how do you prove that the altitudes of a triangle are concurrent using coordinate geometry? prose proof

Question

accessdenied · Answer

You can define arbitrary points as the vertices of a triangle (I find it easiest to use a vertex at the origin, a vertex on the positive x-axis, and one in the first quadrant).

Using the fact that an altitude is the perpendicular line segment from the vertex to the line containing the opposite side, we can define equations that represent the altitudes of our arbitrary triangle (find the slope of each side, use the perpendicular slope and point-slope form with a vertex).  Then, we just solve the system of equations we get to find that all three lines do in fact have the same solution.

anonymous · Answer

Okay so say I have A at (0,0) B (b,0) and C (a,c). Just find the slopes of each altitude? How do you know the end points of the altitudes?

accessdenied · Answer

You do not need to know the endpoints of the altitudes because we need only the vertex to define their linear equations, and we only need to show that they are concurrent.  :)

Basically, you want to find the equations of the altitudes from A to BC, B to AC, and C to AB.  Once you know the slopes of the altitudes, we use point-slope form with the vertex and that slope.

anonymous · Answer

I still do not see it.

accessdenied · Answer

What part exactly are you unsure of?

anonymous · Answer

How do get the slopes, we have always needed the end point. For the points I mentioned earlier the the slope of AB would be b, BC (a-b)/c, and CA a/c. So I am unsure how you can find the slopes for the altitudes with out that other point.

accessdenied · Answer

Oh, okay.  The altitudes are perpendicular to the sides so we would use the slope of the side to indirectly get the altitudes' slopes.  You use the fact that the perpendicular slope is the opposite reciprocal of the slopes of the sides.

So, if the slope of a side is (a-b)/c, then the opposite reciprocal is -c/(a-b).
Although, I don't think b is one of the slopes of the sides.  You may want to recheck AB.

accessdenied · Answer

|dw:1335433562226:dw|
More of a visual representation, may be easier to see visually.

anonymous · Answer

I suppose that since 2 of the altitudes  always meet, one only needs to check the last one.

anonymous · Answer

You're right about the slopes I had it flipped. So once you get the slopes of the altitudes you just put it into an equation.

accessdenied · Answer

you'll have three equations, one for each altitude but yeah
then solve the system, which really isn't too hard since one line is just a vertical one so its just a simple substitution into the others to find that y=y for the x-value.

accessdenied · Answer

Sorry, I gotta run off to sleep now.  I will post all the work I did so you can see the full thing, although I used a few different variables that I am more accustomed to using (your variables are fine, you don't need to use mine -- just be aware of that).

accessdenied · Answer

Suppose we have the vertices: A(0,0), B(b,0), and C(c,d).

We need to find the perpendicular slopes to our sides.  So, let's first find the slope of each side:
  Slope of AB: $ \frac{0-0}{b-0} = 0 $
  Slope of BC: $ \frac{d-0}{c-b} = \frac{d}{c-b} $
  Slope of AC: $ \frac{d-0}{c-0} = \frac{d}{c} $

The perpendicular slopes are the opposite reciprocals:
  Perpendicular to AB: $ undefined $
  Perpendicular to BC: $ -\frac{c-b}{d} $
  Perpendicular to AC: $ -\frac{c}{d} $

Then, we use point slope form ($ y - y_1 = m(x - x_1) $) with the slope and the opposite vertex for the linear equations.
  Altitude from C to AB: $ x = c $  (Vertical line, use x-value of C)
  Altitude from A to BC: $ y - 0 = \left(-\frac{c-b}{d}\right)(x - 0) $
     $ y = \left(-\frac{c-b}{d}\right)x $
  Altitude from B to AC: $ y - 0 = \left(-\frac{c}{d}\right)(x - b) $
     $ y = \left(-\frac{c}{d}\right)(x-b) $

To solve, we may use substitution of the first equation into the second and the third to show that they are both equivalent when $ x=  c$.
$$ x=c;\ \begin{split}
 (-\frac{c-b}{d})(c) &= (-\frac{c}{d})(c-b)\
 -\frac{c(c-b)}{d} &= -\frac{c(c-b)}{d}~~~\text{True!}
\end{split} $$

accessdenied · Answer

$ \text{So, the altitudes of the triangle must be concurrent.}$
I hope I have helped!  :)
If you are uncertain of anything there, you'll have to bump the question and somebody else may assist you.  :)

anonymous · Answer

No that last one helped a ton, thank you. :)

accessdenied · Answer

You're welcome!  Then I can rest easy knowing I helped.  :P