## Monday, October 18, 2010

### Geometrical Proofs

Introduction

Intercept theorem
There are 2 ways to apply the intercept theorem

In the first scenario, you will be able to identify a pair of parallel lines from the question
Upon identifying the pair of parallel lines,
we conclude that the ratio of AD:DB is the same as the ratio of AE:EC

In the second scenario, you identify that that the ratio of AD:DB is the same as the ratio of AE:EC
Subsequently you conclude that the lines DE and BC are parallel

Midpoint Theorem
The Midpoint theorem is a specific case of the Intercept theorem when the ratio of the lengths is 1:1

There are 2 ways to apply the Midpoint Theorem

The first way is to use Midpoint Theorem to show that 2 lines are parallel
From the information given in the question, we will be able to identify that D is the midpoint of AB and E is the midpoint of AC
Given that D and E are midpoints, we will be able to conclude that DE is parallel to BC

The second way is to use Midpoint Theorem to show that E is a midpoint
From the information given in the question, we know that DE is parallel to BC and D is a midpoint
Using the Midpoint theorem, we can conclude that E is the midpoint of AC

Question

Given that:
O is centre of circle, AB is diameter, C lies on circle
AD is tangent to circle at A, EC is tangent to circle at C

i) Prove that triangles AEO and CEO are congruent
Strategy: make use of information given in the question
AD is tangent to circle --> angle OAE is 90 degrees
EC is tangent to circle --> angle OCE is 90 degrees
O is centre of circle --> OA and OC are radii of the circle, hence their lengths are equal
From diagram OE is the common length
Using the RHS property, we can show that triangles AEO and CEO are congruent

## Monday, September 20, 2010

### Integration techniques Part III

Integrating fractions

Revision of anti chain rule
"Increase power by 1 and divide by the new power"

Recall the differentiation of ln (ax + b)

## Monday, September 6, 2010

### Equation of Circles

Formula for Equation of Circle

## Thursday, August 26, 2010

### Completing the Square

Completing the square is a technique required to find equations of circle

Example 1

## Wednesday, August 18, 2010

### Coordinate Geometry Part II

Example 1

A typical coordinate geometry question from the O Levels

Given that AC = 3AB, find the coordinates of point B.

Remarks:
A lot of students will try to use the distance formula to form simultaneous equations in x and y to find the coordinates of B. This is a very long and tedious method. A much simpler method is to use ratio and proportion to find the coordinates of B

To find the x coordinate of B

We can use the same method to find the y coordinate of point B

Therefore the coordinates of point B is (4,4)

## Sunday, August 15, 2010

### Coordinate Geometry Part I

Technique 1: Finding the gradient of a line

Parallel lines
Parallel lines have the same gradient
Perpendicular lines
If a line has a gradient of 1/2 , the gradient of the line perpendicular to it is -2.
-2 is the negative reciprocal of 1/2. In other words, we "flip" the fraction 1/2 to get 2 and attach a negative sign

Technique 2: Finding equation of a straight line
The general format of a straight line is y = mx + c
where m is the gradient and,
c is the y intercept

Example 2
Find the equation of a line that contains the points (1,2) and (3,4).

From the example above we know that gradient of the line is 1.
Hence the equation of the line is: y = 1x + c

To find the y intercept, c, we sub in one of the points
I choose the point (1,2) and sub x =1, y = 2
2 = 1 + c
Hence c = 1

The equation of the line is y = x + 1

Let us combine the techniques we have learnt so far to solve a typical O Level question

Question 1

The figure above shows a trapezium where AB is parallel to CD. Both the lines AB and CD are perpendicular to AD. Find the equation of line AB

Solution

$\fn_jvn \text{Gradient of AD } = \frac{6 - (-2)}{0 - 2} = -\frac{8}{2} = -\frac{4}{1}$

Find gradient of line AB
Since AB is perpendicular to AD, the gradient of line AB is the negative reciprocal of -4 which is 1/4

Find the y intercept
From the figure we can see that the y intercept, c of the line AB is 6

Find equation of line
Therefore the equation of line AB is
$\fn_jvn y = \frac{1}{4}x + 6$

## Friday, August 6, 2010

### Roots of Quadratic Equation Part II

$\fn_jvn \text{Find the values of k for which the line x + 3y = k} \\ \text{and the curve } y^2 = 2x + 3 \text{ do not intersect} \\ \\ \text{Step 1: Combine the 2 equations together using substitution}\\ x + 3y &= k \\ x &= k - 3y \\ \\ y^2 &= 2x + 3 \\ y^2 &= 2(k - 3y)+ 3 \\ \\ \text{Step 2: Rearrange the equation to the form } ax^2 + bx + c = 0 \\ y^2 &= 2k -6y + 3 \\ y^2 + 6y -2k -3 &=0\\ \\ \text{Step 3: Since line and curve do not intersect } b^2 - 4ac < 0 \\ 6^2 - 4(1)(-2k-3) < 0 \\ 36 + 8k + 12 < 0 \\ 8k < -48 \\ k < -6 \\$