But since the line must also pass through the point a"y", this equation must be satisfied when the co-ordinates x", y′′, are substituted for x and y; hence Substituting this value of m, the equation of the line becomes y - y_y" - y x" - x' 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, = 0. (y' – y′′) x − (x − x′′)y + x'y′′ – y'x": COR.-The equation of the line joining the point x'y' to the origin is yx = x'y. It will sometimes happen that we can write down, without calculation, the equation of the line joining two points. If we happen to know beforehand that the co-ordinates of both points are connected by the relations Ax+By+C = 0 and Ax" + By" + C = 0, then it is evident that the equation of the line joining them is Ax+ By + C = 0, for it is the equation of a right line, and is satisfied by the co-ordinates of both points. Ex. 1. Form the equations of the sides of a triangle, the co-ordinates of whose vertices are (2, 1), (3,2), (− 4, − 1). Ans. 3x + y = 7, x + 7y + 11 = 0, 3y - x = 1. Ex. 2. Find the lengths of the perpendiculars from each vertex of this triangle on the opposite side. Ans. 2/2, V10, 210, and the origin is within the triangle. Ex. 3. Form the equations of the sides of the triangle formed by Ans. (y+y2y)x-(x" + x" - 2x)y + x'y' - y'x + x"y-y"x' = 0. Ex. 6. Form the equations of the bisectors of the sides of the triangle described in Ex. 3. Ans. 5x 6y= 21; 17x-3y= 25; 7x + 9y + 17 = 0. - n Ans. x{(m-n)y' + m(n − 1)y" + n(l — m)y" } − y {l(m − n)x' + m(n − 1)x" + n(l — m)y" } = Im(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 co-ordinates of the third will satisfy this equation. The condition, therefore, is (y1 y2)x3 - (X1 X2) Yз + (X1Y2 - X2Y1) = 0, which can be put into the more symmetrical form, Y1 (X2 − X3) + Y2 (X3 − X1) + Y3 (X1 − X2) = 0.* 31. To find the area of the triangle formed by three points: If we multiply the length of the line joining two of the points, by the perpendicular on that line from the third point, we shall have double the area. Now the length of the perpendicular from xзyз on the line joining x1y1, x2Y2, the axes being rectangular, is (Art. 27) (Y1 − Y2) X3 − (X1 − X2)Y3 + X1Y2 and the denominator of this fraction is the length of the line joining x11, x2y2, hence Y1 (X2 − X3) + Y2 (X3 − X1) + Y3 (X1 − X2) represents double the area formed by the three points. If the axes be oblique, it will be found on repeating the investigation with the formula for oblique axes, that the only change that will occur is that the expression just given is to be multiplied by sin w. * In using this and other similar formula, which we shall afterwards have occasion to employ, the learner must be careful to take the co-ordinates in a fixed order (see engraving). For instance, in the second member of the formula just given y2 takes the place of y1, x3 of x2, and x1 of x3. Then, in the third member, we advance from y2 to yз, from xз to x1, and from 1 to 2, always proceeding in the order just indicated. COR. 1.-Double the area of the triangle formed by the lines joining the points xyi, xay to the origin, is y12 - Y21, as appears by making 3 = 0, y = 0, in the preceding formula. COR. 2. The condition that three points should be on one right line, when interpreted geometrically, asserts that the area of the triangle formed by the three points becomes = 0. 32. To express the area of a polygon in terms of the co-ordinates of its angular points. Take any point xy within the polygon, and connect it with all the vertices ay1, x2Y2,... Xnyn; then evidently the area of the polygon is the sum of the areas of all the triangles into which the figure is thus divided. But by the last Article double these areas are respectively x (yn - Y1) y (xn− x1) + Xn Yi - X ̧ Yn• When we add these together, the parts which multiply x and y vanish, as they evidently ought to do, since the value of the total area must be independent of the manner in which we divide it into triangles; and we have for double the area (X1 Y2 − X2 Y1) + (X2 Y3 − X3 Y2) + (X3 Y▲ − X1 Y3) + This may be otherwise written, Xn (Y1 — Yn-1), Yn (Xn-1 − x1). Ans. 10. Ans. 29. Ex. 1. Find the area of the triangle (2, 1), (3, − 2), (− 4, − 1). Ex. 2. Find the area of the triangle (2, 3), (4, 5), (− 3, − 6). Ans. 4. 33. To find the co-ordinates of the point of intersection of two right lines whose equations are given. Each equation expresses a relation which must be satisfied by the co-ordinates of the point required; we find its co-ordinates, E therefore, by solving for the two unknown quantities x and y, from the two given equations. Let the equations be given in the most general form, We said (Art. 14) that the position of a point was determined, being given two equations between its co-ordinates. The reader will now perceive that each equation represents a locus on which the point must lie, and that the point is the intersection of the two loci represented by the equations. Even the simplest equations to represent a point, viz., x = a, y = b, are the equations of two parallels to the axes of co-ordinates, the intersection of which is the required point. The reader will also now understand why two equations of the first degree only represent one point, and why two equations of higher degree represent more points than one (Art. 15). In the first case each equation represents a right line, and two right lines can only intersect in one point. In the more general case, the loci represented by the equations are curves of higher dimensions, which will intersect each other in more points than one. 34. To find the condition that three right lines shall meet in a point. Let their equations be Ax + By + C = 0, A'x + B'y + C′ = 0, A′′x + B′′y + C′′ = 0. If they intersect, the co-ordinates of the intersection of two of them must satisfy the third equation; and using the values found in the last article, we get, for the required condition, A′′ (BC′ – B′C) + B′′ (CA′ – C'A) + C” (AB′ – BA) = 0, which may be also written in either of the forms A (B'C′′ – B′′C') + B (C'A′′ – C′′A') + C (A′B′′ – A′′B') = 0, A (B′C” – B′′C′) + A′ (B′′C – BC′′) + A′′(BC′ – B′C) = 0. Ex. 1. To find the co-ordinates of the vertices of the triangle the equations of whose Ex. 4. Find the co-ordinates of the vertices, and the equations of the diagonals, of the quadrilateral the equations of whose sides are Ex. 5. Find the intersections of opposite sides of the same quadrilateral and the equation of the line joining them. Ans. 83, 162y-199x=4462. Ex. 6. Find the diagonals of the parallelogram formed by x=a, x = a', y = b, y = b'. Ans. (b − b') x − (a − a) y = a'b − ab' ; (b − b') x + (a − a) y = ab – a'b'. Ex. 7. The axes of co-ordinates being the base of a triangle and the bisector of the base, form the equations of the two bisectors of sides, and find the co-ordinates of their intersection. Let the co-ordinates of the vertex be 0, y', those of the base angles x', 0; and - x', 0. Ans. 3x'y - y'x − x'y' = 0; 3x'y + y'x − x'y' = 0; Ex. 8. The equations of the sides of a quadrilateral are 0, find the co-ordinates of the intersections of opposite sides and of the middle point of the line joining them. X. Xz 42 ці bb' (a + a') (aa' (b+b) bb' (a + a')} 1 ba' - ab' Ex. 9. Find the equation of the line joining the middle points of the diagonals of the same quadrilateral. Ex. 10. Verify that the co-ordinates of the middle point found in Ex. 8 satisfy this equation. * 35. To find the area of the triangle formed by the three lines Ax+ By + C = 0, A'x + B'y + C′ = 0, A′′x + By + C" = 0. We find the co-ordinates of the vertices by Art. 33, and substituting in the formula of Art. 31, we obtain for the double area the expression |