This has been bothering me for a while.
Given a vertex $A(2;-4)$ and the line equations of two medians ($2x-3y-2=0$ and $5x+3y-12=0$), find the line equations on which the triangle sides are.
I've tried to approach this several ways, e.g. finding the reflections of point A about the medians, but the answers always seem to be off the mark.
I'd appreciate some guidance!
edit:
So I don't seem to be able to comment on other posts (or this post for that matter) so I'll just reply to hhsaffar's answer here:
So I did find the intersection point of the medians previously $\left(2, \frac{2}{3}\right)$ and the equation for the third median ($x=2$). I also have the coordinate (referring to hhsaffar's triangle here) $F(2, 3)$. However I have trouble figuring out the next step.
Best Answer
Refer to the figure below.
Let A be (2, -4) and B be (p, q).
L(1) : 2x-3y-2=0
L(2): 5x+3y-12=0
Testing showed that A is not on either L(1) or L(2).
Note: the next 6 lines can be skipped.
Solving the above, we have G, the centroid is at (2. 2/3).
This means A and G are on the same vertical line x = 2.
AG = (2/3) + 4 = 14/3
Let A(h) and C(h) be the feet of the medians through A onto BC and C onto AB respectively.
By the properties of centroid, GA(h) = 0.5*AG = 7/3
A(h) is then at (2, 3).
The midpoint of AB = C(h) = ([p + 2]/2, [q – 4]/2).
It lies on L(2). Therefore, 5[[p + 2]/2] + 3[[q – 4]/2] -12 = 0………(1)
B(p, q) lies on L(1). Therefore, 2(p) – 3(q) – 2 = 0……………………..(2)
Solving (1), and (2), we have B = (4,2).
The co-ordinates of C can be similarly obtained.
The required equations can be obtained by applying two-point form.