31. To find the coordinates of the point of intersection of two right lines whose equations are given. Each equation expresses a relation which must be satisfied by the coordinates of the point required; we find its coordinates, 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 coordinates. 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 coordinates, 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 coordinates of the vertices of the triangle the equations of whose sides are x + y = 2; x − 3y = 4; 3x + 5y + 7 = 0. Ans. (-1, -11), (Y, — Y), (4, − 1). Ex. 2. To find the coordinates of the intersections of Ex. 3. Find the coordinates of the intersections of Ans. (4, 4), (- A, 1{), (− 1, Y). 2x + 3y = 13; 5x − y = 7; x − 4y+ 10 = 0. Ans. They meet in the point (2, 3). Ex. 4. Find the coordinates of the vertices, and the equations of the diagonals, of the quadrilateral the equations of whose sides are 2y-3x=10, 2y + x = 6, 16x − 10y= 33, 12x+14y+ 29 = 0. Ans. (-1, 1), (3, 1), (4, — 4), (−3, }); 6y − x = 6, 8x + 2y + 1 = 0. In using this and other similar formulæ, which we shall afterwards have occasion to employ, the learner must be careful to take the coordinates in a fixed order (see engraving). For instance, in the second member of the formula just given, y, takes the place of y1, 3 of x2, and x, of z. Then, in the third member, we advance from y1⁄2 to y, from z to z, and from r, to r, always proceeding in the order just indicated. E Ex. 5. Find the intersections of opposite sides of the same quadrilateral, and the equation of the line joining them. Ans. (83, 25o), (— Y, W), 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 coordinates being the base of a triangle and the bisector of the base, form the equations of the two bisectors of sides, and find the coordinates of their intersection. Let the coordinates 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, 1). Ex. 8. Two opposite sides of a quadrilateral are taken for axes, and the other two are z y =1; find the coordinates of the middle points of diagonals. Ans. (a, b), (a', b). Ex. 9. In the same case find the coordinates of the middle point of the line joining the intersections of opposite sides. 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 ገቢ 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 A (y-y') B(x − x'), = 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 side of the same triangle. The coordinates of the middle points being (− 1, − 4), (− 1, 0), (2, − 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 + 25 = 0, x - 4y = 21; intersecting in (§3, – 14). 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 + 16 = 0, x − 4y = 7; intersecting in (—, — §}). Ex. 5. To find in general the equations of the perpendiculars from the vertices on the opposite sides of a triangle, the coordinates of whose vertices are given. Ans. (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. − y"") y = }} (x''2 − x'''2) + § (y''2 − y'''?), Ans. (x" - x") x + (y′′ 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 coordinates of their intersection. The coordinates of the vertex are now (0, y'), and of the base angles (x", 0), (− x'', 0). Ans. x” (x − x′′) + y'y = 0, x′′ (x + x) — s'y = 0, (0, ""). Ex. 8. Using the same axes, find the equations of the perpendiculars at the middle points of sides, and the coordinates of their intersection. Ans. 2 (x'”x+y'y)=y′2—x'''2, 2 (x′′x—y'y)=x" — y'2, 2x=x” — x′′, y'2-x'x 2 2y' Ex. 9. Form the equation of the perpendicular from x'y' on the line x cos a+ysin a=p; and find the coordinates of the intersection of this perpendicular with the given line. Ans. (x + cosa (p − x' cosa - y' sin a), y' + sin a (p-x' cos a - y' sin a)}. -- Ex. 10. Find the distance between the latter point and x'y'. Ans. (px' cos ay sina). 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 coordinates 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. N R P T We have already indicated (Ex. 9 and 10, Art. 32) one way of solving this question, and we wish now to shew 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 = x' cosa. Again, since SQKB, and QK=y', hence RT=QS=y' cosß; x' cosa+y' cos3 = OR. Subtract OP, the perpendicular from the origin, and K M 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' cos B; 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 cos a + y cos B-p=0, · and the length of the perpendicular from any point x'y' = Ax' + By + C (Ax' + By + C) sin w or " 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 coordinates 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 √5 Ans. 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 cos a + y sin a-p=0, x cos B+ y sin B-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 cos ay sina-p=±(x cos B + y sin ẞ −p'), since each side of this equation denotes the length of one of those perpendiculars (Art. 34). |