Let A, B, and C be the points in the Cartesian plane that are given by
| A = (0, -1) , B = (3, 4) , and C = (-1, 1) . |
The point where the perpendicular bisectors of Delta ABC meet is the point uA + vB + wC = (43/22, 27/22) where
| u = {175}/{242} , v = {135}/{242} , and w = - {34}/{121} . |
This coefficient vector is the unique vector (u, v, w) with u + v + w = 1 that is parallel to the vector (sinalpha cosalpha, sinbeta cosbeta, singamma cosgamma) where alpha, beta, and gamma are the vertex angles in Delta ABC. The common distance of the intersection point from the three vertices is {5SQRT{170}}/{22}.
The point where the angle bisectors of Delta ABC meet is the point uA + vB + wC where
| (u, v, w) = |
|
with a, b, and c the lengths of the sides of Delta ABC, which, respectively, are 5, SQRT{5}, and SQRT{34}. Therefore,
| uA + vB + wC = |
| . |
Note that by the law of sines this coefficient vector, which is parallel to the vector of lengths of sides is also parallel to the vector (sinalpha, sinbeta, singamma). The common distance of the intersection point from the three sides is {11}/{5 + SQRT{5} + SQRT{34}}.
The first two of the following exercises will be important for later work.
Prove the fulcrum principle: If A and B are different points, l the length of the segment AB, and P = uA + vB with u + v = 1, then the length of the segment AP is |v| l and the length of BP is |u| l.
Prove the principle of preservation of proportionality in barycentric coordinates: If A, B, and C are three non-collinear points and P = uA + vB + wC with u + v + w = 1, then the line AP meets the line BC at the point (1/(v+w))(vB + wC) provided that v + w <> 0 or, equivalently, P <> A and the line AP is not parallel to the line BC.
Let A, B, and C be three non-collinear points, D a point of the line segment AC, and E a point of the line segment BC. Use barycentric coordinates relative to A, B, and C to show that the line segments AE and BD meet.