Ex. 5. PA drawn parallel to OC, as before, meets any number of fixed lines in points B, B', B", &c., and PA is taken proportional to the sum of all the ordinates BA, B'A, &c., find the locus of P.

Ans. If the equations of the lines be

y = mx, y = mx + n', y = mx + n", &c.,

the equation of the locus is

ky:
' = mx + (m'x + n') + (m"x + n') + &c.

Ex. 6. Given bases and sum of areas of any number of triangles having a common vertex, to find its locus.

Let the equations of the bases be

x cos a + y sin a -p

=

0, x cosẞ+ y sin ß-p1 = 0, &c.,

and their lengths, a, b, c, &c.; and let the given sum = m2; then, since (Art. 34) x cos a + y sin a -p denotes the perpendicular from the point xy on the first line, a (x cos a + y sin a − p) will be double the area of the first triangle, &c., and the equation of the locus will be

a (x cos a + y sin a− p) + b (x cosẞ+y sinẞ-p1)+c (x cos y + y sin y − P2) +&c. =2m2, which, since it contains x and y only in the first degree, will represent a right line.

Ex. 7. Given vertical angle and sum of sides of a triangle, find the locus of the point where the base is cut in a given ratio.

The sides of the triangle are taken for axes; and the ratio PK: PL is given =n: m. Then by similar triangles,

L

N

P

Ex. 8. Find the locus of P, if when perpendiculars PM, PN are let fall on two fixed lines, OM + ON is given.

The co-ordinates of the middle point expressed in terms of the co-ordinates of P are 2 (x + y cos w), ≥ (y + x cos ∞); and since these satisfy the equation of the given line, the co-ordinates of P satisfy the equation

y + x cosw = m (x + y cos w) + 2n.

Ex. 11. P moves along a given line y = mx + n, find the locus of the middle point of MN. If the co-ordinates of P be a, ẞ, and those of the middle point x, y, it has just been proved that 2x = a + ẞ cosw, 2y = B+ a cosw. Whence solving for a, ß, a sin2 w = 2x-2y cosw, ẞ sin2 w = 2y- 2x cos w.

But a, ẞ are connected by the relation ẞ = ma + n, hence

2- 2x cos x = m (2x - 2y cos w) + n sin2w.

47. It is customary to denote by x and y the co-ordinates of a variable point which describes a locus, and the co-ordinates of fixed points by accented letters. Accordingly in the preceding examples we have from the first denoted by x and y the coordinates of the point whose locus we seek. But frequently in finding a locus it is necessary to form the equations of lines connected with the figure; and there is danger of confusion between the x and y, which are the running co-ordinates of a point on one of these lines, and the x and y of the point whose locus we seek. In such cases it is convenient at first to denote the co-ordinates of the latter point by other letters such as a, ß, until we have succeeded in obtaining a relation connecting these co-ordinates. Having thus found the equation of the locus, we may if we please replace a, ẞ by x and y, so as to write the equation in the ordinary form in which the letters x and y are used to denote the co-ordinates of the point which describes the locus.

P

Ex. 1. Find the locus of the vertex of a triangle, given the base CD, and the ratio AM: NB of the parts into which the sides divide a fixed line AB parallel to the base. Take AB and a perpendicular to it through A for axes, and it is necessary to express AM, NB in terms of the co-ordinates of P. Let these co-ordinates be aß, and let the co-ordinates of C, D be x'y', x'y', the y' of both being the same since CD is parallel to AB. Then the equation of PC joining the points aß, x'y' is (Art. 29)

A M

C

D

B

This equation being satisfied by the x and y of every point on the line PC, is satisfied by the point M, whose y = 0 and whose x =

AM. Making then y = 0 in this equa

We have now expressed the conditions of the problem in terms of the co-ordinates of the point P; and now that there is no further danger of confusion, we may replace a, ẞ by x,y; when the equation of the locus, cleared of fractions, becomes

yx' - xy' = k {c (y — y') — (yx” — xy')}.

Ex. 2. Two vertices of a triangle ABC move on fixed right lines LM, LN, and the three sides pass through three fixed points O, P, Q which lie on a right line; find the locus of the third vertex.

G

Take for axis of x the right line OP, containing the three fixed points, and for

axis of y the line OL joining the inter-
section of the two fixed lines to the point
O through which the base passes. Let the
co-ordinates of C be a, ß, and let

OL=b, OM=a, ON=a', OP=c, OQ=c'.
Then obviously the equations of LM, LN

A

B

M

PQN

The co-ordinates of B are found by simply accentuating the letters in the preceding: a'b (a c') + a'c'ß b (a− c') + a'ß

b (a'-c') B

b (a — c') + a'ß'

two points x11, x¿1⁄2 shall lie on a right line passing through

X2 =

Now the condition that

the origin, is (Art. 30)

We have now derived from the conditions of the problem a relation which must be satisfied by aẞ the co-ordinates of C: and if we replace a, ẞ by x, y we have the equation of the locus written in its ordinary form. Clearing of fractions, we have (a — c) [a'b (x − c′) + a'c'y] = (a′ — c′) [ab (x − c) + acy],

or

(ac' - a'c) x

cc′ (a — a') — aa' (c — c′)

the equation of a right line through the point L.

+ = 1,

Ex. 3. If in the last example the points P, Q lie on a right line passing not through O but through L, find the locus of vertex.

We shall first solve the general problem in which the points P, Q have any position. We take the fixed lines LM, LN for axes. Let the co-ordinates of P, Q, O, C be respectively x'y', x"y", x""y"", aß; and the condition which we want to express is that if we join CP, CQ and then join the points A, B, in which these lines meet the axes, the line AB shall pass through 0. The equation of CP is (ẞ — y') x − (a — x'′) y = ẞx' — ay'.

In like manner the intercept which CQ makes on the axis of y is

And the condition of the problem is that this equation shall be satisfied by the co-ordinates ""y"". In order then that the point C may fulfil the conditions of the problem, its co-ordinates aß must be connected by the relation

When this equation is cleared of fractions, it in general involves the co-ordinates aß in the second degree. But suppose that the points x'y', x"y" lie on the same line passing through the origin y = mx, so that we have y' = mx', y'' = mx", the

equation may be written

x" (B-y'), y'" (a — x")

+

x' (B- am) x" (am - B)

1.

Clearing of fractions and replacing a, ẞ by x and y, the locus is a right line, viz., x'""'x'' (y — y′) — y'"'x' (x − x'') = x'x′′ (mx − y).

48. It is often convenient, instead of expressing the conditions of the problem directly in terms of the co-ordinates of the point whose locus we are seeking, to express them in the first instance in terms of some other lines of the figure; we must then obtain as many relations as are necessary in order to eliminate the indeterminate quantities thus introduced, so as to have remaining a relation between the co-ordinates of the point whose locus is sought. The following Examples will sufficiently illustrate this method.

Ex. 1. To find the locus of the middle points of rectangles inscribed in a given triangle.

Let us take for axes CR and AB; let CR = p, RB =s, AR = s'. The equations of AC and BC are

C

Having the abscissæ of F and S, we have (by Art. 7) the abscissa of the middle

point of FS, viz., x = (1-4). This is evidently the abscissa of the middle

82

point of the rectangle. But its ordinate is y = k. Now we want to find a relation which will subsist between this ordinate and abscissa whatever k be. We have only then to eliminate k between these equations, by substituting in the first the value of k (= 2y), derived from the second, when we have

This is the equation of the locus which we seek. It obviously represents a right line, and if we examine the intercepts which it cuts off on the axes, we shall find it to be the line joining the middle point of the perpendicular CR to the middle point of the base.

Ex. 2. A line is drawn parallel to the base of a triangle, and the points where it meets the sides joined to any two fixed points on the base; to find the locus of the point of intersection of the joining lines.

We shall preserve the same axes, &c., as in Ex. 1, and let the co-ordinates of the fixed points, T and V, on the base, be for T (m, 0), and for V (n, 0).

Now since the point whose locus we are seeking lies on both the lines FT, SV, each of the equations just written expresses a relation which must be satisfied by its coordinates. Still, since these equations involve k, they express relations which are only true for that particular point of the locus which corresponds to the case where the parallel FS is drawn at a height k above the base. If, however, between the equations, we eliminate the indeterminate k, we shall obtain a relation involving only the co-ordinates and known quantities, and which, since it must be satisfied whatever be the position of the parallel FS, will be the required equation of the locus. In order, then, to eliminate k between the equations, put them into the form

But this is the equation of a right line, since x and y are only in the first degree.

Ex. 3. A line is drawn parallel to the base of a triangle, and its extremities joined transversely to those of the base; to find the locus of the point of intersection of the joining lines.

This is a particular case of the foregoing, but admits of a simple solution by choosing for axes the sides of the triangle AC and CB. Let the lengths of those lines be a, b, and let the lengths of the proportional intercepts made by the parallel be μa, μb. Then the equations of the transversals will be

Subtract one from the other; divide by the constant 1

equation of the locus

which we have elsewhere found (see p. 34) to be the equation of the bisector of the base of the triangle.

Ex. 4. Given two fixed points A and B, one on each of the axes; if A' and B' be taken on the axes so that OA' + OB′ = OA + OB; find the locus of the intersection of AB', A'B.