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

When this equation is aß in the second degree. line passing through the equation may be written

cleared of fractions, it in general involves the coordinates But suppose that the points x'y', x"y" lie on the same origin y = mx, so that we have y' = mx', y′′ = mx", the

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 coordinates 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 coordinates 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

Thus we get from

= 1; .. x or RK = – - (1 − 1);

and from the second equation

k

s'

-

+ 2 = 1; .. x or RL = s ( 1 − 1).

p 8

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

point of FS, viz. z = (1-2). This is evidently the abscissa of the middle

x=

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 coordinates 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 coordinates 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, ub. Then the equations of the transversals will be

Subtract one from the other, divide by the constant 1

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.

Let OA = a, OB = b, OA′ = a +k, then, from the conditions of the problem, OB'b-k. The equations of AB', A'B are respectively

Subtracting, we eliminate k, and find for the equation of the locus

x + y = a + b.

Ex. 5. If on the base of a triangle we take any portion AT, and on the other side of the base another portion BS, in a fixed ratio to AT and draw ET and FS parallel to a fixed line CR; to find the locus of 0, the point of intersection of EB and FA. Take AB and CR for axes; let AT = k, BR = 8, AR s', CR = p, let the fixed ratio be m, then BS will=mk; the coordinates of S will be (s-mk, 0), and of T - (s' - k), 0}.

C

F

E

Now form the equations of the transverse lines, and the equation of EB is

To eliminate k, subtract one equation from the other, and the result, divided by k, will be

Ex. 6. PP' and QQ' are any two parallels to the sides of a parallelogram; to find the locus of the intersection of the lines PQ and P'Q'.

Let us take two of the sides for our axes, and let the lengths of the sides be a

sible to eliminate them from two equations. However, if we add the above equations, it will be found that both vanish together, and we get for our locus

bx - ay = 0,

the equation of the diagonal of the parallelogram.

Ex. 7. Given a point and two fixed lines; draw any two lines through the fixed point, and join transversely the points where they meet the fixed lines; to find the locus of intersection of the transverse lines.

Take the fixed lines for axes, and let the equations of the lines through the fixed point be

The conditions that these lines should pass through the fixed point x'y' give us

Now from this and the equation just found we can eliminate

Ex. 8. At any point of the base of a triangle is drawn a line of given length, parallel to a given one, and so as to be cut in a given ratio by the base; find the locus of the intersection of the lines joining its extremities to those of the base.

49. The fundamental idea of Analytic Geometry is that every geometrical condition to be fulfilled by a point leads to an equation which must be satisfied by its coordinates. It is important that the beginner should quickly make himself expert in applying this idea, so as to be able to express by an equation any given geometrical condition. We add, therefore, for his further exercise, some examples of loci which lead to equations of degrees higher than the first. The interpretation of such equations will be the subject of future chapters, but the method of arriving at the equations, which is all with which we are here concerned, is precisely the same as when the locus is a right line. In fact, until the problem has been solved, we do not know what will be the degree of the resulting equation. The examples that follow are purposely chosen so as to admit of treatment similar to that pursued in former examples, according to the order of which they are arranged. In each of the answers given it is supposed that the same axes are chosen, and that the letters have the same meaning as in the corresponding previous example.

Ex. 1. Find the locus of vertex of a triangle, given base and sum of squares of sides. Ans. x2+ y2 = {m2 — c2.

Ex. 2. Given base and m squares of one side + n squares of the other.
Ans. (m ±n) (x2 + y2) + 2 (m + n) cx + (m ± n) c2 = p2.

Ex. 3. Given base and ratio of sides.

Ex. 4. Given base and product of tangents of base angles. In this and the Examples next following, the learner will use the values of the tangents of the base angles given Ex. 2, Art. 46. Ans. y2 + m2x2 = m2c2.

Ex. 5. Given base and vertical angle or, in other words, base and sum of base angles.

Ex. 6. Given base and difference of base angles.

Ans. x2 + y2 - 2cy cot Co2. Ans. x2 - y2+ 2xy cot D = c2.

Ex. 7. Given base, and that one base angle is double the other.

Ans. 3x2
Ans. m (x2 + y2 — c2)

-

y2 + 2cx = c2. =2c (c-x).

Ex. 8. Given base, and tan C=m tan B. Ex. 9. PA is drawn parallel to OC, as in Ex. 4, p. 39, meeting two fixed lines in points B, B'; and PA2 is taken = PB. PB', find the locus of P.

Ans. mx (m'x + n' ) = y (mx + m'x + n').

Ex. 10. PA is taken the harmonic mean between AB and AB'.

Ans. 2mx (m'x + n') = y (mx + m'x + n').

Ex. 11. Given vertical angle of a triangle, find the locus of the point where the base is cut in a given ratio, if the area also is given.

Ex. 12. If the base is given.

2

32

Ans.

Ans. ry 2xy cos w

=

constant.

b2 (m + n)2 *

Ex. 14. Find the locus of P [Ex. 8, p. 40] if MN is constant.

Ex. 16. If MN pass through a fixed point, find the locus of the intersection of 1.

Ex. 17. Find the locus of P [Ex. 1, p. 41] if the line CD be not parallel to AB. Ex. 18. Given base CD of a triangle, find the locus of vertex, if the intercept AB on a given line is constant.

Ans. (x'y-y'x) (y - y′′) - (x′′y - y'x) (y-y') = c(y-y') (y-y').

50. Problems where it is required to prove that a moveable right line passes through a fixed point.

We have seen (Art. 40) that the line

Ax+ By +C+k (A'x + By + C') = 0;

or, what is the same thing,

(A + kA') x + (B+ kB') y + C + kC' = 0,