the upper or lower sign being taken according as (x, y) and the origin are on different sides, or on the same side of the line (p, a). We may also arrive at the result imperfectly, thus. We may first prove, as in Art. 47, that the perpendicular must always be equal to one of the two expressions +(x cos a+ y sin a—p), and may then proceed to distinguish the cases. Now the expression x cos a + y sin a-p is negative when the point (x, y) is the origin, because it becomes then -p; also this expression cannot change its sign so long as (x, y) is taken on the same side of the line (p, a) as the origin because it cannot change its sign without passing through the value zero, and it cannot vanish until the point (x, y) is on the line. Hence the expression remains negative so long as (x, y) is on the same side of the line (p, a) as the origin. Similarly, if the expression is positive when the point (x, y) has any one position on the other side of the line (p, a), it will continue positive so long as (x, y) is on that side of the line; and it may be easily shewn that the expression can be made positive by suitable values of x and y; hence it is always positive while (x, y) is on the opposite side from the origin. We call this an imperfect method, because the sentence in italics on which the method depends, has probably not sufficiently attracted the student's attention up to this period of his studies to produce perfect conviction. 55. If the equation to a line be x cos a+y sin a = 0, so that p=0, we shall still have + (x cos a + y sin a) as the length of the perpendicular from the point (x, y) on it. We may discriminate as follows, let the equation be so written that the coefficient of y is positive; then for points on the same side of the line as the positive part of the axis of y, the perpendicular is x cos a+ y sin a; for points on the other side it is-(x cos a + y sin a). This is easily shewn by comparing a few figures, or as in Art. 54. Oblique Axes. 56. The results in Arts. 32-40 hold whether the axes are rectangular or oblique; in Art. 33, however, m must have that meaning which is required when the axes are oblique. To find the angle between two straight lines referred to oblique axes. Let be the angle between the axes; y=mx+c, the equation to one line; y=mx+c, the equation to the other. Let a,, a, be the angles which these lines make with the axis of x; and ẞ the angle between them. Hence the condition that the lines may be at right angles is 1+ (m,+m) cos w+m,m,= 0. 57. To find the length of the perpendicular drawn from a given point on a given straight line. We shall proceed as in the latter part of Art. 47; the student may also obtain the result by the method in the former part of that article. Let AB be the given straight line; D the given point; h, k its co-ordinates. Let the equation to AB be then y = mx + c. Draw DHM parallel to OY, and DE perpendicular to AB; DE= DHsin DHE. DH=DM_HM = k − (mh + c) = k — mh — c. BAX=a, then DHE or AHM=w-α, sin a Now Let and = m (Art. 24); If a line be drawn from D to meet AB at an angle ß, its length will be DE cosec ß, and will therefore be known since DE is known. If the equation to a straight line be in the form given in Art. 26, namely, x cos a + y cos ẞ-p=0, the length of the perpendicular on it from the point (x', y') will be ± (x'cos a + y'cos ß—p). This may be deduced from the preceding expression, or it may be obtained in the manner of Art. 51. Polar Co-ordinates. 58. To find the polar equation to the straight line which passes through two given points. Өз Let r,, 0, be the co-ordinates of one point; and r2, 0, those of the other; and suppose the equation to the line 1 r cos (0 − a) = P, that is, r cos e cos a +r sin 0 sin a = =p.. ........ Since this line passes through the two points, we have .(1). From (1) and (2) 2 (r cos 0-r, cos 0) cos a + (r sin 0 — r1 sin 01) sin a = 1 From (2) and (3) a=0...(5), (r1 cos 0 ̧ — r2 cos 02) cos a + (r, sin 0 ̧−r, sin 02) sin a = 2 2 rr ̧ sin (0 ̧−0) +ˆ‚ˆ‚ sin (0 ̧ — 0 ̧) +ˆ‚r sin (0 – 0) = 0...(6). 1 2 This equation has a simple geometrical interpretation; for if we draw a figure and take O for the origin, and A, B, P for the points (r,, ), (2, 2), (r, 0), respectively, we see that equation (6) is the expression of the fact that one of the triangles OAP, OBP, ÓAB, is equal in area to the sum of the other two. 59. We have seen that a straight line is the locus of an equation of the first degree; as we proceed it will appear that if an equation be of a degree higher than the first, the corresponding locus will be generally some curve; we may notice here some exceptional cases. Suppose the equation x2-4ax+4a2+ y2=0 be proposed; this equation may be written Thus the corresponding locus consists only of a single point on the axis of x at a distance 2a from the origin. Again, suppose the equation to be х x2 + y2+1=0. No real values of x and y will satisfy this equation; in this case then there is no corresponding locus, or as it is usually expressed, the locus is impossible. Thus, the locus corresponding to a given equation may reduce to a single point, or it may be impossible. 60. We have seen that the equation to a single straight line is always of the first degree; an equation of a higher degree than the first may however represent a locus consisting of two or more straight lines. For example, suppose y2 - x2=0 .. y=x ......... (2), or y=-x (1); (3). If the co-ordinates of a point satisfy either (2) or (3), they will satisfy (1); that is, every point which is comprised in the locus (2) is comprised in (1), and every point which is comprised in (3) is also comprised in (1). Hence (1) represents two straight lines which pass through the origin, and make respectively angles of 45° and 135° with the axis of x. 61. An equation which only involves one of the variables, represents a series of lines parallel to one of the axes. Thus, if there be an equation f(x)=0, we obtain by solving it a series of values for x, as xa, or xa,,...... and each of these equations represents a line parallel to the axis of y. Similarly f(y) = 0 represents a series of lines parallel to the axis of x. An equation of the form f 0 represents a series of lines passing through the origin; for by solving the equation y y y = m12 = M2,••• and we obtain a series of values for 2, as x = x each of these equations represents a line passing through the |