Distance Between Two Points, Division and Midpoint of Line Segment - Analytic Geometry

* Distance between Two Points

- if line 1 and line 2 are parallel with x and y-axis respectively, x₂ minus x₁ will be the distance parallel to the x-axis while y₂ minus y₁ will be the distance parallel to the y-axis. We will then form a right triangle whose hypotenuse is the distance between the two points :( x₁, y₁) and (x₂, y₂).

Formula:  
(Derived from Pythagorean Theorem) 

Where:

 
d = distance between the two points.
x₁ = x-coordinate of point 1
x₂ = x-coordinate of point 2
y₁ = y-coordinate of point 1
y₂ = y-coordinate of point 2

Examples:

* Find the distance between the points (1,3) and (5,3).

Given:

Let:
P1(1,3)
P2(5,2)

Required: d

Solution:

d = √[(x₂ - x1)2 - (y₂ - y1)2]

d = √[(5 - 1)2 - (2 - 3)2]

d = 4.12 units

* Division of a Line Segment

            - Simply finding the coordinates of the points that divides a line segment into parts (ratio).

Formula:

x = x₁ + r(x₂ - x₁)
y = y₁ + r(y₂ - y₁)

Where:

x and y = coordinate of the point that divides the line segment.
x₁ and x₂ = x-coordinates of the line segment's end points
y₁ and y₂ = y-coordinates of the line segment's end points
r = ratio

Example:

* Find the coordinates of the points that divide the line into 3 parts. The end points of the line are (1, 3) and (5, 8)

Given:
Let:
P1(1,3)
P2(5,8)

Required: points that divides the line (A and B)

Solution:

(solving for 1st point: A)

* ratio: 1/3 

x_A = x₁ + r(x₂ - x₁)

= 1 + (1/3)(5 - 1)

= 1 + 1.33

x_A = 1.33

y_A = y₁ + r(y₂ - y₁)

= 3 + (1/3)(8 - 3)

= 3 + 1.67

y_A= 4.67


A(2.33,4.67)

* ratio: 2/3

x_B = x₁ + r(x₂ - x₁)

= 1 + (2/3)(5 - 1)

= 1 + 2.67

x_B= 3.67

y_B = y₁ + r(y₂ - y₁)

= 3 +(2/3)(8 - 3)

= 3 + 3.33

y_B = 6.33

B(3.67,6.33)

Note: 
- If you are dividing a line. The number of points is always less than the number of division.
 

    no. points = no. division - 1

    e.g.
    no. divisions = 5
    no. points = 4

- to check if your answer is correct, solve the distance of each part of the line segment by using distance formula. The sum of the distances of the parts must be equal with the distance between the endpoints of the line segment itself.

* Midpoint of a Line Segment
            - a point that divides a line segment into two equal parts.

Formula:

x_M = (x_{2 +} x1)/2

y_M = (y_{2 +} y1)/2

Where:

x_M and y_M = coordinates of midpoint
x₁ and x₂ = x-coordinates of the endpoints of the line segment.
y₁ and y₂ = y-coordinates of the endpoints of the line segment.

Example:

* Find the midpoint of the line. Endpoints: (3,4) and (4,6)

Given:
Let:
P1(3,4)
P2(4,6)

Required: M

Solution:

x_M = (x_{2 +} x1)/2

= (3 + 4)/2 

= 7/2

x_M = 3.5



y_M = (x_{2 +} x1)/2

= (4 + 6)/2

= 10/2

y_M = 5

M(3.5,5)

Ratio, Proportion and Variation - Algebra

* Ratio

            - When a number x is to y denoted as x:y where y is not equal to zero. 
            - Simply the fraction of two numbers.

            x:y = x/y

Examples: 

1. What is the ratio of 5 ft to 85 in?

If 1ft = 12 in, then 5 ft is equal to 60. Hence,

60ft/85ft = 12/17 = 12:17

2. What is the ratio of the mass of the blood to the mass of the body of a 75 kg man? If the mass of blood of the man is 5 kg.
5kg/75kg = 1/15 = 1:15

* Proportion
            - is the equality of two ratios or fraction.

a:b = c:d or  a/b = c/d 

Where:

- a and d are called extremes
- b and c are called means

We have Laws in Proportion and here are the following:

1. The product of extremes is equal to the product of means.

From a/b = c/d we will get,

ad = bc

Example:

1/4 = 2/8  

(1)(8) = (4)(2)

        8 = 8

2. The reciprocals of both ratio in a proportion are equal.
b/a = d/c

Example:

1/4 = 5/20

4 = 20/5

4 = 4

3. Interchanging the mean of the first ratio with the extreme of the other will not affect the proportion and vice versa.
a/c = b/d     or    d/c = b/a 

Example:

1/2 = 4/8 by interchanging the mean and extreme,

1/4 = 2/8

1/4 = 1/4

4. Adding the mean on the extreme of the left-side of the equation while adding the extreme on the mean of the right-side of the equation (Vise-versa) will not change the equality.  

(a+b)/b = (c+d)/d 

Example:

1/3 =  5/15

(1+3)/3 = (5+15)/15

4/3 = 20/15

4/3 = 4/3

5. Subtracting the mean on the extreme of the left-side of the equation while subtracting the extreme on the mean of the right-side of the equation (Vise-versa) will not change the equality.

 
(a - b)/b = (c - d)/d 

Example:

1/5 = 5/25

(1-5)/5 = (5-25)/25

- 4/5 = -20/25 

-4/5 = - 4/5

6. (a+b)/(a-b) = (c+d)/(c-d)

Example:
2/7 = 6/21

(2+7)/(2-7) = (6+21)/(6-21)

-9/5 = -27/15

-9/5 = -9/5 

* Variation

3 Principles of Variation:

1. If y varies directly as x then, y = kx.

2. If y varies inversely as x then, y = k/x

3. If y varies jointly as x and z then, y = kxz

Note:

 
- k is the constant of variation.

Example:

* If y varies inversely as x and directly as z and if y = 4 when x = 3 and z = 5. Find z when x = 4 and y = 6.

First, solve for k:(Substitute the first given where x,y and z are known)

y = kz/x

k = xy/z

k = (3 * 4)/5

k = 12/5 or 2.4

Now that we have the value of k, we can solve for z in the second set of values of x and y.

y = kz/x

z = xy/k

z = (4 * 6 )/ 2.4

z = 24/2.4

z = 10

Hence, the value of z is 10 when x is 4 and y is 6.

Functions - Algebra

* Operations of Functions

(f + g)(x) = f(x) + g(x)

(f - g)(x) = f(x) - g(x)

(f . g)(x) = f(x) . g(x)

                                    Note: g(x) is not equal to zero

Where: 

f(x) and g(x) are functions and read as "f of x and g of x."

Example:

Given:

f(x) = 3x + 2
g(x) = 5x - 4

Required: (f + g)(x), (f - g)(x), (f . g)(x), (f/g)(x)

Solution:
* (f + g)(x) = f(x) + g(x)

= 3x + 2 + (5x - 4)

= 3x + 2 + 5x - 4

= 8x - 2

* (f - g)(x) = f(x) - g(x)

= 3x + 2 - (5x - 4)

= 3x + 2 - 5x + 4

= -2x + 6

* (f . g)(x) = f(x) . g(x)

= (3x + 2)(5x - 4)  by using the FOIL method

= (3x)(5x) + (3x)(-4) + (2)(5x) + (2)(-4)

= 15x² - 12x + 10x - 8

= 15x² - 2x - 8

 

* Compositions of Functions

f[g(x)] = fog composition of f of g on x
g[f(x)] = gof composition of g of f on x


Examples:

Given:

f(x) = 3x - 2
g(x) = 2x + 3

Required: fog and gof

Solution: 

* fog
f[g(x)] = 3(2x + 3) - 2

= 6x + 9 - 2

= 6x + 7

* gof
g[f(x)] = 2(3x - 2) + 3

= 6x - 4 + 3

= 6x - 1

Given:

f(x) = x² - 3
g(x) = x + 2

Required: fog and gof

Solution:

* f[g(x)] = (x + 2)² – 3 by using Square of Binomials

= x² + 4x + 4 - 3

= x² + 4x + 1

* g[f(x)] = (x² - 3) + 2

= x² - 1

- it is just like substituting the value of one function inside the other.

* Inverse of a Function

To find the inverse of a given function we need to follow these 4 steps:

1. Let f(x) be equal to y
2. Interchange x and y and vise-versa.
3. Solve for y. 
4. Change y to f^-1(x).

Examples: 

Given:

f(x) = 4x - 5 

Required: f^-1(x)

Solution: 

Step 1:

f(x) = 4x - 5

y = 4x - 5

Step 2:

y = 4x - 5

x = 4y - 5

Step 3:

x = 4y - 5
4y = x + 5

Step 4:

Checking: 

f[f ^-1(x)] = x

x + 5 - 5 = x
x = x


Note:
- If we get the composition of a function and its inverse, the answer will be equal to x. This can be used for checking if the inverse that we have solved is correct.