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 √(A2+B3), we have A B B2) с x + √ (A* + B*) Y + 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, is the length of this √(A* + B*) *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 w) = sin'w, and the equation reduced to the required form is √(A2 + B* − 2AB cos w)' √(A2+ B2 – 2AB cos ∞) ' * 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 axes of x and y; and that of this perpendicular. C sin w is the length √(A2 + B2 – 2AB cos w) This length may be also easily calculated by dividing the double area of the triangle NOM, (ON. OM sin ∞) 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 xcos a+y cosẞ-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 BA-AB-0 (Art 21), 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 y=mx+b, y=m'x + b' ; 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, consequently, Hence A sin w ́√(A2 + B2 − 2AB cos w) B'A' cos w √(A'2 + B′′ – 2A'B' cos w)' (BA' — AB') sin w √(A2 + B2 − 2AB cos w) √(A” + B” – 2A'B' cos ∞) BB'+AA' - (AB'+ A'B) cos w √(A2 + B2 − 2AB cos w) √√(A”2 + B'2 − 2A'B' cos w) ' (BA' AB') sin o ·a')= 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 stants is Ax + By + C=0; but in this case we are concerned not with the absolute magnitudes, but only with the mutual ratios. of the quantities A, B, C. For if we multiply or divide the equation by any constant it will still represent the same line: we may divide therefore by C, when the equation will only A B contain the two constants. Choosing, then, any of these forms, such as y=mx+b, to represent a line in general, we may consider m and b as two unknown quantities to be determined. And when any two conditions are given we are able to find the values of m and b, corresponding to the particular line which satisfies these conditions. This is sufficiently illustrated by the examples in Arts. 28, 29, 32, 33. 28. To find the equation of a right line parallel to a given one, and passing through a given point x'y'. If the line y=mx+b be parallel to a given one, the constant m is known (Cor., Art. 21). And if it pass through a fixed point, the equation, being true for every point on the line, is true for the point x'y', and therefore we have y' = mx' +b, which determines b. The required equation then is y = = mx + y' — mx', or y — y' = m (x — x′). If in this equation we consider m as indeterminate, we have the general equation of a right line passing through the point x'y'. 29. To find the equation of a right line passing through two fixed points x'y', x"y". We found, in the last article, that the general equation of a right line passing through x'y' is one which may be written in the form where m is indeterminate. But since the line must also pass through the point x"y", this equation must be satisfied when the coordinates x", y", are substituted for x and y; hence Substituting this value of m, the equation of the line becomes In this form the equation can be easily remembered, but, clearing it of fractions, we obtain it in a form which is sometimes more convenient, (y' — y′′) x — (x′ — x") y + x'y" — y'x" = 0. The equation may also be written in the form (x − x') ( y − y′′) = (x − x') (y — y'). For this is the equation of a right line, since the terms xy, which appear on both sides, destroy each other; and it is satisfied either by making = x', y=y', or x=x", y=y". Expanding it, we find the same result as before. COR. The equation of the line joining the point x'y' to the origin is y'x=x'y. Ex. 1. Form the equations of the sides of a triangle, the coordinates of whose vertices are (2, 1), (3, − 2), (— 4, − 1). Ans. x + 7y + 11 = 0, 3y − x = 1, 3x + y = 7. sides of the triangle formed by (2, 3), (4, — 5), Ans. x7y=39, 9x - 5y = 3, 4x + y = 11. Ex. 2. Form the equations of the (-3, -6). Ex. 3. Form the equation of the line joining the points mx' + nx" my + ny” x'y' and Ex. 4. Form the equation of the Ans. (y'-y") x − (x' — x") y + x'y′′ — y'x" = 0. line joining x"+x" y′′+y”” x'y' and " 2 Ans. (y"+y"" — 2y') x (x" + x'” — 2x′) y + x'y' — y′′x' + x""y' — y'"'x' = 0. Ex. 5. Form the equations of the bisectors of the sides of the triangle described in Ex. 2. Ans. 17x-3y = 25, 7x + 9y + 17 = 0, 5x – 6y = 21. Ex. 6. Form the equation of the line joining lx' - mx" ly' - my" lx' - nx"" ly' - ny" 7-m -m to Ans. x {lm-n) y+m (n − 1) y′′+ n (l—m) y""} — y {l (m—n) x'+ m (n − 1) x" + n (l—m) x"} = lm (y'x' — x'y'') + mn (y′′x'" — x'y'"') + nl (y'"'x′ — y'x””). 30. To find the condition that three points shall lie on one right line. We found (in Art. 29) the equation of the line joining two of them, and we have only to see if the coordinates of the third will satisfy this equation. The condition, therefore, is |