22. To express the equation of a line MN in terms of the intercepts OM-a, ON=b which it cuts off on the axes. We can derive this from the form already considered A B Ax+ By + C = 0, or 4x + 2y +1 = 0. This equation must be satisfied by the coordinates of every point on MN, and there fore by those of M, which (see Art. 2) are x =α, y=0. Hence we have #a+1=0, /= 1 a N M Substituting which values in the general form, it becomes This equation holds whether the axes be oblique or rectangular. It is plain that the position of the line will vary with the signs of the quantities a and b For example, the equation 818 y + b =1, which cuts off positive intercepts on both axes, re presents the line MN on the preceding figure; 2 a У = 1, cutting off a positive intercept on the axis of x, and a negative intercept on the axis of y, represents MN'. By dividing by the constant term, any equation of the first degree can evidently be reduced to some one of these four forms. Ex. 1. Examine the position of the following lines, and find the intercepts they make on the axes: 2x-3y=7; 3x+4y+9=0; 23. To express the equation of a right line in terms of the length of the perpendicular on it from the origin, and of the angles which this perpendicular makes with the axes. Let the length of the perpendicular OP=p, the angle POM which it makes with the axis of x = a, PON=B, OM = a, ON=b. We saw (Art. 22) that the equation of the right line MN was X У + = 1. a b Multiply this equation by p, and we have But? a P M =cosa,=cos; therefore the equation of the line is x cosa+y cosẞ=p. In rectangular coordinates, which we shall generally use, we have ẞ=90° - a; and the equation becomes x cosa + y sina=p. This equation will include the four cases of Art. 22, if we suppose that a may take any value from 0 to 360°. Thus, for the position NM', a is between 90° and 180°, and the coefficient of x is negative. For the position M'N', a is between 180° and 270°, and has both sine and cosine negative. For MN', a is between 270° and 360°, and has a negative sine and positive cosine. In the last two cases, however, it is more convenient to write the formula x cosa + y sin a=-p, and consider a to denote the angle, ranging between 0 and 180°, made with the positive direction of the axis of x, by the perpendicular produced. In using, then, the formula x cosa + y sin a=p, we suppose p to be capable of a double sign, and a to denote the angle, not exceeding 180°, made with the axis of x either by the perpendicular or its production. The general form Ax+ By +C=0, can easily be reduced to the form x cos a + y sin a=p; for, dividing it by √(A+B), we have since the sum of squares of these two quantities = 1. Hence we learn that A and B are re √(A2 + B2) √(A2 + B2) spectively the cosine and sine of the angle which the perpendicular from the origin on the line (Ax + By + C=0) makes with the axis of x, and that perpendicular. c √ (A2 + B*) is the length of this *24. To reduce the equation Ax+ By + C=0 (re,erred to oblique coordinates) to the form x cosa + y cosẞ=p. Let us suppose that the given equation when multiplied by a certain factor R is reduced to the required form, then RA= cosa, RB = cos ß. But it can easily be proved that, if a and be any two angles whose sum is w, we shall have Hence R* (A + B-2AB cos ∞) = sin'w, and the equation reduced to the required form is √(A2 + B2 − 2AB cosw)' √(A2+ B* − 2AB cos w) ' * Articles and Chapters marked with an asterisk may be omitted on a first reading. are respectively the cosines of the angles that the perpendicular from the origin on the line Ax + By + C=0 makes with the C sin w axes of x and y; and that √(4+ B - 2AB cos w) is the length √(A2 of this perpendicular. This length may be also easily calculated by dividing the double area of the triangle NOM, (ON. OM sin w) by the length of MN, expressions for which are easily found. The square root in the denominators is, of course, susceptible of a double sign, since the equation may be reduced to either of the forms x cos a+ y cos ẞ− p = 0, x cos (a + 180°) + y cos (B + 180°) + p = 0. 25. To find the angle between two lines whose equations with regard to rectangular axes are given. The angle between the lines is manifestly equal to the angle between the perpendiculars on the lines from the origin; if therefore these perpendiculars make with the axis of x the angles a, a', we have (Art. 23) COR. 1. The two lines are parallel to each other when since then the angle between them vanishes. COR. 2. The two lines are perpendicular to each other when AA'+ BB'=0, since then the tangent of the angle between them becomes infinite. If the equations of the lines had been given in the form since the angle between the lines is the difference of the angles they make with the axis of x, and since (Art. 21) the tangents of these angles are m and m', it follows that the tangent of the ; that the lines are parallel if m=m' ; required angle is m-m' 1 + mm' and perpendicular to each other if mm' + 1 = 0. *26. To find the angle between two lines, the coordinates being oblique. We proceed as in the last article, using the expressions of Art. 24, A sin w cos a' = consequently, sin α = √(A + B -2AB cos w) (BA' — AB') sin w 12 sin (a — a') = √(4' + Ba — 2AB cos ∞) √(4” + B* — 2A′B′ cos ∞) tan (a-a')= = √(A2 + B2 − 2AB cos w) √(A” + B'2 — 2A'B' cos ∞)' (BA'AB') sin w AA' + BB' – (AB' + BA') cos w COR. 1. The lines are parallel if BA' = AB'. COR. 2. The lines are perpendicular to each other if 27. A right line can be found to satisfy any two conditions. Each of the forms that we have given of the general equation of a right line includes two constants. Thus the forms y=mx+b, x cos a + y sin a=p, involve the constants m and b, The only form which appears to contain more con |