MODERN ANALYTICAL GEOMETRY.

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1 TRILINEAR COORDINATES AND OTHER METHODS OF MODERN ANALYTICAL GEOMETRY. CHAPTER I. OF PERPENDICULAR COOEDINATES REFERRED TO TWO AXES. BEFORE proceeding to explain the method of trilinear coordinates, we will introduce a system which may be regarded as a connecting link between the usual Cartesian system, and the trilinear system: it is the method of perpendicular coordinates referred to two oblique axes or lines of reference. 1. In the ordinary system of oblique coordinates, the position of any point P with reference to a pair of axes CX, GY, is determined by the lengths of two lines PM, PN measured parallel to either axis to meet the other : that is, if x, y be the oblique coordinates of P referred to CX, GY, then x = PM, and y = PK It is obvious that the position of the point would be equally well determined, if it were agreed to take as coordinates the perpendicular distances of the points from each axis, instead of measuring the distance from each axis parallel to the other. Thus, if we let fall the perpendiculars PM' and PN' on GB and GA, we observe that PM' and PN' might be used, as lawfully w. 1

2 2 OF PERPENDICULAR COORDINATES as PM and PN, to determine the position of P, provided it be specified beforehand which system of measurement is intended. We shall speak of PJ/and iwas the oblique coordinates, and PM and PN f as the perpendicular coordinates of the point P referred to the same axes CX and CY. We shall use a and /3 to denote perpendicular coordinates, reserving x and y to denote, as usual, the oblique coordinates. Thus for the point P x = PM, y = PN: a = PM', /3 = PiV. When the axes are rectangular these two systems of measurement will indeed coincide, but in the case we have introduced of oblique axes they will be distinct. Y Fig. 1. C N N' A X The same convention as to the algebraical signs of the coordinates will hold equally in the perpendicular as in the oblique coordinates. Thus we shall consider as positive the distances from CX of all points lying on the same side with Y, and consequently the distances of all points on the opposite side will be negative. So also the positive side of CY will be that on which X lies, and the negative side the side remote from X.

3 REFERRED TO TWO AXES If the angle XGYbe denoted by C, then we have PM' PN' - ^ = sin<7 and - ^ = sino. Hence if a, /3 be the perpendicular coordinates of any point whose oblique coordinates are x, y, we shall have a = x sin G and /3 = y sin G; or x = a cosec G and # = /3 cosec (7. Consequently, if we have any relation holding good between the oblique coordinates of all points on a locus, we can, by the substitution of x = a cosec G, y~^ cosec G, obtain a relation holding good between the perpendicular coordinates of the same locus. In other words, we may transform by this substitution the oblique equation of any locus into an equation in perpendicular coordinates representing the same locus. For example, the equation in oblique coordinates to the straight line AB cutting off intercepts GA = b and CB= a from the axes is known to be x b y a Hence the equation to the same line in perpendicular coordinates will be a cosec G /3 cosec G _ - j j^ b a or aa + 5/3 = ah sin C; and in a similar way any other equation might be transformed. 3. Or, instead of taking an equation, we might by the same substitution transform any function whatever of the oblique coordinates into an equivalent function of the perpendicular coordi- 1 2

4 4 OP PERPENDICULAR COORDINATES nates. For example, writing the equation to the same straight line AB in the form ab ax by = 0, we can at once write down the expression for the perpendicular distance of any point (x, y) from it, viz. ab ax by. ~ ± -. * sin G. J a ab cos 0 Hence if a', y8' be the perpendicular coordinates of the same point, we shall have as the expression for the distance a&sin C ao! b$ "v / a a^osC' or if c denote the distance AB, and A the area of the triangle ABC so that c 2 = a 2 +& 2-2a&cosC, and 2 A = ab sin (7, then the expression (for the perpendicular distance from AB of the point whose coordinates are a and /3') becomes ~ c This is given here merely as an example of transformation of coordinates from the one system to the other, but the result is one which will be seen hereafter to have an important bearing on trilinear coordinates. 4. The student will do well to examine at this stage of the subject the interpretation of some of the simpler equations connecting a and ft. (1) Consider the equation a = 0. It is evidently satisfied by all points on the line CY and by no other: it is therefore the equation to this axis.

5 REFERRED TO TWO AXES. 5 (2) Consider the equation a = d, where d is a constant. It is equally obvious that this is satisfied at any point on a line parallel to GB at a distance d from it on the side towards X. Similarly the equation a = d is satisfied at any point on a parallel line at an equal distance on the other side of GY. (3) Consider the equation a = /3. The equation speaks to us of points whose perpendicular distance from GX is equal to the perpendicular distance from GY, and of the same algebraical sign. The locus of such points is the interior bisector of the angle XGY, which is therefore the locus of the equation. (4) Consider the equation This speaks of points whose perpendicular distances from the axes are equal, but of opposite sign. These points will be seen to lie on the exterior bisector of the angle XGY 7 which will therefore be the locus of the equation. (5) Consider the equation a = m/3. Suppose P to be a point on this line, join PG and draw PM', PN' perpendiculars on 0 Y, OX, then (by the equation) and therefore PM' = m.pn', PM' PN' ~pr r = m ' ~PP' sin PCY=m. sin PCX,

6 6 OF PERPENDICULAR COORDINATES or (in words), Plies on a straight line dividing the angle XOY into two parts PGY 9 PCX such that their sines are in the ratio mil. This straight line will therefore be the locus of the equation. 5. To find the area of a triangle the perpendicular coordinates of whose angular points are given with respect to a pair of oblique axes. Y Fig. 2. B/ / P C Q' P' Rf AX Let PQR be the triangle and (a 19 &), (a 2, /3J, (a 8, /3 3 ) the perpendicular coordinates of the angular points P, Q, R referred to the oblique axes CX, GY. Parallel to GY draw PP\ QQ\ RE to meet CX in P, Q\ R'. Then A PQR = trapezium PQ Q'P + trapezium PRRP' trapezium QRR'Q'. But the trapezium PQQ'P' stands on a base PQ' equal to \ a \ a 2) c sec G and it has a mean altitude equal to J {fi x + /3 2 ).

7 REFERRED TO TWO AXES. 7 Hence its area = - cosec G {ft 1 + ft 2 ) (a t a 2 ). Similarly, area PRR'F = - cosec G (/3 3 + /3J (a 3 aj, and area #.#.72' </ = - cosec G (ft 2 + /3 3 ) (? 3 - a 2 ); therefore the area of the triangle PQE = \ cosec (7{(ft+ft)(«,-«,) + (ft+ ft) (0,-0,) -(ft+ft) (a 3 -«,)J = \ cosec (7 a x (ft - ft) + a 2 (ft - ft) + a 3 (ft - ft) J, or with determinant notation J 1, 1, 1 I EXERCISES ON CHAPTER I. (1) If (a, /3) be the perpendicular coordinates of the point whose oblique coordinates (referred to the same axes) are (x, #), and (a', ft') the perpendicular coordinates of the point whose oblique coordinates are (x\ y), shew that (a + KOL, ft + K/3') will be the perpendicular coordinates of the point whose oblique coordinates are (x + xx, y + icy). (2) Shew that the points whose perpendicular coordinates are (a, ft), (a, /3'), (a-a', /3-/3') and (a + a, ft + ft f ) lie all on one straight line, provided aft' = aft.

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0 Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y