# Thread: Coordinate of a point in a triangle

1. ## Coordinate of a point in a triangle

Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

"Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!

2. Originally Posted by juanma101285
Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

"Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!
Drop a perpendicular from P to AB, let the foot of the perpendicular be M.
AM = the x - coordinate of P.

Now use similar triangles:

triangle ABC similar with triangle PMB:

$\displaystyle \frac{BP}{BC}=\frac{BM}{AB}$
$\displaystyle \frac{1}{1+s}=\frac{1-x}{1}$

$\displaystyle 1-x=\frac{1}{1+s}$

and find x.

In the same way, use similar triangles to find the y coordinate of P.

3. Originally Posted by juanma101285
Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

"Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!
The horizontal co-ordinate of P is

$\displaystyle \displaystyle\frac{s}{1+s}$

since if you drop a perpendicular from P to [OB], you will get the horizontal co-ordinate.

You get the vertical co-ordinate if you draw a horizontal from P to [OC].

This is

$\displaystyle \displaystyle\frac{1}{1+s}$

4. Cool, thanks a lot! I have other questions in the same exercise but I think I can solve them by myself now

5. OK... I am stuck again . How do you solve these parametric equations? (they need to be in terms of t and u). I'd just like to have a tip so that I can work on it myself and learn...

So, I got for BQ:

(x, y)=(1,0)+m(-1, t/(1+t))

and for CR I got:

(x, y)=(0,1)+n(1/(1+u), -1)

I know I have to equate the x and y, but I still can't get rid of m and n...

So I got

1-m=n/(1+u) and mt/(t+1)=1-n

So I substituted n=1+u-m-mu into

mt/(t+1)=1-n, but I get mt=m+mu-u-tu+mut+mt

How am I supposed to get rid of n and m then? :-/