If the three lines meet in a point, this expression for the area vanishes (Art. 38); if any two of them are parallel, it becomes infinite (Art. 25). 40. Given the equations of two right lines, to find the equation of a third through their point of intersection. The method of solving this question, which will first occur to the reader, is to obtain the coordinates of the point of intersection by Art 31, and then to substitute these values for x'y' in the equation of Art. 28, viz., y − y' = m (x − x'). The question, however, admits of an easier solution by the help of the following important principle: If S=0, S' = 0, be the equations of any two loci, then the locus represented by the equation S+kS' = 0 (where k is any constant) passes through every point common to the two given loci. For it is plain that any coordinates which satisfy the equation S=0, and also satisfy the equation S'=0, must likewise satisfy the equation S+kS = 0. Thus, then, the equation (Ax+ By + C) + l (A'x + B'y + C′) = 0, which is obviously the equation of a right line, denotes one passing through the intersection of the right lines Ax+By+C=0, A'x + B'y + C' = 0, for if the coordinates of the point common to them both be substituted in the equation (Ax + By + C) + k (A'x + By + C') = 0, they will satisfy it, since they make each member of the equation separately = 0. Ex. 1. To find the equation of the line joining to the origin the intersection of Multiply the first by C', the second by C, and subtract, and the equation of the required line is (AC′ – A'C) x + (BC' — CB′) y = 0; for it passes through the origin (Art. 18), and by the present article it passes through the intersection of the given lines. Ex. 2. To find the equation of the line drawn through the intersection of the same lines, parallel to the axis of x. Ans. (BA' AB') y + CA' — AC'′ = 0. -- Ex. 3. To find the equation of the line joining the intersection of the same lines to the point x'y'. Writing down by this article the general equation of a line through the intersection of the given lines, we determine k from the consideration that it must be satisfied by the coordinates x'y', and find for the required equation (Ax+ By + C) (A'x' + B'y' + C′′) = (Ax′ + By' + C) (A'x + B'y + C"). Ex. 4. Find the equation of the line joining the point (2, 3) to the intersection of 2x + 3y+ 1 = 0,3x-4y = 5. Ans. 11 (2x + 3y + 1) + 14 3x − 4y-5)=0; or 64x-23y=59. 41. The principle established in the last article gives us a test for three lines intersecting in the same point, often more convenient in practice than that given in Art 38. Three right lines will pass through the same point if their equations being multiplied each by any constant quantity, and added together, the sum is identically = 0; that is to say, if the following relation be true, no matter what x and y are: 1 (Ax+ By + C) + m (A'x + B'y + C') + n (A′′x + B′′y + C′′) = 0. For then those values of the coordinates which make the first two members severally 0 must also make the third = 0. = Ex. 1. The three bisectors of the sides of a triangle meet in a point. Their equations are (Art. 29, Ex. 4) x − (x" + x′′ x − (x'"' + x′ – 2x′ ) y + (x'y' - y′′x′ ) + (x""'y' (y" + y"" — 2y′ ) − y'"x′ ) = 0, (y'"' + y′ − 2y" ) - 2x′′ ) y + (x'''y′′ — y'"x") + (x'y′′ — y'x′′ ) = 0, (y′+y" - 2y""') x − (x′ +x′′ – 2x"") y + (x'y'"' — y'x""') + (x"y"" — y′′x"") = 0. And since the three equations when added together vanish identically, the lines represented by them meet in a point. Its coordinates are found, by solving between any two, to be} (x′ + x" + x'''), } (y' + y'' + y"). Ex. 2. Prove the same thing, taking for axes two sides of the triangle whose lengths are a and b. 2x a a Ex. 3. The three perpendiculars of a triangle, and the three perpendiculars at middle points of sides respectively meet in a point. For the equations of Ex. 5 and 6, Art. 32, when added together, vanish identically. Ex. 4. The three bisectors of the angles of a triangle meet in a point. For their equations are (x cos a + y sin a − p) – (x cos ẞ + y sin ẞ — p') = 0, *42. To find the coordinates of the intersection of the line joining the points x'y', x"y", with the right line Ax+By+ C=0. We give this example in order to illustrate a method (which we shall frequently have occasion to employ) of determining the point in which the line joining two given points is met by a given locus. We know (Art. 7) that the coordinates of any point on the line joining the given points must be of the form which the line joining the points is cut by the given locus; and we determine this unknown quantity from the condition, that the coordinates just written shall satisfy the equation of the locus. Thus, in the present example, we have and consequently the coordinates of the required point are (Ax' + By' + C) x" − (Ax" + By" + C) x' x= (Ax' + By + C) This value for the ratio mn with a similar expression for y. might also have been deduced geometrically from the consideration that the ratio in which the line joining x'y', x"y" is cut, is equal to the ratio of the perpendiculars from these points upon the given line; but (Art. 34) these perpendiculars are Ax' + By + C Ax" + By" + C √ (A* + B*) and √(A+B) The negative sign in the preceding value arises from the fact that, in the case of internal section to which the positive sign of m: n corresponds (Art. 7), the perpendiculars fall on opposite sides of the given line, and must, therefore, be understood as having different signs (Art. 34). If a right line cut the sides of a triangle BC, CA, AB, in the points LMN, then Let the coordinates of the vertices be x'y', x"y", x"y", then *43. To find the ratio in which the line joining two points x11, xy, is cut by the line joining two other points xy11 XY1 The equation of this latter line is (Art. 29) (y,—y1) x — (x ̧-x)y + x12y-x1y2 = 0. It is plain (by Art. 36) that this is the ratio of the two triangles whose vertices are x,y, Y7 XAY 47 and geometrically evident, X2Y 21 X3Y 31 X4Y4, as is also 37 If the lines connecting any assumed point with the vertices of a triangle meet the opposite sides BC, CA, AB respectively, in D, E, F, then Let the assumed point be xy, and the vertices x11, xy *y, then L3Y 37 44. To find the polar equation of a right line (see Art. 12). Suppose we take, as our fixed axis, OP the perpendicular on If the fixed axis be 04 making an angle a with the perpendicular, then ROA = 0, and the equation is p cos (0 - a) = p. This equation may also be obtained by transforming the equation with regard to rectangular coordinates, x cosa + y sin a = p. Rectangular coordinates are transformed to polar by writing x, p cose, and for se, and for y, p sin (see Art. 12); hence the equation for becomes p(cose cosa + sin sina) = p; or, as we got before, An equation of the form p cos (0 - a) = p. P(A cose + B sin 0) C = can be (as in Art. 23) reduced to the form p cos ( − a)=p, by dividing by √(42 + B'); we shall then have Ex. 2. Find the polar coordinates of the intersection of the following lines, and also the angle between them: p cos (0 - Ex. 3. Find the polar equation of the line passing through the points whose polar coordinates are p', '; p', 0′′. Ans. p'p" sin (0'- 0′′) + p′′p sin (0′′ — 0) + pp' sin (0 – 0') = Q |