Could someone help me with some math?

Question

1. There are four points on a line in the order of A, D, C and B. AD=(5/9)*DB, AC= (9/5)*CB and CD=4, what's the length of the whole line being AB.

2. A line segment AB is divided into 3 parts, the ratio of the length of the three parts is 3:4:5. The distance between the midpoint of the first part and the midpoint of the first part and the midpoint of the second part is 1.75 cm. What is the length in centimeters of AB.

3. There are 6 points on a plane, and exactly 3 points are in the same line. How many lines can be drawn through these points?

Answer 1

Qu.1 :

AB = AD + DB
= (5/9)*DB + DB (as AD is given)
= (14/9)*DB ( = Equ.1)

Now, AC = AD + DC
That is : (9/5)*CB = (5/9)*DB + 4

Multiplying through by 45 :
81*CB = 25*DB + 180
So, CB = (25*DB + 180) / 81

Now, find AB a different way :
AB = AD + DC + CB

= (5/9)*DB + 4 + (25*DB + 180) / 81 (= Equ.2)

Now we can set Equ.1 equal to Equ.2.

(14/9)*DB = (5/9)*DB + 4 + (25*DB + 180) / 81

Multiplying through by 81 :
126*DB = 45*DB + 324 + 25*DB + 180

56*DB = 504
DB = 504 / 56 = 9

Therefore, AD = (5/9)*DB = (5/9)*9 = 5

Thus, AB = AD + DB = 5 + 9 = 14

Qu.2 :

Let x = length of the first part.
If in the ratio 3:4:5, then second part = 4x/3
and third part = 5x/3.

Distance to end of first part = x.
Distance to end of second part = x + 4x/3 = 7x/3.

Now midpoint of first part = (x + 0) / 2 = x/2.
And midpoint of second part = (x + 7x/3) / 2 = 5x/3.

Therefore, 5x/3 - x/2 = 1.75
from which we get : x = 1.5.

Thus AB = x + 4x/3 + 5x/3 = 12x/3 = 4x
= 4*1.5 = 6cm

Qu.3 :

Not sure - running out of time - but I count 13 lines
in total without delving too deeply.

Answer 2

3. There are 13 lines. There are three points on a line; call these points A, B, and C. Label the other points D, E, F. Notice that except for A, B, C, no other three points are on a line.

Count the line thru A, B, C as ONE. Now DA, DB, and DC count TWO, THREE, FOUR. Similarly for E and F to each of A, B, C we count FIVE, SIX, SEVEN for E, and EIGHT, NINE, TEN for F.

Now we only have to count the lines connecting D, E, and F. We know they cannot lie on a line, so consider a triangle with vertices D, E, F; there are three sides--extend the sides of this triangle into lines and count these ELEVEN, TWELVE, and THIRTEEN. None of these has been counted previously, because if any one of them touched A, B, or C, we would have another set of three points on a line.