questions are clearer and more simple, the algebraical method proceeds with more uniformity, and reaches its end with greater certainty. It should be the student's aim to make himself master of both instruments of investigation, so as to be able to apply either, according as the nature of the subject demands. We shall give examples of some of the classes of problems which are of most frequent occurrence: the student who has mastered these will find no difficulty in applying the same method to any others that may present themselves. 46. Problems where it is required to prove that three lines meet in a point. It seems unnecessary to add any illustrations to those given in the last chapter, on this subject. The process we pursue is as follows: We form the equations of the three lines: it may then happen that we observe at once that the three equations vanish identically when added together (multiplied, it may be, by suitable constants): if this be the case, we know, by Art. 37, that the lines represented by the equations meet in a point. Otherwise, we find the co-ordinates of the intersection of two of them, and examine whether they satisfy the equation of the third; or else we apply to the equations the test of Art. 34. In the solution of this and every other class of geometrical problems, our equations may generally be much simplified by a judicious choice of axes of co-ordinates: since, by choosing for axes two of the most remarkable lines on the figure, several of our expressions will often be much shortened. On the other hand, it will sometimes happen that by choosing axes unconnected with the figure, the equations will gain in symmetry more than an equivalent for what they lose in simplicity. The reader may compare the two solutions of the same question, given Ex. 1 and 2, Art. 37, where, though the first solution is the longest, it has the advantage that the equation of one bisector being formed, those of the others can be written down without further calculation. Since expressions containing angles become more complicated by the use of oblique co-ordinates, it will be generally advisable to use rectangular axes in any question in which the consideration of angles is involved. Problems where it is required to prove that three points lie in one right line. It may happen that we observe that, the co-ordinates of two of them being x'y', a"y", those of the third are of the form in which case it is obvious, Art. 7, that the three points are in one right line. Otherwise we form the equation of the line joining two of them, and examine whether it is satisfied by the co-ordinates of the third. 47. Loci.-Analytic geometry adapts itself with peculiar readiness to the investigation of loci. We have only to find what relation the conditions of the question assign between the co-ordinates of the point whose locus we seek, and then the statement of this relation in algebraical language gives us at once the equation of the required locus. Ex. 1. Given base and difference of squares of sides of a triangle, to find the locus of vertex. Take the base for axis of x, and a perpendicu lar through one extremity A for axis of y. Call the length of base c, and let the co-ordinates of vertex be x, y. Then BC2 = CR2 + RB2 = y2 + (c − x)2. C P D E is the equation of the locus of vertex ; but this is (Art. 15) the equation of a line per Ex. 2. Given base and sum of sides of a triangle, if the perpendicular be produced beyond the vertex until its whole length is equal to one of the sides, to find the locus of the extremity of the perpendicular. Take the same axes, and let us inquire what relation exists between the co-ordinates of the point whose locus we are seeking. The x of this point plainly is AR, and the y is, by hypothesis, AC; and if m be the given sum of sides, = BC= my. Ex. 3. Given two fixed lines, OA and OB, if any line be drawn to intersect them parallel to a third fixed line, OC, to find the locus of the point where AB is cut in a given ratio. We may here employ oblique axes, since angles are not concerned (Art. 46). Let us take the fixed line OA for axis of x, and the fixed line OC for axis of y, then the equation of OB must be of the form y = mx, and it is required to find the locus of the point P cutting AB, so that AP may, for instance, = nAB. C Since the point B lies on the line whose equation is y = mx, we have A but AP is the y of the point P, and OA its x, therefore the locus of P is expressed by the equation y = mnx, and is, therefore, a right line through the point O. Ex. 4. 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 cosẞy sinẞ - P1 = 0, and their lengths, a, b, c, &c.; and let the given sum = m2; then, since (Art. 27) x cos ay sin a p denotes the perpendicular from the point ry 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 cosa + y sin a− p) + b (x cos ß + y sin ß − p1) + c(x cos y + y sin y − p2) + &c. which, since it contains x and y only in the first degree, will represent a right line. 2m3, Ex. 5. 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. The co-ordinates of B are found by simply accentuating the letters in the preceding: Now the condition that two points, 11, 22, shall lie on a right line passing through This being a relation then which must always be satisfied by the co-ordinates x'y', the equation of the locus is obtained by simply removing the accents from x'y'; and clearing of fractions, we have or (ac) [ab (x-c') + a'c'y] = (a' — c') [ab (x − c) + acy], (ac' - a'c) x cc' (aa) aa' (cc) the equation of a right line through the point L. Ex. 6. 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. Take for axis of a the line LP, and for axis of y the line LO. Let LP = a, LQ = a', LO = b, and let the equations of LM, LN be y = mx and y = m'x. The equation of CP through x'y' and (a, 0) is M B N amy' We must now express that the line joining these points passes through O. Substitute the co-ordinates x = 0, yb, in the equation of the line joining two points x,y1, x2y2 (Art. 29), and it becomes Substitute in this the values just obtained for x1, y1, X2, y2, and clear of fractions; the equation becomes divisible by y', and we have for the relation to be satisfied by the point x'y', a {(a'm' — b) y + m'b (x − a')} = a' {(am − b) y + mb (x − a)}, the equation of a right line. It passes (Art. 36) through the intersection of the lines found by equating each side of the equation separately to 0. It will be found that these are the lines joining P and Q to the points where a parallel to LQ through O meets LM, LN. 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, BR: =s, AR = s'. The equations of AC and BC are which the line FS meets AC and BC, by substituting in the equations of AC and BC this value, y=k. Thus we get from the first equation Having the abscissæ of F and S, we have (by Art. 7) the abscissa of the middle point of 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 2x = (8~5) 2y p |