we may write the preceding equation Σ (mx') cos a + Σ (my') sin a − pΣ (m) = = 0. Substituting in the original equation the value of p hence obtained, we get for the equation of the moveable line or xΣ (m) cos a + yΣ (m) sin a − Σ (mx') cos a Σ (my) sin a = 0, Now as this equation involves the indeterminate tan a in the first degree, the line passes through the fixed point determined by the equations This point has sometimes been called the centre of mean position of the given points. 51. If the equation of any line involve the coordinates of a certain point a'y' in the first degree, thus, (Ax' + By + C) x + (A'x' + B'y' + C') y + (A′′x' + B'y' + C") = 0 ; then if the point x'y' move along a right line, the line whose equation has just been written will always pass through a fixed point. For, suppose the point always to lie on the line Lx' + My' + N=0, then if, by the help of this relation, we eliminate x' from the given equation, the indeterminate y' will remain in it of the first degree, therefore the line will pass through a fixed point. Or, again, if the coefficients in the equation Ax+ By + C=0 be connected by the relation aA+bB+cC=0 (where a, b, c are constant and A, B, C may vary), the line represented by this equation will always pass through a fixed point. For by the help of the given relation we can eliminate C, and write the equation a right line passing through the point (a= 52. Polar Coordinates.—It is, in general, convenient to use this method, if the question be to find the locus of the extremities of lines drawn through a fixed point according to any given law. Ex. 1. A and B are two fixed points; draw through B any line, and let fall on a perpendicular from A, AP; produce AP so that the rectangle AP.AQ may be constant; to find the locus of the point Q. Let us Take A for the pole, and AB for the fixed axis, then AQ is our radius vector, designated by p, and the angle QAB = 0, and our object is to find the relation existing between p and 0. call the constant length AB = c, and from the right-angled triangle APB we have AP-c cos 0, but AP.AQ= const,= k2: therefore but we have seen (Art. 44) that this is the equation of a right k2 line perpendicular to AB, and at a distance from A = Ex. 2. Given the angles of a triangle; one vertex A is fixed, another B moves along a fixed right line: to find the locus of the third. Take the fixed vertex A for pole, and AP perpendicular to the fixed line for axis, then AC p, CAP=0. Now since the angles of ABC are given, AB is in a fixed ratio to AC(=mAC) and BAP=0-a ; but AP = AB cos BAP; therefore, if we call AP, a, we have C B B mp cos (0 - a) = a, A which (Art. 44) is the equation of a right line, making an angle a with the given line, and at a distance from a Ex. 3. Given base and sum of sides of a triangle, if at either extremity of the base B a perpendicular be erected to the conterminous side BC; to find the locus of P the point where it meets CP the external bisector of vertical angle. Let us take the point B for our pole, then BP will be our radius vector p; and let us take the base produced for our fixed axis, then PBD = 0, and our object is to express p in terms of 0. Let us designate the sides and opposite angles of the triangle a, b, c, A, B, C, then it is easy to see that the angle BCP = 90° – C, and from the triangle PCB that a=p tan C. Hence it is evident that if we could express a and tan express p in terms of 0. A C in terms of 0, we cou b2 = a2 + c2 2ac cos B, P B D but if the given sum of sides be m, we may substitute for b, m — a; and cos B plainly sin ; hence = Thus we have expressed a in terms of and constants, and it only remains to find an expression for tan C. Now but hence b sin C = c sin B = c cos 0, and b cos Cac cos B = a - c sin 0; We are now able to express p in terms of 0, for, snbstitute in the equation a=p tan C, the values we have found for a and tan C, and we get pc cos 0 Hence the locus is a line perpendicular to the base of the triangle at a distance m2 - c2 from B = The student may exercise himself with the corresponding locus, if CP had been the internal bisector, and if the difference of sides had been given. Ex. 4. Given n fixed right lines and a fixed point 0; if through this point any radius vector be drawn meeting the right lines in the points 71, 72, 73...rn, and on this a point R be taken such that locus of R. n 1 1 Let the equations of the right lines be p cos (0 - a) = P1; p cos (0 - ẞ) = P2, &c. Then it is easy to see that the equation of the locus is the equation of a right line (Art. 44). This theorem is only a particular case of a general one, which we shall prove afterwards. We add, as in Art. 49, a few examples leading to equations of higher degree. Ex. 5. BP is a fixed line whose equation is p cos 0 = m, and on each radius vector is taken a constant length PQ; to find the locus of Q [see fig., Ex. 1]. Ex. 6. Find the locus of Q, if P describe any locus whose polar equation is given, p = (0). We are by hypothesis given AP in terms of 0, but AP is the P of the d; we have therefore only to substitute in the given equation p locus d for p. Ex. 7. If AQ be produced so that AQ may be double AP, then AP is half the p of the locus, and we must substitute half p for p in the given equation. Ex. 8. If the angle PAB were bisected, and on the bisector a portion AP' be taken so that AP'2 = mAP, find the locus of P' when P describes the right line o cos 0=m. PAB is now twice the 0 of the locus, and therefore AP = the equation of the locus is p2 cos 20 = m2. m cos 20' and *CHAPTER IV. APPLICATION OF ABRIDGED NOTATION TO THE EQUATION OF THE RIGHT LINE. 53. We have seen (Art. 40) that the line (x cos a + y sin a-p) - k (x cos B+ y sin B-p')=0 denotes a line passing through the intersection of the lines x cos ay sina-p=0, x cos ẞ+ y sin B-p' = 0. We shall often find it convenient to use abbreviations for these quantities. Let us call x cos a+ y sin a − p, a; x cos ß + y sin ß —p', B. Then the theorem just stated may be more briefly expressed; the equation a— kß = 0 denotes a line passing through the intersection of the two lines denoted by a=0, B=0. We shall for brevity call these the lines a, B, and their point of intersection the point aß. We shall, too, have occasion often to use abbreviations for the equations of lines in the form Ax + By + C = 0. We shall in these cases make use of Roman letters, reserving the letters of the Greek alphabet to intimate that the equation is in the form x cos ay sina-p=0. A P B 54. We proceed to examine the meaning of the coefficient k in the equation a-kß=0. We saw (Art. 34) that the quantity a (that is, x cos a + y sin a − p) denotes the length of the perpendicular PA let fall from any point xy on the line OA (which we suppose represented by a). Similarly, that ẞ is the O length of the perpendicular PB from the point zy on the line OB, represented by B. Hence the equation a-kß=0 asserts that if, from any point of the locus represented by it, perpendiculars be let fall on the lines OA, OB, the ratio of these perpendiculars (that is, PA : PB) will be constant and = k. Hence the locus represented by a - kß=0 is a right line through O, and It follows from the conventions concerning signs (Art. 34) that a + kB = 0 denotes a right line dividing externally the angle AOB into parts such that sin POA = k. It is, of course, assumed in what we have said that the perpendiculars PA, PB are those which we agree to consider positive; those on the opposite sides of a, ß being regarded as negative. Ex. 1. To express in this notation the proof that the three bisectors of the angles of a triangle meet in a point. α The equations of the three bisectors are obviously (see Arts. 35, 54) aß = 0, B − y = 0, y — a = 0, which, added together, vanish identically. Ex. 2. Any two of the external bisectors of the angles of a triangle meet on the third internal bisector. Attending to the convention about signs, it is easy to see that the equations of two external bisectors are a + B = 0, a + y = 0, and subtracting one from the other we get By = 0, the equation of the third internal bisector. Ex. 3. The three perpendiculars of a triangle meet in a point. Let the angles opposite to the sides a, ẞ, y be A, B, C respectively. Then since the perpendicular divides any angle of the triangle into parts, which are the complements of the remaining two angles, therefore (by Art. 54) the equations of the perpendiculars are a cos A - ẞ cos B = 0, ẞ cos B − y cos C = 0, y cos C· which obviously meet in a point. — a cosД = 0, Ex. 4. The three bisectors of the sides of a triangle meet in a point. The ratio of the perpendiculars on the sides from the point where the bisector meets the base plainly is sin A: sin B. Hence the equations of the three bisectors are a sin A - ẞ sin B = 0, ß sin B − y sin C = 0, y sin C- a sin A = 0. Ex. 5. The lengths of the sides of a quadrilateral are a, b, c, d; find the equation of the line joining middle points of diagonals. Ans. aa – b toy do 0; for this line evidently passes through the intersection of aa - bẞ, and cy - do; but, by the last example, these are the bisectors of the base of two triangles having one diagonal for their common base. In like dò, bẞcy intersect in the middle point of the other diagonal. manner aa Ex. 6 To form the equation of a perpendicular to the base of a triangle at its extremity. Ans. a + y cos B = 0. Ex. 7. If there be two triangles such that the perpendiculars from the vertices of one on the sides of the other meet in a point, then, vice versâ, the perpendiculars from the vertices of the second on the sides of the first will meet in a point. Let the sides be a, ß, y, a', ß', y', and let us denote by (aß) the angle between a and B. Then the equation of the perpendicular |