I have $A(3,0)$, $B(k,2)$, $C(k,5)$, $k$ is a variable.
I need to find the locus of the intersection between AC and BO. (O is the origin).
What I tried is to find the equation of both AC and BO in terms of $k$.
I got
$$y_{AC} = \frac{5}{k-3}x - \frac{15}{k-3}$$ $$ y_{BO} = \frac2kx$$
I found the intersection between both lines (i.e. $y_{BO} = y_{AC}$).
I found that
$$x = \frac{15k}{3k+6}$$
In other words, the locus would be in that case
$$y = \frac{15x}{3x+6}$$
This however is not a line (and it should be). It is not the answer either.
What did I do wrong? I can show more steps of my solution if you need.
Thanks!
Your error is that $$\tag{1}x = \frac{15k}{3k+6}$$
[or simpler
$$\tag{1.5}x = \frac{5k}{k+2} \ \ ]$$
doe not imply $y = \frac{15x}{3x+6}$ because you can not say that $x$ and $k$ are the same.
You have to plug the expression of $x$ in (1.5) into relationship $y=(2/k)x$, in order to get
$$\tag{2}y = \frac{10}{k+2}$$
(1.5)+(2) constitutes a parametric representation of the locus.
If you desire a cartesian representation, you need to eliminate parameter $k$, which is very simple by adding (1.5)+(2), giving:
$$x+y=5$$
which is the equation of a straight line.