geometry
So, given a triangle with points A, B, C.
$A(1,3)$
ecuations of two medians are.
$x-2y+1=0$
and $y-1=0$
I found that all medians intersect at point $(1,1)$ , so i can find equation for third median that goes through $A$ point.
But how to find three equations of each side of the ABC triangle?
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.
Let $A'$ be the midpoint of $BC$; let $B'$ be the midpoint of $AC$. Let $M$ be the intersection point of the medians. We will find the lengths of the medians $AA'$ and $BB'$ from the fact that the medians are divided by the intersection point $M$ forming the ratio $2/1$, that is, $$ {AM\over A'M} = {BM\over B'M} = 2. $$ Suppose that $${1\over3}AA'=MA'=x, \qquad {1\over3}BB'=MB'=y.$$
By the Pythagorean theorem, from the right triangles $AMB'$ and $BMA'$ we have the system of equations $$ 4x^2 + y^2 = 9, $$ $$ x^2 + 4y^2 = 16. $$ Solving these equations we find $$ x = {2\over\sqrt3}, \qquad y = \sqrt{11\over3}. $$ Therefore, the medians are $$ AA' = 3x = {2\sqrt3}, \qquad BB' = 3y = \sqrt{33}. $$ We can find the third side, $AB$, from the right triangle $AMB$: $$ AB^2 = AM^2 +BM^2 = 4x^2 + 4y^2 = 20. $$ So we now know all tree sides of triangle $ABC$: $$ a = BC = 8, \qquad b=AC=6, \qquad c = AB = \sqrt{20}. $$ Finally, we can find the area of $ABC$ using an alternative form of Heron's formula: $$ \mbox{Area}(ABC)=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =4\sqrt{11}\approx13.266. $$
Best Answer
I'll give you an idea on how to construct the lines. Suppose we have one of those lines. Then we know it intersects one median at one of the vertices of the triangle, let's just say it will be $B$, and the other median at the midpoint between $A$ and $B$. By the way, let's call the intersection of medians $M$.
So after looking at it a bit you can see that if you draw a parallel to the second median at $A$ and intersect it with the second median, you'll get some point $P$. If you draw a line through $P$ orthogonal to the second median, you'll get the point $B$. Why? Because $AMBP$ is a rectangle (green in the picture), so the first median, going through $M$ and $P$, goes through the midpoint $M_1$ of $AB$.
You can proceed to find $C$ by mimicking the first approach, however now you won't get a rectangle but a parallelogram.