JOLY IS, 1884.] 



♦ KNOWLEDGE ♦ 



63 



:=0 



v/A' + B^ 



This is of the required form 



B 



= 1 



57. — The student will find it a useful exercise to trans)iose an 

 equation to a fixed, straight lino into the different forms we 

 have been examining. He will find that when proper attention is 

 paid to the directions in which lines and ann;les are measured, no 

 difficulty can occur in interpretinff the constants which appear in 

 different forms of the equation. On one point comment ia neces- 

 sary. The angle which a straight linn makes with either axis is not 

 necessarily the acute angle between the line and that axis. Thus the 

 line B C (B on O X. C on O V) makes with X is not the angle C B O 

 but the angle C B X. since this is the angle through which a line 

 coincident with O X would have to revolve in the positive direction, 

 about the point B, to become coincident with C B. We might also 

 say that the line C B is inclined to <) X at a negative angle C B 0. 

 <ir at a positive angle — + CBX,or2T + CBX,orin fine, generally, 

 at an angle nn- + C B X where n is any integer positive or negative. 

 This would in no way affect the interpretation of the particular 

 eqaatioB involving this angle, that is of the equation 



y = m ii- + c 

 where m is the tangent of the angle we are considering ; since we 

 ^earn from trigonometry that tan a = tan (n w* a) whatever inte- 

 gral value 71 may have. 



58. In Figure"53, Z A B = Z C B X - Z A B ; that is the angle 

 « in the equation 



a- cos a -t- V sin o = J) 

 is less by one right angle than the angle at which the line it repre- 

 sents is inclined to the axis of .r. Proper consideration of the 

 directions in which angles are measured will show that this relation 

 must always subsist. Thus since the general expression for the 

 angles at which a line is inclined to the axis of x is (n 1^-^l) the 

 torresponding general expression for the angle at which the per- 

 pendicular upon the line is inclined to the axis of x, is 

 , / 7r\ (2 n — \)rr 

 '"^ + ('-2). = that is ^ 2 +'• 



59. — Prop. To find the equation to a straight line in terms of the eo- 

 ^rdinates of a point on the line, and of the angle at u'hich the line is 

 inclined to the axis ofx. 



Let A P be a straight line, passing through .^ point A whose co- 

 ordinates are h, A-. Let a be the angle at whii !i A P is inclined to 

 OX. Draw P M, A N parallel to V, A L parallel to O X. Then 



PL 



= tan n. 



that is 



AL 



v-l; 

 x-h 



(i) 



the required equation. 



It may be noted that (i.) is clearly the equation obtained from 



y 



the form - = ni of 30, by transformation of co-ordinates. 



Equation (i) may be written into the form. 

 w — t x — h 



PL 



sma cos a 



(ii) 



and since -r-p = sin. <i, we have, putting A P = r 



y-k 



x — h 



a mode of writing the equation which is often found convenient in 

 practice. 



60. Oblique Co-orpin-.\tes.— The equations of forms considered 

 in 57 to 59 can be readih- obtained for oblique axes. Thus, if we 

 suppose Z X Y = u and' Z A C ( = w-a) = ;j it would follow, as 

 in 44, that 



OA = OK + PX = OMcosa + PMcos/3 

 that is 



X cos o + i( COS ;5 --• p 

 the required equation, in which ii=ii) — a. 



In place of equation (i) of 59 we should clearly obtain 

 '/ — '■ sin a 



which may be written 



x~ It (sin uf — a) 



y-k 

 sin a 



x-h 



sin (w — u) 

 {To he continued). 



EASY RIDERS OX EUCLIDS FIRST BOOK. 



WrrH Suggestions. 



{Continued from p. 41.) 



174. In the parallelogram ABC D, the angle ADB is equal to 

 one-third part of the angle A E B ; also A C and B D intersect at 

 an angle equal to one-third part of two right angles. Show that 

 one of the diagonals is at right angles to opposite sides of the 

 parallelogram. 



175. If the angle between two adjacent sides of a parallelogram 

 be increased, while their lengths remain unchanged the diagonal 

 through the point of intersection will be diminished. 



176. If two opposite sides of a parallelogram be bisected, the 

 lines drawn from the points of bisection to the opposite sides trisect 

 the diagonal. 



177. If A B a side of the parallelogram A B C D be divided into 

 n equal parts, show that a line drawB from C to the division point 



nearest to B cuts off from the diagonal BD one (»^-^l)th part, 



measured from B. 



Take for i\ any convenient number — say 7. Diii'de A B and CD 

 i>i(o 7 equal parts, and join C with the division nearest to B, the 

 division nearest to C with the next division from B and so on. It 

 will then be easy, in the inanner nf the preceding example, to show 

 that any one of the 8 parts into tt'hich the diagonal is thus divided is 

 equal to any other part — or, in other words, that the diagonal is 

 dividf^d into 8 equal parts. 



178. In the straight line A B C, A B is equal to BC. Show that 

 perpendiculars drawn from the points A, B, and C upon any straight 

 line mset it in eqni-distant points. 



(i.) When the line passes between A and C. 



(ii.) When the line does not pass between A and C. 



179. In case (ii.) of Example 178, show that the perpendicular 

 from A and C are together double of the perpendicular from B. 



180. Incase (i.) of Example 178, show that the difference of the 

 perpendiculars from A and C is double of the perpendicular 

 from B. 



181. If straight lines be drawn from the angles of any parallelo- 

 gram perpendicular to a straight line which is outside the paral- 

 lelogram, the sum of those from one pair of opposite ano-les is 

 equal to the sum of those from the other pair of opposite angles. 



182 Determine a point in the base of a triangle from which 

 lines drawn parallel to the sides, to meet them, are equal. 



183. If an hexagonal figure admits of division into three paral- 

 lelograms each pair of opposite sides are equal and parallel. 



Siiow that in general such an hexagonal figure admits of beino- 

 divided into three parnllelograms in two different ways. 



184. If each pair of opposite sides of a hexagon are parallel, and 

 one pair equal, the other pairs are also equal. 



185. If each pair of opposite sides of a hexagon be equal and 

 parallel, the three straight lines joining opposite angles will meet 

 in a point. 



ISO. If each pair of opposite sides of a rectilinear figure havin"- 

 an even number of sides be equal and parallel, all the lines joining 

 opposite angles meet in a point. 



187. Describe a rhombus within a given parallelogram, so that 

 one of the angular points may occupy a given point on the peri- 

 meter of the parallelogram. 



188. Describe a rectangle within a given parallelogram, so that 

 one of the angular points may occupy a given point on the peri- 

 meter of the parallelogram. 



In examples 187 and 189 if suffices that the angles of the con- 

 structed figures should lie on the sides or the sides produced of the 

 parallelogram. Prei-ious examples shoio the relations which hold 

 tvhen a parallelogram is a rhombus or rectangular, and these icill be 

 found sufficient for the solution of Examples 1S7 and 180. 



189. The three sides of a triangle are together less than the three 

 lines drawn from the angles to the bisections of the opposite sides. 



Complete a parallelogram having two sides of the triangle as adja- 

 cent sides. Then show that these sides are together greater than the 

 diagonal which passes through the bisection of the 6a.se, ^c. 



