21y=213.5=6.0Correct value of y is 6.0.


Tips To Reduce Non-Linear Function To Linear Function

Tips:
(1)  The equation must have one constant (without x and y).
(2)  X and Y cannot have constant, but can have the variables (for example x and y).
(3)  m and c can only have the constant (for example a and b), cannot have the variables x and y.


Examples
(1)
X and Y cannot have constant, but can have the variables (for example x and y)



(2)
 m and c can only have the constant (for example a and b), cannot have the variables x and y





SPM Practice 2 (Linear Law) – Question 4

Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation y=qx+pqx , where p and q are constants.


One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against x2  .  Hence, draw the line of best fit

(b) Use your graph in (a) to answer the following questions:
(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.

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

(b) (i) State the values of y which is incorrectly recorded and determine its actual value.


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

Distance Between Two Points

Distance between point A(x1,y1) and point is B(x2,y2) given by

(x1x2)2+(y1y2)2


 

SPM Practice 3 (Linear Law) – Question 2


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 y2x against 1x is plotted.


(a) Based on Table, construct a table for the values of 1x and y2x.  
(b) Plot y2x against 1x, using a scale of 2 cm to 0.1 unit on the 1x-axis  and 2cm to 2 units on the y2x-axis.  Hence, draw the line of best fit.

(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)
When x=2.7, 1x=0.37From graph,y2x=5.2y22.7=5.2y=3.75


(c)(ii)

Form graph, y-intercept, c = –4gradient, m=16(4)0.80=25Y=mX+cy2x=25(1x)4y=254x