29 the co-ordinates of P, and xy, those of Q. Draw PR parallel to OX. Then, by Trigonometry, PQ2 = PR2 + QR2-2PR. QR cos PRQ = PR2 + QR2+2PR. QR cos w. But PR = x, x1, and QR=y,-y1; therefore PQ2 = (x, − x ̧)2 + (Y2 − y1)2 + 2 (x ̧ — x1) (y2 — y1) cos w...(1), 2 and thus the distance PQ is determined. If the axes are rectangular, we have PQ2 = (x ̧ − x ̧)2 + (Y2—Y1)2.. (2). The student should draw figures placing P and Q in the different compartments and in different positions; the equations (1) and (2) will be found universally true. From the equation (2) we have 2 2 2 2 .....(3). PQ2 = x2+y12 + x‚ ̧2 + y22 − 2 (x ̧¤ ̧+Y,Y,) The following particular cases may be noted. PQ2 = x2+y2. If P be on the axis of x and Q on the axis of y, y1 = 0 and x=0; thus 2 PQ2 = x2+Y2 Let 0,,,, be the polar co-ordinates of P, and 0,, r,, those of Q; then, by Art. 8, Substitute these values in (3) and we have This result can also be obtained immediately from the triangle POQ formed by drawing lines from P and Q to the origin. 10. To find the co-ordinates of the point which divides in a given ratio the line joining two given points. Let A and B be the given points, x, y, the co-ordinates of A, and x, y, those of B; and let the required ratio be that of n to n Suppose C the required point, so that AC: CB: nn. Draw the ordinates AL, BM, CN; and AR parallel to OX meeting CN in D. Let x, y, be the co-ordinates of C. It is obvious from the figure that LN AD AC In this article the axes may be oblique or rectangular. A simple case is that in which we require the co-ordinates of a point midway between two given points; then n1 = n2 and y = 1 (y1 + y2). x = 1 (x1 + x2), 11. To express the area of a triangle in terms of the coordinates of its angular points. Let ABC be a triangle; let x, y, be the co-ordinates of A; x, y, those of B; x, y, those of C. Draw the ordinates AL, BM, CN. The area of the triangle is equal to the trapezium ABML+trapezium BCNM – trapezium ACNL. The area of the trapezium ABLM is LM (AL+BM). This is obvious, because if we join BL we divide the trapezium into two triangles, one having AL for its base and the other BM, and each having LM for its height; = } {(x,− x1) (Y2+Y2) + (X3−X2) (Y2+Y ̧) −(x ̧−x ̧) (ÿ1+Y1)}• This expression may be written more symmetrically thus; } {(x,−x1) (Y2+Y1) + (X3−X2) (Y3+Y2) + (x ̧ − x ̧) (Y1 +Y ̧)}.....(1). By reducing it, we shall find the area of the triangle If the axes be oblique and inclined at an angle w, the area of the trapezium ABML = } LM (AL + BM) sin o, and similarly for the other trapeziums. Thus the area of the triangle will be found by multiplying the expressions given above by sin w. However the relative situations of A, B, C, may be changed, the student will always find for the area of the triangle the expression (2), or that expression with the sign of every term changed. Hence we conclude, that we shall always obtain the area of the triangle by calculating the value of the expression (2), and changing the sign of the result if it should prove negative. Locus of an equation. Equation to a curve. ... 12. Suppose an equation to be given between two unknown quantities, for example, y-x-2=0. We see that this equation has an indefinite number of solutions, for we may assign to x any value we please, and from the equation determine the corresponding value of y. Thus corresponding to the values 1, 2, 3, of x, we have the values 3, 4, 5, ... of y. Now suppose a line, straight or curved, such that it passes through every point determined by giving to x and y values that satisfy the equation y -x-2=0; such a line is called the locus of the equation. It will be shewn in the next chapter that the locus of the equation in question is a straight line. We shall see as we proceed that generally every equation between the quantities x and y has a corresponding locus. But instead of starting with an equation and investigating what locus it represents, we may give a geometrical definition of a curve and deduce from that definition an appropriate equation; this will likewise appear as we proceed; we shall take successively different curves, define them, deduce their equations, and then investigate the properties of these curves by means of their equations. We shall in the next chapter begin with the equation to a straight line. The connexion between a locus and an equation is the fundamental idea of the subject and must therefore be carefully considered; we shall place here a formal definition which we shall illustrate in the next chapter by applying it to a straight line. DEF. The equation which expresses the invariable relation which exists between the co-ordinates of every point of a curve is called the equation to the curve; and the curve, the co-ordinates of every point of which satisfy a given equation, is called the locus of that equation. 13. The student has probably already become familiar with the division of algebraical equations into equations. of the first, second, third... degree. When we speak of an equation of the nth degree between two variables we mean that every term is of the form Axy where a and B are zero or positive integers such that a +ẞ is not greater than n, A is a constant numerical quantity, and the equation is formed by connecting a series of such terms by the signs + and -, and putting the result = 0. EXAMPLES. 1. Find the polar co-ordinates of the points whose rectangular co-ordinates are and indicate the points in a figure. 2. Find the rectangular co-ordinates of the points whose polar co-ordinates are and indicate the points in a figure. 3. The co-ordinates of P are -1 and 4, and those of Q are 3 and 7; find the length of PQ. 4. Find the area of the triangle formed by joining the first three points in question 1. 5. A is a point on the axis of x and B a point on the axis of y; express the co-ordinates of the middle point of AB in terms of the abscissa of A and the ordinate of B; shew also that the distance of this point from the origin = AB. |