GRAPHICAL CALCULUS CHAPTER I. INTRODUCTORY.—CURVES AND THEIR EQUATIONS. § 1. CO-ORDINATES OF A Point. The exact position of a point in a plane is completely known if its perpendicular distances from two intersecting lines in that plane are known. Thus, suppose the lines OX, OY (Fig. 1) represent to some scale two hedges of a field meeting at right angles, and we are told that an article is buried in the field at a given depth at a point A, whose perpendicular distance from the hedge OY is 20 yards, and from OX 30 yards. It is clear that the position of the article could be immediately found by measuring 20 yards from O along OX to L, and 30 yards along LA perpendicular to OX. Definitions. The lines OL, LA, would be called the coordinates of the point A, with reference to the axes OX, OY; OL is the abscissa, and LA the ordinate ; and the point A would be described in mathematical language as the point whose abscissa is 20, and whose ordinate 30, or shortly as “the point (20, 30)." $ 2. EQUATION TO A LINE. Suppose, however, we are told that the distance of the point A from OX is half (its distance from OY) + 20 yards. From this condition alone we could not find the exact B position of the point, for there are many points in the field, in addition to the point A, of which the statement would be equally true. Thus if we take OM = 50 yards along Ox, and MB at right angles = ( x 50 + 20 yards = 45 yards, we should find a point B which would also “ satisfy the condition that its distance from OX was half its distance from OY + 20 yards. Or we might have taken ON = 80 and NC = 60; or, indeed, any arbitrary (or, as it is called, “independent ") distance along Ox, and calculated and measured the corresponding distance perpendicular to OX. We could thus find any number of points “satisfying the given condition.” All these points would be found to lie on a certain straight line in the field, and no point which is not on the straight line would be found to satisfy the condition. And, further, any point which is on the line will satisfy it. We should then be sure of finding the buried article, if we were to dig a trench of the given depth along a line represented by AC. Now let us attempt to discover the position of the article by an algebraical process. 1 This statement should be tested by plotting the points to scale on a plan of the field. Let x be the perpendicular distance of the buried article from OY; and y the perpeadicular distance from OX. Then we have 2 2 2 This is the only equation we can obtain from the data, and we are here met with the same difficulty as before, namely, that there are an infinite number of possible solutions to the equation, each solution corresponding to one particular point on the plan. Now, just as (x = 20, y = 30) may be taken to represent the point A, so the equation y = + 20 may be taken to represent the line AC. In other words, the line AC is a picture or geometrical representation of the equation y = + 20. To put it in still another way, the line AC shows the relation between the value of x and that of + 20), or y, corresponding to all values of x or y. Thus suppose we wish to find from the diagram what is the value of y or (+ 20) when « is 59'2, say, or any other arbitrary value, we measure off to scale 59'2 along OX, and erect a perpendicular to OX from the point so found. The length of this perpendicular cut off by the line AC gives the required value. The algebraical counterpart of this process is as follows : In the equation y = *+20, find the value of y when x is 59'2. To solve this we have merely to substitute 59'2 for x in the equation, and solve the resulting simple equation in y; we thus find the value of y corresponding to x = 59'2. This is easier and more accurate than the graphical process. Another example of the same thing is, “ Find where the line y = * + 20 cuts the axis of y.” At the required point it is obvious that |