31. 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, therefore, by solving for the two unknown quantities x and y, from the two given equations. 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. When the equations are both of the first degree they denote but one point; for 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. Ex. 1. To find the co-ordinates of the vertices of the triangle the equations of whose sides are x + y = 2; x-3y=4; 3x+5y + 7 = 0. Ans. (14,11), (1, − y), (4, − 1). Ex. 2. To find the co-ordinates of the intersections of 3x + y 2 =0; x + 2y = 5; 2x − 3y+7 = 0. Ans. (4, 4), (− 11, 21), (− 1, 4). Ex. 3. Find the co-ordinates of the intersections of 2x + 3y = 13; 5x - y = 7; x - 4y+ 10 = 0. Ans. They meet in the point (2, 3). Ex. 4. Find the co-ordinates of the vertices, and the equations of the diagonals, of the quadrilateral the equations of whose sides are 16x - 10y = 33, 12x + 14y + 29 = 0. 2y-3x= 10, 2y + x = 6, - (1, − 1⁄2), (−3, 1); by — x = 6, 8x + 2y + 1 = 0. Ans. (— 1, 2), (3, 1), * In using this and other similar formulæ, 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, y, takes the place of 1, 3 of X21 of x. Then, in the third member, we advance from y1⁄2 to y, from 3 to 1, and from x, to x2, always proceeding in the order just indicated. and x1 E Ex. 5. Find the intersections of opposite sides of the same quadrilateral, and the equation of the line joining them. Ans. (83, 25), (— U, 0), 162y-199x4462. 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; (0, 2) . Ex. 8. Two opposite sides of a quadrilateral are taken for axes, and the other Ex. 9. In the same case find the co-ordinates of the middle point of the line joining the intersections of opposite sides. Ans. a'b.aab'. a' a'b.b' - ab'.b and the form of the result shows (Art. 7) that this point divides externally, in the ratio a'b: ab', the line joining the two middle points (a, b′), (a', b). 32. To find the equation to rectangular axes of a right line passing through a given point, and perpendicular to a given line, y = mx + b. The condition that two lines should be perpendicular, being mm'-1 (Art. 25), we have at once for the equation of the required perpendicular 1 It is easy, from the above, to see that the equation of the perpendicular from the point x'y' on the line Ax+ By +C=0 is that is to say, we interchange the coefficients of x and y, and alter the sign of one of them. Ex. 1. To find the equations of the perpendiculars from each vertex on the opposite side of the triangle (2, 1), (3, − 2), (− 4, − 1). The equations of the sides are (Art. 29, Ex. 1) x + 7y+11=0, 3y − x = 1, 3x+y=7; and the equations of the perpendiculars 7x - y = 13, 3x + y = 7, 3y − x = 1. The triangle is consequently right-angled. Ex. 2. To find the equations of the perpendiculars at the middle points of the sides of the same triangle. The co-ordinates of the middle points being (− 1, − 2), (− 1, 0), (§, − 1). The perpendiculars are 7x-y+2=0, 3x + y + 3 = 0, 3y-x+4=0, intersecting in (− 1, − 1). Ex. 3. Find the equations of the perpendiculars from the vertices of the triangle (2, 3), (4, — 5), (— 3, — 6) (see Art. 29, Ex. 2). Ans. 7x+y= 17, 5x + 9y + 250, x-4y = 21; intersecting in (99, — 13). Ex. 4. Find the equations of the perpendiculars at the middle points of the sides of the same triangle. Ans. 7x+y+2 = 0, 5x+9y+160, x-4y=7; intersecting in (-2, - §3). Ex. 5. To find in general the equations of the perpendiculars from the vertices on the opposite sides of a triangle, the co-ordinates of whose vertices are given. Ans. (x" — x") x + (y′′ − y"") y + (x'x'"' + y'y'") - (x'x' + y'y′′ ) = 0, (x” — x′ ) x + (y''' (x" - x") x + (y′ - y′′) y + (x""x" +y""y") - (x""x" +y""y'′ ) = 0. Ex. 6. Find the equations of the perpendiculars at the middle points of the sides. Ans. (x" - x''') x + (y" — y"") y = ≥ (x''2 '— x'''2) + † (y''2 — y'''2), Ex. 7. Taking for axes the base of a triangle and the perpendicular on it from the vertex, find the equations of the other two perpendiculars, and the co-ordinates of their intersection. The co-ordinates of the vertex are now (0, y'), and of the base angles (x′′, 0), (~ x'', 0). 2 Ans. x' (x − x′′) + y'y = 0, x′′ (x + x'"') − y'y = 0, (0, 2"x"). y' Ex. 8. Using the same axes, find the equations of the perpendiculars at the middle points of sides, and the co-ordinates of their intersection. Ans. 2(x'"'x+y'y)=y'2—x'''2, 2(x′′x—y'′y)=x"2—y'2, 2x=x”—x"", 2 y'2-x"x" 2y' Ex. 9. Form the equation of the perpendicular from x'y' on the line x cosa+y sina=p; and find the co-ordinates of the intersection of this perpendicular with the given line. Ans. (x + cosa (p-x' cosa - y' sin a), y' + sin a (p-x' cosa - y' sin a)}. Ex. 10. Find the distance between the latter point and x'y'. Ans. (px' cos a-y' sin a). 33. To find the equation of a line passing through a given · point and making a given angle 4, with a given line y=mx+b (the axes of co-ordinates being rectangular). Let the equation of the required line be 34. To find the length of the perpendicular from any point x'y', on the line whose equation is x cosa+y cosß – p=0. We have already indicated (Ex. 9 and 10, Art. 32) one way of solving this question, and we wish now to show how the same result may be obtained geometrically. From the given point draw QR parallel to the given line, and QS perpendicular. Then OK=x', and OT will be a' cosa. Again, since SQK=8, and QK=y', hence = RT=QS=y' cosß; x' cosa +y' cosẞ=OR. N R Subtract OP, the perpendicular from the origin, and A x' cosa+y' cosẞ− p = PR = the perpendicular QV. But if in the figure the point Q had been taken on the side of the line next the origin, OR would have been less than OP, and we should have obtained for the perpendicular the expression p-x' cosa-y' cosß; and we see that the perpendicular changes sign as we pass from one side of the line to the other. If we were only concerned with one perpendicular, we should only look to its absolute magnitude, and it would be unmeaning to prefix any sign. But if we were comparing the perpendiculars from two points, such as Q and S, it is evident (Art. 6) that the distances QV, SV, being measured in opposite directions must be taken with opposite signs. We may then at pleasure choose for the expression for the length of the perpendicular either (p-x' cosa-y' cos B). If we choose that form in which the absolute term is positive, this is equivalent to saying that the perpendiculars which fall on the side of the line next the origin are to be regarded as positive, and those on the other side as negative; and vice versa if we choose the other form. If the equation of the line had been given in the form Ax+ By +C=0, we have only (Art. 24) to reduce it to the form x cosa+y cosẞ − p = 0, and the length of the perpendicular from any point x'y', according as the axes are rectangular or oblique. By comparing the expression for the perpendicular from x'y' with that for the perpendicular from the origin, we see that x'y' lies on the same side of the line as the origin when Ax' + By +C has the same sign as C, and vice versa. The condition that any point x'y' should be on the right line Ax+By+C=0, is, of course, that the co-ordinates x'y' should satisfy the given equation, or Ax' + By +C=0. And the present Article shows that this condition is merely the algebraical statement of the fact, that the perpendicular from the point x'y' on the given line is = = 0. Ex. 1. Find the length of the perpendicular from the origin on the line the axes being rectangular. 3x+4y+20=0, Ans. 4. Ex. 2. Find the length of the perpendicular from the point (2, 3) on 2x + y −4=0. 3 Ans. 1: 15 and the given point is on the side remote from the origin. Ex. 3. Find the lengths of the perpendiculars from each vertex on the opposite side of the triangle (2, 1), (3, − 2), (— 4, − 1). Ans. 2 √(2), √(10), 2 √(10), and the origin is within the triangle. Ex. 4. Find the length of the perpendicular from (3, − 4) on 4x + 2y − 7, the angle between the axes being 60°. Ans. and the point is on the side next the origin. Ex. 5. Find the length of the perpendicular from the origin on 35. To find the equation of a line bisecting the angle between two lines, x cosa+y x cosay sina-p=0, x cosẞ+y sinẞ-p=0. We find the equation of this line most simply by expressing algebraically the property that the perpendiculars let fall from any point xy of the bisector on the two lines are equal. This immediately gives us the equation x cosa+y sina-p=±(x cosß + y sinß—p'), since each side of this equation denotes the length of one of those perpendiculars (Art. 34). |