Calculus: Analytic Geometry and Calculus, with Vectors

22 Analytic geometry in two dimensions

center is called the circumcenter of thetriangle. The answer can be put in the

forms

yl(xa - xz)(xa + X2) + y2(x1 - xa)(x1 + Si) +Y3(x2 - x1)(x2 + x1)

+ (y2 - yi)(ya - y2)(Y1 - y3)

2[yl(x3 - x2) + y2(x1 - XS) + y3(x2 - x1)]

x1(y3 - y2)(Y3 + Y2) + x2(y1 - Ya)(y1 + ya) + x8(y2 - y1)(Y2 + Y1)

+ (X2 - XI) (X3 - x2) (x l- Si)

2[x1(ya - y2) + x2(y1 - ya) + x3(y2 - y1)]

(1) x =

(2) Y=

and

y1(xa + y2 - x2 - Y2) + y2(xi + Y2 -- 2 2 x2- Ya) +Ya(x2 + Y2 - xi - yi)

(3) x 2[yl(xa - x2) + Y2(x1 - Si) + ya(x2 - xl)]

( =xl(xa + y3 - x2 - y2) + x2(x1 + Y1 - x3 - Y3) +^222 xa(x2 + Y2 -XI- Yi)

4) y

2[x1(y3 - y2) + x2(yl - ya) + x3(y2 - y1)]

and in still other forms which look quite different.

Figure 1.391

23 The triangle in Figure 1.391 has vertices at Pi(xi,yi), P2(x2,y2), and

Pa(x,,y8). The mid-points M1, M2, Ma of the sides of this triangle are the ver-

tices of the mid-triangle of the given triangle. With or without making use of

the ideas and results of the preceding problem, find the coordinates of the cir-

cumcenter of this mid-triangle. Remark: The answer can be put in the form

yl(xa - x2)(2x1 + x2 + XS) + y2(x1 - x3)(2x2 + xa + Si)

(1) x = + ya(xs - x1)(2xa + Si + Si) - (y2 - y1)(ya - y2)(y1 - ya)

4[Y1(xa - x2) + Y2(x1 - Si) + ya(x2 - Si)]

x1(ys - y2)(2y1 + y2 + ya) + x2(yi - ya)(2y2 + ya + yi)

(2) = + x3(Y2 - y1)(2ys + y1 + y2) - (x2 - x1)(x8 - x2)(x1 - x

y- 4[xi(ya - ys) + x2(y1 - ya) + x3(y2 - y1)]