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### Find the Orthocenter

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Date: 11/04/98 at 19:29:56
From: Phil
Subject: Given 3 points, find the orthocenter

Hi,

I was given this problem:

(-2,4) (7,2) (3,8)  Find the orthocenter

First I think you are supposed to take two points and find the slope
of one line beween two points, then invert to find the slope of a
perpendicular. You put the points and slope into a Y = mx+B form and
find the intercept, then put the line into an x+y = constant form.
Repeat for another vertex. Once you have 2 equations, cancel out the
X to find the Y, then cancel out the Y to find the X. That point (X,Y)
should be the intersection of the altitudes.

I have worked this problem for a long time and still keep coming up
with answers that don't match if I construct it on a grid.

```

```
Date: 11/05/98 at 15:26:15
From: Doctor Floor
Subject: Re: Given 3 points, find the orthocenter

Hi Phil,

I think your plan to solve the problem is good. The only thing is that
it seems fuzzy to me how you find the slope of the perpendicular and
how you convert to x+y = constant.

So let me try to find the orthocenter of A(-2,4), B(7,2) and C(3,8).

The slope of AB is m = (2-4)/(7--2) = - 2/9.

The slope m' of a line perpendicular to AB should be such that

m * m' = -1

So that gives m'= 9/2 = 4.5.  So the C-altitude is of the form:

y = 4.5x + n

Inserting C's y and x gives:

8 = 4.5*3 + n
8 = 13.5  + n

So the y-intercept is n = -5.5. That delivers the first formula, of
the C-altitude:

y = 4.5x - 5.5     [1]

Now consider AC. The slope is m = (8-4)/(3-(-2)) = 4/5. So the slope
for a line perpendicular to AC is m' = -5/4 = -1.25.

So the B-altitude is of the form y = -1.25x + n. Inserting B(7,2)
gives:

2 = -1.25*7 + n
2 = -8.75   + n

So the y-intercept is n = 10.75 and the formula for the B-altitude is

y = -1.25x + 10.75 [2]

Saying that y in equation [1] must equal y in equation [2] gives:

4.5x - 5.5 = -1.25x + 10.75
5.75x      = 16.25
x          = 65/23

Inserting this x in [2] gives

y = -1.25 * 65/23 + 10.75
= 166/23

So the orthocenter should be (65/23, 166/23).

I hope I didn't make any computation errors!

If you have a math question again, please send it to Dr. Math.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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