Calculus: Analytic Geometry and Calculus, with Vectors

32 Analytic geometry in two dimensions

29 From the first two of the equations

(1) x2+y2+ajx+bjy+c1 =0

(2) x2+y2+a2x+b2y+c2=0

(3) (a2 - a1)x + (b2 - bi)y + (c2 - c1) = 0

we can obtain the third by equating the left members of (1) and (2) and simpli-
fying the result. Supposing that the graphs of (1) and (2) are nonconcentric
circles, show that the graph of (3) is a line perpendicular to the line containing
the centers of the circles. Show also that if these circles intersect in one or two
points, then the line contains the point or points of intersection. Remark: The
line is called the radical axis of the circles, it being named because some people
want to talk about it.
30 The points PI(xl,yl), P2(x2,ys), and Ps(xa,ya) are, in positive or counter-
clockwise order, the vertices of an equilateral triangle. Find formulas which
express xs and ya in terms of the coordinates of PI and P2. Solution: While the
problem can be attacked in other ways, we eliminate difficulties involving order
relations by observing that if the half-line extending from PI through P2 makes
the angle 0 with the positive x axis, then the half-line extending from PI through
Ps makes the angle 9 + it/3 with the positive x axis. Let a be the lengths of
the sides of the equilateral triangle. The definitions of the trigonometric func-
tions then give

(1)

(2)

x2-xl=acos0, y2-yl=asin0

xa-x1 =acos(B+3)=Zacos8-23asin6

(3) ys - yI =asin (0+3)=^2 acos0+asin0.

In obtaining the latter formulas, we use the "addition formulas"

(4)

(5)

cos (0 + 95) = cos 0 cos 0 - sin 0 sin 4,

sin (0 + 0) = sin 0 cos ¢ + cos 0 sin 4)

and the values of the sine and cosine of r/3 which can be determined with the

aid of Figures 1.493. From (2), (3), and (1), we obtain the

answers

1

\\ (6) xs = xI +

2

(x2 - xl) - 2 (y2 - yI)

(7) y3 =y1+ 2 (x2-XI)+(y2-y1).

Figure 1.493

Remark: The points in the xy plane for which both coordi-

nates are integers are called lattice points. Our results

enable us to show very easily that triangles having vertices

at lattice points cannot be equilateral. To prove this, let an

equilateral triangle have two of its vertices, say PI and P2, at lattice points.

Then x1, x2, y1, y2 are integers and, since Nr3 is irrational, (6) shows thatxs can-

not be an integer unless y2 = y1. If y2 = yI, we must have x2 0 x1, and then