# Building a Shower Equidistant to Campsites Geometry Problems

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Henry Prosser
Unit 5 Project
5/27/2020

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To build a shower equidistant to the campsites, the shower would have to be placed in the incenter of
the triangle. To do this, first you must find the perpendicular bisectors of lines BP, BH, and PH. Line BP is
(3,6), line PH is (8,6) and line BH is (7,2). The point where they meet is in the incenter. This is (6, 4.6) and
is where the shower should be placed.
A. The orthocenter of ∆ ABC is (5,1). To find the orthocenter I had to find the equation of the line
containing the altitude of AC. X= 2 because it is a horizontal line. The line with the altitude of BC
has a y value of -3. (2 5)/(4 5)
The orthocenter of ∆ ABO is the same as point a (2,2). The orthocenter of ∆ ACO is point B (5,5)
and the orthocenter of ∆ BCO is point c (2,2). What I found is that the orthocenter becomes the
vertex of the other triangles.
B. No it will not be the same for acute and right triangles. To find an orthocenter you need three
altitudes. An orthocenter of a right triangle is the vertex of the right angle. This means that the
point O will be the same as point B. ( I used triangle ABC with A (2,2), B (2,1) and C (3,1).) You
cannot find the orthocenter since the points are the same.
For an acute triangle, it also does not work. The orthocenter of an acute triangle is inside the
triangle. I used coordinates A (2,3) B (0,4) and C (1,2) with an orthocenter of (1.3, 2.6). The
orthocenter became a coordinate outside of the triangle when I used O as one of the triangle’s
vertexes instead of another existing point.

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Henry Prosser Unit 5 Project 5/27/2020 Task 1: Please reference above graph. To build a shower equidistant to the campsites, the shower would have to be placed in the incenter of the triangle. To do this, first you must find the perpendicular bisectors of lines BP, BH, and PH. Line BP is (3,6), line PH is (8,6) and line BH is (7,2). The point where they meet is in the incenter. This is (6, 4.6) and is where the shower should be placed. Task 2: A. The orthocenter of ∆ ABC is (5,1). To find the orthocenter I had to find the equation of the line containing the altitude of AC. X= 2 because it is a horizontal line. The line with the altitude of BC has a y value of -3. (2 − 5)/(4 − 5) The orthocenter of ∆ ABO is the same as point a (2,2). The orthocenter of ∆ ACO is point B (5,5) and the orthocenter of ∆ BCO is point c (2,2). What I found is that the orthocenter becomes the ve ...
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