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Concurrent Triangle Altitudes

triconn.html

Exploration

Show that the three altitudes of any ΔABC are concurrent in a single point (x,y) whose coordinates are given by

where

and

and the coordinates of the vertices are , and . This point is called the orthocenter of the triangle.

Approach

From the exploration trialt.html the altitude from vertex A of ΔABC to side BC (or the extension of BC) is the line

where the coordinates of the vertices are , and . Use this formula to show that the three altitudes are concurrent and intersect one pair of the altitudes to find the coordinates of the orthocenter.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Define a function for constructing the altitude line of a triangle (the altitude from the first vertex).

Altitude2D[Triangle2D[{x1_,y1_},{x2_,y2_},{x3_,y3_}]]:=

Line2D[x3-x2,y3-y2,-x1*(x3-x2)-y1*(y3-y2)]

Define a function for constructing the orthocenter.

Orthocenter2D[Triangle2D[{x1_,y1_},{x2_,y2_},{x3_,y3_}]]:=

Module[{X1,Y1,D1},

Point2D[X1/D1,Y1/D1] //. {

X1->-(y1-y2)(x1*x2+y3^2)

+(y1-y3)(x1*x3+y2^2)

-(y2-y3)(x2*x3+y1^2),

Y1-> (x1-x2)(y1*y2+x3^2)

-(x1-x3)(y1*y3+x2^2)

+(x2-x3)(y2*y3+x1^2),

D1->Det[{{x1,y1,1},{x2,y2,1},{x3,y3,1}}]} ];

Create the three vertex points.

Clear[x1,y1,x2,y2,x3,y3];

A1=Point2D[x1,y1];

B1=Point2D[x2,y2];

C1=Point2D[x3,y3];

Construct the three altitudes.

lnA=Altitude2D[Triangle2D[A1,B1,C1]];

lnB=Altitude2D[Triangle2D[B1,C1,A1]];

lnC=Altitude2D[Triangle2D[C1,A1,B1]];

Show that they are concurrent by showing that the determinant of their coefficients is zero.

Det[{List @@ lnA,

List @@ lnB,

List @@ lnC}]

Find the coordinates of the orthocenter by intersecting a pair of altitudes.

p1=Point2D[lnA,lnB]

Compute the orthocenter using the formula.

The x- and y-coordinates of the intersection point and the orthocenter are identical.

Discussion

This is the plot of a numerical example.

Sketch2D[{Triangle2D[A1,B1,C1],lnA,lnB,lnC,p2} //.

{x1->-1,y1->-2,x2->4,y2->0,x3->-2,y3->3}]

Graphics saved as "tricon01.eps".

www.Descarta2D.com