Vertices of a median through R? i am studying for a test and seem to have hit yet another obstacle. Hint/Answers are highly appreciated.
The question: The vertices of a Triangle PQR are P (2,1), Q(-2,3), R(4,5). Find equation of the median through the vertex R.
I don't know how to solve this problem. At first i thought about taking the intercept, but still ended up with with a extra unknown variable. In my book the answer is given as :"3x-4y+8=0".
Note: Please provide a answer in layman's terms, i only have a beginner level education on the subject.
 A: The equation you're looking for is the equation for the line through $R$ and through the midpoint betwenn $Q$ and $P$. The midpoint between $Q$ and $P$ is $$\frac 12 (Q+P) = (0,2).$$ 
So you're looking for the equation of a line through $(4,5)$ and $(0,2)$. You know that any line has the form $ax+by+c=0$ for some constants $a,b$. Plugging in $(4,5)$ and $(0,2)$ for $x,y$, we get two equations:
$$ 4a+5b+c=0$$ and $$2b + c= 0 $$
Now comes the point where we have only two equations, but three unknowns! The solution is that the equation is only unique up to multiplication by non-zero numbers. That is, if your equation is $ax+by+c=0$, then for $d \neq 0$, also $dax+dby+dc=0$ will also describe the same line.
Since this is not a line through the origin, you can assume that $c \neq 0$, so you can divide by $c$. That way, you get the equation $a^\prime x + b^\prime y + 1 = 0$, describing the same line (here $a^\prime = a/c$ and $b^\prime = b/c$). Now you have two unknowns and two equations! This can be solved!
