Chance to brag about my latex

So this is a writing problem that I had to do in one of my classes.

Ok. This looks hard to solve but if we break it step by step. We know that to find one of the lines, we need to connect one of the midpoints to the opposite vertex. Lets do one of them:

$$(4,8)$$

$$(2,-6)$$

To find the midpoint, we need to find the average of the $x$ and $y$.

$$x=\frac{4+2}{2}$$

$$\Longrightarrow x=3$$

$$y=\frac{8+-6}{2}$$

$$\Longrightarrow y=1$$

So the midpoint for one of the sides is $(3,1)$

Now, we need to find the other point opposite of the midpoint of one of the sides, which is simply just $(-4,4)$

Now, we need to find the slope of the line because we want to see if the three lines intersect at [b]one[/b] point.

Plugging it in for the formula $\frac{y_1-y_2}{x_1-x_2}$ will be $\frac{4-1}{-4-3}$

$$\frac{4-1}{-4-3}$$

$$\Longrightarrow \frac{3}{-7}$$

So the slope of the line is $-\frac{3}{7}$

Now lets find the equation.

$$y=mx+b$$

$$y=-\frac{3}{7}\cdot x+b$$

$$1=-\frac{3}{7}\cdot 3+b$$

$$1=-\frac{9}{7}+b$$

$$\frac{16}{7}=b$$

Plug that in for the original equation.

$$y=-\frac{3}{7}x+\frac{16}{7}$$

Moving on to the second point. Lets do the same thing we did for the first line.

Let's find the midpoint of $(4,8)$ and $(-4,4)$ by taking the average of the $x$'s and the $y$'s.

$$x=\frac{4+-4}{2}$$

$$\Longrightarrow x=0$$

$$y=\frac{8+4}{2}$$

$$\Longrightarrow y=6$$

So the midpoint is $(0,6)$

Now, we need to find the opposite point(or the leftover point we didn't use. Thats what I call it :D) which is just $(2,-6)$

Now, like always, we need to find the slope.

$$\frac{y_1-y_2}{x_1-x_2}$$

$$\Longrightarrow \frac{-6-6}{2-0}$$

$$\Longrightarrow -6$$

So negative six is the slope.

Now lets find the equation.

$$y=mx+b$$

$$y=-6x+b$$

$$6=-6\cdot 0+b$$

$$6=b$$

Plugging $B$ and $M$ in for the original equation:

$$y=mx+b$$

$$y=-6x+6$$

Moving on to the last but not least line. We first find the midpoint of two points like always. In this case, those two points are $(-4,4)$ and $(2,-6)$

To find the midpoint, we have to get the average of the $x$ and $y$ coordinates. :|

So lets do it!

$$x=\frac{-4+2}{2}$$

$$\Longrightarrow x=-1$$

$$y=\frac{4+-6}{2}$$

$$\Longrightarrow y=-1$$

The midpoint is $(-1,-1)$

Let's use the third point $(4,8)$ that we didn't use to find the slope of the line;

$$\frac{y_1-y_2}{x_1-x_2}$$

$$\frac{8--1}{4--1}$$

$$\frac{9}{5}$$

so the slope of the last line is $\frac{9}{5}$

Now lets make an equation for this one:

$$y=mx+b$$

$$y=\frac{9}{5}x+b$$

$$y=\frac{9}{5}x+b$$

$$-1=\frac{9}{5}\cdot -1+b$$

$$-1=-\frac{9}{5}+b$$

$$\frac{4}{5}=b$$

Plugging it in for the original equation will give us:

$$y=\frac{9}{5}x+\frac{4}{5}$$

So now lets gather all of our equations that we found:

$$y=-\frac{3}{7}x+\frac{16}{7}$$

$$y=-6x+6$$

$$y=\frac{9}{5}x+\frac{4}{5}$$

Those equations look like a mess! Lets name variables to make things neater.

$$a=y=-\frac{3}{7}x+\frac{16}{7}$$

$$b=y=-6x+6$$

$$c=y=\frac{9}{5}x+\frac{4}{5}$$

To prove that three lines intersect at one point, we need to prove three things. $(x,y)$ are the same for $a$ and $b$, $b$ and $c$, and $a$ and $c$.

Lets do this one by one.

[size=200]________________________________A=B__________________________________[/size]

$$-y=\frac{3}{7}x-\frac{16}{7}$$

$$y=-6x+6$$

(adding the two equations)

$$0=-\frac{39}{7}x+\frac{26}{7}$$

$$-\frac{26}{7}=-\frac{39}{7}x$$

$$\rightarrow \cdot -7$$

$$26=39x$$

$$\rightarrow \div 39$$

$$\frac{26}{39}=x$$

$$\frac{2}{3}=x$$

Now, plugging in $x$ for the equation $y=-6x+6$ will give us:

$$y=-6\cdot \frac{26}{39}+6$$

$$y=-\frac{52}{13}+6$$

$$y=\frac{26}{13}$$

$$y=2$$

So using $A=B$ the ordered point is $(\frac{2}{3},2)$

[size=200]________________________________B=C__________________________________[/size]

Now its time to prove that $B=C$

Lets apply the same system of equations methods we used for the previous one.

$$-y=6x-6$$

$$y=\frac{9}{5}x+\frac{4}{5}$$

(add the two equations)

$$0=\frac{39}{5}x-\frac{26}{5}$$

$$rightarrow +\frac{26}{5}$$

$$\frac{26}{5}=\frac{39}{5}x$$

$$\rightarrow \cdot 5$$

$$26=39x$$

$$\rightarrow \div 39$$

$$\frac{26}{39}=x$$

$$\frac{2}{3}=x$$

Now, we plug it in into the equation $y=-6x+6$ just like we did when we tried to prove that $A=B$

$$y=-6\cdot \frac{26}{39}+6$$

$$y=-\frac{52}{13}+6$$

$$y=\frac{26}{13}$$

$$y=2$$

So the ordered pair for $B=C$ is also $(\frac{2}{3},2)$

Yippee! it works! But we're not done. The ordered pair might be different.

[size=200]________________________________A=C__________________________________[/size]

Last, but definitely not least is $A=C$

$$-y=\frac{3}{7}x-\frac{16}{7}$$

$$y=\frac{9}{5}x+\frac{4}{5}$$

(adding the two equations)

$$0=\frac{78}{35}x-\frac{52}{35}$$

$$\frac{52}{35}=\frac{78}{35}x$$

$$\rightarrow \cdot 35$$

$$52=78x$$

$$\rightarrow \div 78$$

$$\frac{52}{78}=x$$

$$\frac{2}{3}=x$$

Time to plug $x$ in for the easiest equation to deal with.

$$y=\frac{9}{5}x+\frac{4}{5}$$

$$y=\frac{9}{5}\cdot \frac{2}{3}+\frac{4}{5}$$

$$y=\frac{6}{5}+\frac{4}{5}$$

$$y=\frac{10}{5}$$

$$y=2$$

Now lets compare answers.

$$(\frac{2}{3},2)$$

$$(\frac{2}{3},2)$$

$$(\frac{2}{3},2)$$

hmmmmmmmm. Do you notice how THEY ARE ALL THE SAME!

That means their is a median to the triangle. And the median is $\boxed{(\frac{2}{3},2)}$

(PS This took me a [b]long[/b](about two and a half hours in total) time to write all of this out. Especially the $\LaTeX$. Can you rate how well I used my $\LaTeX$ 1-10? Thanks!)