# show that `|bar(RX)|=4/13|bar(RA)|`.  `X` is the point of intersection between `bar(RA)` and `bar(HS)`. http://img338.imageshack.us/img338/3783/20120606165646.jpg Given the diagram on this link,...

show that `|bar(RX)|=4/13|bar(RA)|`.  `X` is the point of intersection between `bar(RA)` and `bar(HS)`.

http://img338.imageshack.us/img338/3783/20120606165646.jpg

Given the diagram on this link, the parallelogram called `MATH`````````, show that `|bar(RX)|=4/13|bar(RA)|`.  `X` is the point of intersection between `bar(RA)` and `bar(HS)`.

Some important data:

Point `S` divides side `bar(AT)` in this ratio: 3:4

Point `R` divides side `bar(HT)` in this ratio: 1:2

mlehuzzah | Certified Educator

You can do this problem by imposing a coordinate system:
H=(0,0)
M=(a,b)
T=(c,0)
A=(a+c,b)

R divides HT in the ratio 1:2 means that:
R=(c/3,0)
Similarly, S divides AT in the ratio 3:4 means that:
S=(c+4a/7,4b/7)

Now find equations for the lines HS and RA.  Their intersection will give the location (in coordinates) of X

HS is given by:

`y=(4b/7)/(c+(4a/7))x`

which simplifies to:

`y=(4b)/(7c+4a)x`

and RA is given by:

`y=(b)/((2/3)c+a) (x-(1)/(3)c)`

` ` which simplifies to:

`y=(3b)/(2c+3a) (x-(1)/(3) c)`

Setting these equations equal to one another and solving for x gives:

`(4b)/(7c+4a) x = (3b)/(2c+3a) x - (bc)/(2c+3a)`

`x((3)/(2c+3a) - (4)/(7c+4a) )= (c)/(2c+3a)`

`x ( (3(7c+4a) - 4(2c+3a))/((2c+3a)(7c+4a)))=(c)/(2c+3a)`

`x(13)/(7c+4a)=1`

`x=(7c+4a)/13`

From here we can conclude that RX is 4/13 RA, (although it becomes more obvious with the y-coordinate):

R is at (c/3,0), A is at (a+c,b)

Thus from R to A you must travel 2c/3+a in the x direction.

So, start at c/3 and travel (4/13) of (2c/3 + a):

`(c)/(3)+(4)/(13)((2c)/(3)+a)=(7)/(13)c+(4)/(13)a=x`

Alternatively, take the answer for x and plug it into either of the line equations to get y:

`y=((4b)/(7c+4a))((7c+4a)/13) = (4b)/13`

The y coordinate of R is 0

The y coordinate of A is b

The y coordinate of X is 4b/13

Thus

|RA| `(4)/(13)`= |RX|