Revision on March term-exam 2008 for Sc1 *****

This section will be updated whenever I have an idea about term-exam examples, revision on each chapter and formula … etc

**Make sure you fully understand all the revision examples here :

1-Mar-2008 (Sat) : (1) Equation of a straight line and its relation with coordinates points

e.g. A straight line passes through a point (1,3). This line cuts x-axis at (b,0) and cuts y-axis at (0,b). Find the equation of this straight line.

Solution:

Explanation: we know that any two points can form a straight line. That is to say any two points will make a straight line. Meaning we can find the equation of a line by a given of two points!!!!!!

So, you can use any two points out of the given three points in the above example. Try to use the points on x-axis and on y-axis to form a straight line …{ (b,0) & (0,b) } . Next substitute (1,3) into the calculated equation. Finally, you can obtain the value of b. After that, substitute b into the equation to make something like y=mx+c or ax+by+c=0.

Try it out yourself now…..!!!!!!!!!!!!!!!!!

Ans : x+y=4

1-Mar-2008 (Sat) : (2) Parametric equation

Pg 38 (Ex3 Q19) : All these are called parametric equation.

You must understand the meaning of “locus”. A parametric equation can generate an equation of the locus as the 3rd variable varies.

This means a function (or an equation) is given indirectly by another variable. Then, this 3rd variable is related by x and y. Go and look at the examples (pg38, Q19).

e.g. Given a parametric equation is in the form of x=t+1, y=3t-2. This equation meets with 4x+5y=12. Find the point of intersection between them.

Solution:

from the given parametric equation, I can write it as:

x=t+1 , therefore t=x-1 Substitute into y=3t-2

I get y=3(x-1)-2 -> y=3x-3-2

I have y=3x-5 (the locus generated by the parametric equation.)

Now, this line meets with 4x+5y=12.

Try to solve these two equations.

Ans : Pt. of intersection is (37/19, 16/19)

Another way to do this question :

step 1: substitute x=t-1 and y=3t-2 into 4x+5y=12.

step 2:solve for the value t.

step 3:substitute back value of t into x=t-1 and y=3t-2.

Try it now………..!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1-Mar-2008 (Sat) : (3) Perpendicular distance from a point to a line

Given a straight line of the form ax+by+c=0. The distance from a point (p,q) to this line is :

D=absolute value of {(a(p)+b(q)+c)/square root of (a^2+b^2)}

This is the formula you have to memorize!!!

e.g. How to find the distance between two parallel lines?

Solution:

Method (1) : Find the perpendicular distance from the origin to each line and sum up the two results!

Method (2) : Choose any 1 point on 1 line and find the perpendicular distance to another line.

Method (3) : Find one perpendicular equation of a line to the given line. Then find the point of intersection on the two lines. Finally, find the distance between two points. {this is not a good way to do, longer but no need to use the above formula} .

Now, let you try this :

e.g. Find the distance between these two parallel lines x+y=4 and x+y=-4

Ans : 4 X (square root of 2 )

2-Mar-2008 (Sun) : (4) Incentre of a triangle

Incentre of a triangle : This is the intersection point of two equations of the angle bisector lines in a triangle.

Step 1 : Find the equations of a particular angle bisector. You will have two equations.

Step 2 : Determine the correct equation and abandon the other one.

Step 3 : Repeat the same process for step1 & step2 for another angle.

Step 4 : Solve x and y for the two equations found.

Note : The question may give you 3 sides of a triangle as 3 equations OR the question may give you 3 vertices of a triangle

e.g. Find the incentre of a triangle with x=0, y=0 & 4x+3y=5 as the three sides of the triangle.

Solution :

Step 1: sketch the three lines on x-y graph

Step 2: Choose any two angle to do the process

Step 3: Write down the ± equations for that particular angle

Step 4: Determine which is the required equation by looking at the graph (check the + or – gradient)

Step 5: repeat the step 3 & step 4 for the second angle and find the second equation which bisects that angle.

Step 6: solve the two equations found in step 4 & step 5.

You will get the (x,y) coordinates now!

Try it out for yourself!!!!!!!!!!!

Ans : (5/12,5/12)

2-Mar-2008 (Sun) : (5) Equation of bisector of an angle

Explanation : you are given 2 lines which intersected at a point. You may have one of these 2 situations :

(1) an acute angle (<90°) and an obtuse angle (>90°) at that point.

(2) a right angle for both angles at that point.

Base on formula on Ch. 3, each angle will have an equation of bisector to itself. The formula said that each angle bisector equation has 1 ‘+’ and 1 ‘-‘ gradient for that angle. After finding out these 2 equations (straight lines), we have to determine which is the actual equation for the Q.

Refer to your recent test (2) Part II Q3(a).

Also, refer to pg 36 Q5, Q6, pg 62 rev. paper B3 Q2

2-Mar-2008 (Sun) : (6) Perpendicular bisector line to a line connected by two points

e.g. let P(1,4) and Q(-3,-5) be two points. A line is perpendicular bisector into PQ. Find the equation of the line.

Solution :

Step1:Find the gradient of PQ, let m be the gradient, m=(4-(-5))/(1-(-3))=9/4

Step2:Find the gradient that perpendicular to 9/4, which is -4/9

Step3:Find the mid-point of PQ, let M be the mid-point, M=((1-3)/2,(4-5)/2) =(-1,-1/2)

Step4:Find the equation with gradient=-4/9 and passes thru’ (-1,-1/2)

Ans : the equation is 18y+8x+17=0

There is another way to ask you about this Question :

Find the coordinates of Q so that a line 18y+8x+17=0 is perpendicular bisector to the line PQ where P is (1,4)?

Try it out for yourself…..!!!! *****

So, this is the end of our revision for Ch1 to Ch3. Tomorrow I will start the dy/dx … (Ch4 to Ch7)

8 thoughts on “Revision on March term-exam 2008 for Sc1 *****

  1. 老师可以给我们那时你分的ch1-3的练习的答案吗???

    Ok, I will try to post it tonight after 8:30pm

    I’m sorry. I didn’t bring back the answers. It is at the office.

  2. sad = = some of the Question inside is hard.

    If you can understand and solve harder problem, then you can do very fast in the term-exam…
    My advice is now you should concentrate on my revision tips and examples….

  3. Q5, 4x+3y-6+p(2x-5y-16)=0
    Can you group all x together, all y together and all constant together? (p is just a constant!!!)

    So, I can re-write the equation to :
    4x+3y-6+2px-5py-16p=0
    (4+2p)x+(3-5p)y-6-16p=0 ——— (1)

    (a) passes thru’ (0,0). Therefore, you just substitute x=0, y=0

    0+0-6-16p=0
    16p=-6
    p=-6/16=-3/8
    (b) parallel to 5x-6y-11=0
    therefore it must have a gradient of 5/6
    re-arrange equation (1), you have (3-5p)y=-(4+2p)x+6+16p
    gradient=-(4+2p)/(3-5p)
    hence, -(4+2p)/(3-5p)=5/6
    solve this…
    (c) perpendicular to x+4y=0
    gradient for x+4y=0 is -1/4
    hence gradient for (1) must be 4, 4=-(4+2p)/(3-5p)
    solve it…

  4. 4(b) is a bit harder (you haven’t learnt the product rule for differentiation)

    A=2xy ………(1)
    x^2 + y^2 = R^2 ………(2)

    from (2) y=sqrt(R^2 – x^2) sub into (1)
    so A=2(x)(sqrt(R^2 – x^2)
    dA/dx = …
    let dA/dx=0 (because want to find max. of A)
    you can get x=R/sqrt(2), so y=R/sqrt(2) sub into (1)
    you will get max of A=2(R/sqrt(2))(R/sqrt(2))=R^2

    You no need to do this Q!

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