253. If Ax+By+C=0 be the equation of a straight line in Cartesian co-ordinates, the intercepts which the therefore A and B be very small the line will be at a very great distance from the origin. The equation of the line will in the limit, assume the form 0.x +0.y + C = 0. The equation of an infinitely distant straight line, generally called the line at infinity, is therefore 0.x +0.y + C = 0. When the line at infinity is to be combined with other expressions involving x and y it is written C=0. The equation of the line at infinity in trilinear co-ordinates is ax+bB+cy = 0. For if ka, kß, ky be the co-ordinates of any point, the invariable relation gives k(aa+bß + cy) = 2A, or If therefore k become infinitely great, we have in the limit the relation ax + bB+cy = 0. This is a linear relation which is satisfied by finite quantities which are proportional to the co-ordinates of any infinitely distant point, and it is not satisfied by the co-ordinates, or by quantities proportional to the co-ordinates, of any point at a finite distance from the triangle of reference. 254. To find the condition that two given lines may be parallel. Let the equations of the lines be la + mB+ny=0, l'a+m'B+ n'y = 0. If the lines are parallel their point of intersection will be at an infinite distance from the origin and therefore its co-ordinates will satisfy the relation aa+bB+cy = 0. Eliminating a, B, y from the three equations, we have the required condition, viz. 255. To find the equation of a straight line through a given point parallel to a given straight line. Let the equation of the given line be The required line meets this where aa+bB+cy = 0. The equation is therefore of the form la + mß +ny + λ (aa + bß + cy) = 0. If f, g, h be the co-ordinates of the given point, we must also have whence If+mg +nh + λ (af + bg+ ch) = 0, la + mB + ny ax+bB+cy = If + mg + nh ̄af+bg+ch* A useful case is to find the equation of a straight line through an angular point of the triangle of reference parallel to a given straight line. If A be the angular point, its co-ordinates are f, 0, 0, and the equation becomes (ma-lb) B+ (na-lc) y = 0. 256. To find the condition of perpendicularity of two given straight lines. Let the equations of the lines be 1x + mB + ny = 0, l'a+m'B+ n'y = 0. If these be expressed in Cartesian co-ordinates by means of the equations found in Art. 246, they will be x (l cos 02+m cos 02+n cos 0 ̧) + y (l ́sin 01+m sin 02+nsinė ̧) 1 2 2 and 2 2 x(l'cos2+m'cos 2+n'cos § ̧)+y (l'sin 01+m'sin 02+n'sin 0 ̧) t'p1 - m'p12- n'p1 = 0: the lines will therefore be perpendicular [Art. 29] if 1 2 1 2 (l cos 0, +m cos 0, + n cos 0 ̧) (l' cos 0, + m' cos 0, +n' cos 0 ̧) +(l sine,+m sine2+n sine) (l'sin 01+m'sin 02+n'sin0 ̧)=0; that is, if 2 ll' + mm' + nn' + (lm' + l'm) cos (0,0) 2 2 = cos B, +(mn' + m'n) cos (9 ̧ ~ 03) + (nl′+ n'l) cos (0 ̧~ 0 ̧) = 0. But cos (0-0)=-cos A, cos (0,01) and cos (0-0)= cos C; therefore the required condition is ll' + mm' + nn' — (mn' + m'n) cos A 257. − (nľ' + n'l) cos B - (lm' + l'm) cos C' = 0. To find the perpendicular distance of a given point from a given straight line, Let the equation of the straight line be lx + mB + ny = 0, Expressed in Cartesian co-ordinates the equation will be x (l cos§ ̧+m cos 02+n cos 0 ̧) + y (l sin 0 ̧‡m sin 02+n sin 0 ̧) -lp-mp-np1 = 0. 1 1 2 The perpendicular distance of any point from this line is found by substituting the co-ordinates of the point in the expression on the left of the equation and dividing by the square root of the sum of the squares of the coefficients of x and y. If this be again expressed in trilinear coordinates, we shall have, for the length of the perpendicular from f, g, h on the given line, the value If + mg + nh 2 √{(lcose,+m cose2+ncos e ̧)2+(lsin,+msinë ̧+nsinė ̧)*} * The denominator is the square root of l2 + m2 + n2 + 2lm cos (0,0)+2mn cos (0-0) or of 12+ m2 + n2 - 2lm cos C-2mn cos A - 2nl cos B. Hence the length of the perpendicular is equal to √ (l2 + m2 + n2 — 2mn cos A 2nl cos B-2lm cos C) 258. To shew that the co-ordinates of any four points may be expressed in the form ±f, ±g, ±h. Let P, Q, R, S be the four points. The intersection of the line joining two of the points and the line joining the other two is called a diagonalpoint of the quadrangle. There are therefore three diagonal-points, viz. the points A, B, C in the figure. Take ABC for the triangle of reference, and let the co-ordinates of P be ƒ, g, h. Then the equation of AP will be γ h The pencil AB, AS, AC, AP is harmonic [Art. 60], and the equations of AB, AC are y = 0,B=0 respectively, So that the co-ordinates of S are proportional to f, g,-h. Similarly the co-ordinates of R are proportional to -f,g, h. Similarly the co-ordinates of Q are proportional to f,-g,h. 259. To shew that the equations of any four straight lines may be expressed in the form lx ± mß ±ny = 0. Let DEF, DKG, EKH, FGH be the four straight lines. Let ABC be the triangle formed by the diagonals FK, EG, and DH of the quadrilateral, and take ABC for the triangle of reference. K Let the equation of DEF be la + mß + ny = 0. Then the equation of AD is mß+ny = 0. Since the pencil AD, AB, AH, AC is harmonic [Art. 60], and the equations of AD, AB, AC are mß+ny=0, y= 0, B = 0 respectively; therefore [Art. 56] the equation of AH is mẞ-ny = 0. Since E is the point given by ẞ=0, lx + ny = 0; and H is the point given by a=0, mB-ny=0; the equation of HE is lx-m8+ny = 0. |