CHAPTER III. PROBLEMS ON THE STRAIGHT LINE. 32. WE proceed to apply the results of the preceding articles to the solution of some problems. To find the form of the equation to a straight line which passes through a given point. Let x, y, be the co-ordinates of the given point, and suppose y = mx + c.. (1) to represent the straight line. Since the point (x,, y1) is on the line, its co-ordinates must satisfy (1); hence 33. The equation (3) of the preceding article obviously represents what is required, namely, a line passing through the point (x, y). For the equation is of the first degree in the variables x, y, and therefore, by Art. 16, must represent some straight line. Also the equation is obviously satisfied by the values xx1, y=y,; that is, the line which the equation represents does pass through the given point. The constant m is the tangent of the angle which the line makes with the axis of x, and by giving a suitable value to m we may make the equation (3) represent any straight line which passes through the assigned point. = The geometrical meaning of equation (3) is obvious. For let AB be any straight line passing through the given point 2. Let P be any point in the line; x, y, its co-ordinates. Draw the ordinates PM, QN; and QR parallel to OX; then which agrees with equation (3). 34. In Art. 32 we eliminated c between the equations (1) and (2) and retained m; we may if we please eliminate m and retain c. From (2) This equation obviously represents a straight line passing through the given point, because it is an equation of the first degree and is satisfied by the values x = x, y = y1• 35. To find the equation to the straight line which passes through two given points. Let x, y, be the co-ordinates of one given point; x, y, those of the other; suppose the equation to the straight line to be Since the line passes through (x1, y1) and (x, y2), Y1 = mx1 + c.... (1). .(2), .(3). From (1) and (2) by subtraction, y-y1 = m (x-x1) From (2) and (3) by subtraction, Y2-y1 (x-x1). Substitute the value of m in (4) and we have for the required equation As a particular case we may suppose the point (x, y) to be the origin; hence x = 0, and y=0; therefore (5) becomes 36. The equation (5) of the preceding article becomes after multiplying by x-x, and reducing X3 If we compare the expression on the left-hand side of this equation with the expression in brackets in equation (2) of Art. 11, we see the only difference is that we have x and y in the place of x, and y, respectively. Thus the equation informs us that the area of the triangle formed by joining (x, y), (x1, Y1), (x, y) vanishes, as should evidently be the case since the vertex (x, y) falls on the base, that is, on the line joining (x1, Y1) to (x2, Y1⁄2). 37. To find the equation to the straight line which passes through a given point and divides the line joining two other given points in a given ratio. Let (h, k) be the first given point; let (x, y), (x2, y2) be the two other given points; let the given ratio in which the line joining the last two points is to be divided be that of n to n; then, by Art. 10, the co-ordinates of the point of division are Hence by equation (5) of Art. 35 the equation required is 38. To find the form of the equation to a straight line which is parallel to a given straight line. Let the equation to the given straight line be y = m1x + c1 and the equation to the other straight line y = mx + c .(1), (2). Since the lines represented by (1) and (2) are parallel, they must have the same inclination to the axis of x; hence Thus (2) becomes m = m1• y = mx + c. The quantity c remains undetermined since an indefinite. number of straight lines can be drawn parallel to a given straight line. |