Calculus: Analytic Geometry and Calculus, with Vectors

1.3 Lines and linear equations^21

of Ll is the first of the equations

(1)

(2)

(3)

(x3 - x2)x + (YS - y2)y = (xs - X2)XI + (y3 - y2)Yl
(x1 - X3)X + (y1 - y3)y = (x1 - x3)x2 + (Y1 - Y3)Y2
(x2 - x1)x + (y2 - yl)Y = (x2 - x1)x3 + (y2 - yl)Ya.
It is possible to repeat the process to show that the equations of L2 and L3 are
(2) and (3). It is, however, more fun to observe that we can convert the deriva-
tion of the equation of Li into a derivation of the equation of L2 by making a
"cyclic advance" of the subscripts so that 1 goes to (or is replaced by) 2, 2 goes
to 3, and 3 goes to 1. The first cyclic advance converts (1) into (2), and another
cyclic advance converts (2) into (3). The routine way to finish the proof is to
solve (1) and (2) for x and y and show that these numbers (x,y) satisfy (3)
However, if we do not want to obtain and preserve the formulas for x and y, we
can finish the problem by observing that adding the members of (1), (2), and (3)
gives 0 = 0 and shows that the third equation is satisfied whenever the first two
are satisfied. For the record, we note that solving (1) and (2) for x and y gives
the formulas

(4) x =

(5) y =

y1(x3 - x2)xl + y2(xl - x3)x2 + y3(x2 - xi)xa

• (y2 - yl)(ys - y2)(yl - y3)
  yi(xa - x2) + y2(xl - X3) + y3(x2 - XI)
  xl(Y3 - y2)yl + x2(Y1 - ya)y2 + x3(y - yi)y3


• (x2 - xl)(x3 - x2)(xl - x3)
  x1(Y3 - y2) + x2(y1 - YS) + x3(y2 - y1)
  for the coordinates of the orthocenter. With the aid of the identity

(6) (y2 - yl)(ys - y2)(yl - YS) = yl(y3 - y2) + y2(Y1 - ya) + ya(y2 - Yi),

we can put these formulas in the forms

yl[xl(xa - X2) + Y2 - ya] + Y2[x2(xl - x3) + ya -' Yl] 7

(7) x = + y3[xa(x2 - x1) + Yl - y2]

yl(xa - X2) + y2(xi - x3) + y3(x2 - x11)

xi[yi(ya - y2) + x2 - xa] + x2[y2(yi - YS) + xa - x1]

+ xa[y3y2( -Yi) + xi- x2)

(8) Y =

x1(y3 - y2) + x2(yl - y3) + xs(y2 - yl)

and in many other forms which look quite different. Interchanging the x's and

y's in one of these formulas gives the other. Except for sign, the denominators

are equal to each other and (as we shall see later) to twice the area of the triangle.

As is to be expected, a cyclic advance of the subscripts does not alter the triangle

and does not alter the formulas for the coordinates of the orthocenter.

22 Again let the vertices of a triangle be P1(xi,yi), P2(x2,y2), Pa(xs,y3). Find

the coordinates of the point (x,y) of intersection of the three perpendicular bisec-

tors of the sides of the triangle, and write also the result of making a cyclic advance

(see the preceding problem) of the subscripts appearing in the answer. Remark:

Elementary plane geometry shows that the point (x,y) is equidistant from the

three vertices and hence is the center of the circle containing the vertices. The

circle is called the circumcircle (or circumscribed circle) of the triangle, and its