Ex. 19. Show that 3y2 - 8xy-3x2+30x-27=0 represents two straight lines at right angles to one another. If we solve for y we shall find y=' 8x+ (10x-18) or y=3x-3, and y=x+3. And mm, +1=3 (−3)+1=0, the lines are at right angles to each other. The given equation may be written in the form of factors; but it is not easy to determine the factors in this case by simple inspection. Thus we have (y-3x+3) (3y+x-9)=0. Ex. 20. Prove that the equation 2y2 3xy 2x2-y+2x=0 represents two straight lines which are perpendicular to each other; and that one of them passes through the point (1, 2) and the other through the point (2, − 1). We leave this as a simple exercise for the student. Ex. 21. Find what the equations y = mx+b...(1), y - Y1 = m(x-x1)...(2), represent when m is indeterminate. (1) May be written (y-b) - mx=0, and on the given condition will represent a series of straight lines passing through the intersection of the lines y-b=0, x=0; that is, through the point (0,b), which is the intercept on the axis of y, when we consider the form of equation y=mx+b. (2) May be written y-y-m(x-x1)=0, and on the same condition of m being indeterminate will represent a system of straight lines passing through the intersection of the lines y −y1=0, x-x1=0; that is, through the point (x11). THE DETERMINATION OF LOCI. 34. In the above examples the student will observe that we have been engaged in the analysis of certain given equations in order to see what loci they represent. On the other hand, problems are often proposed in which, from given geometrical conditions, we are required to find the equations which will represent the loci implied in those conditions. This is a very important and useful branch of study, and we would earnestly entreat the student to give some attention to the method of representing loci by means of equations. In order that he may understand the kind of problem to which we refer, we will present a few exercises on which he may try his hand after he has made a little progress with his studies. We may, however, previously observe that we are virtually solving problems of this kind when we are finding the common forms of equations to the straight line and the circle. The only difference is that in these cases we assume beforehand what the loci are, and then proceed to find the equations which will represent these loci. For example, instead of saying, Find the equation to the circle, we may say, Find the locus of a point which moves so that its distance from a given point is constant. In this case we already know what the locus is. But suppose we did not; we should then proceed from the given geometrical condition to find in the usual way, but with the wording slightly different, the equation to the given locus. In this case the equation would be, if we used rectangular axes, (x-a)2 + (y = b)2=c2, where a, b, are the co-ordinates of the given point, x, y, the co-ordinates of any point in the required locus, and c the constant given distance. Having obtained the equation, we then determine from its form what it represents. In the above instance we know that it is the equation of a circle. We will give one other illustration: Find the locus of a point the distances of which from two given points are in a constant ratio. We leave the reader to draw the figure for himself. Let O, A be the two given points, P any point on the locus. Taking O as the origin of co-ordinates, and OA as the direction of the axis of x, let OA=a, and let (xy) be the co-ordinates of P. Let m be the constant ratio of OP to AP. Then by definition OP=mAP, and .. OP2= m2. AP2. Now, OP2=x2+y2 and AP2= (x− a)2+y2 ; (1 − m2) (x2 + y2) + 2am2x - m2a2=0. The experienced student can at once perceive that this is the form of equation to a circle, and that the locus is a circle of which the centre is on the axis of x. Observe, if m=1 the locus will be the axis of y. In all such problems there are two things to be done. (1) To find the equation which will represent the data. (2) To ascertain from the simplified form of the equation the locus denoted by it. Ex. 1. AB and BC are two straight lines at right angles to each other; A is a fixed point, B moves along a given straight line, and AB to BC is a given ratio. Determine the locus of C. a By taking the axis of y as the straight line along which B moves, and using rectangular axes, the equation found will be of the form y= -nx + where x and y are the co-ordinates of C, n is the given ratio, and a the distance between A and the origin on the axis of x. n It is obvious that this equation represents a straight line. Hence the locus of C is a straight line. Ex. 2. A line is drawn parallel to the base of a triangle, and from the extremities of this line, lines crossing each other in a point P, are drawn to the extremities of the base. Prove that the locus of P is a straight line from the vertex to the middle of the base. Ex. 3. ABC is a triangle, P a point such that the sum of its distances from the three sides is constant. Find the locus of P. Suppose s the sum of the distances, and that the equations to the sides are xcosa+ysina=p, xcosa' +ysina'=q, xcosa" + ysina"=r. Let x and y be the co-ordinates of P. Then the distances of P from the sides are xcosa + ysina - p, xcosa'+ysina' -q, xcosa"+ ysina" — r. Add these expressions together and put them equal to s, then we get x (cosa + cosa + cosa") + y (sina + sina '+ sina") = p + 9 +r+8. This is obviously the equation to a straight line, which is, therefore, the locus of P. Ex. 4. Given BC, the base of a triangle, =b, and the difference of the squares of its sides=d2. Determine the locus of the vertex. Taking the base as the axis of x, and its middle point as the origin, we shall find x= the equation to a straight line parallel to the axis of y. d2 2a' Ex. 5. The base and vertical angle of a plane triangle being given, to find the locus of the intersections of straight lines from the vertices to bisect the opposite sides. The locus is a circle. Ex. 6. From a given point in the circumference of a circle chords are drawn. Find the locus of their middle points. The locus is a circle of half the diameter. Ex. 7. Given the base and the sum of the squares of the other two sides of a triangle. Find the locus of the vertex. The locus is a circle whose centre is at the middle of the base, and the square of its radius is half the sum of the squares of its sides less the square of half the base. Ex. 8. The base and vertical angle of a triangle being given, prove that the locus of the intersection of the perpendiculars from the vertices to the opposite sides is a circle. Ex. 9. In one plane the locus of the point from which two given unequal circles would appear equally large, that is, subtend equal angles, is a circle. The equation will be of the form x2+y2-2x arri = 0, where r, r ̧ are the radii of the unequal circles, and a the distance between the centres. Ex. 10. To find the locus of the intersection of two straight lines which pass each through a given point and contain a given angle. The equation found will be of the form x2+y2-2 ay cota=a*, where a is half the distance between the two points and a the given angle. Ex. 11. A point moves so that the sum of the squares of its distances from the four sides of a square is constant. Show that the locus of the point is a circle. Ex. 12. A point moves so that the sum of the squares of its distances from the sides of an equilateral triangle is constant. Show that the locus of the point is a circle. Ex. 13. To find the locus of the middle points of a system of parallel chords in a circle. The locus is a straight line through the centre perpendicular to the chords. Ex. 14. The locus of a point the algebraic sum of whose distances from the sides of a polygon is constant is a straight line. Ex. 15. The locus of a point the distances of which from two fixed lines are in a given ratio is a straight line. Ex. 16. ABC is a triangle, AB : AC is a given ratio; if B moves along a straight line so does C. 35. The reader will observe that in order to preclude the necessity of referring again to the subject of the determination of loci, we have anticipated in some of the above problems what is implied in the propositions yet to be given under the head of THE CIRCLE. "In attempting to solve these problems algebraically, the student will find that much depends upon a judicious selection of the origin and axes, and the application of the proper equations and formulæ. He should in every case consider the problem before he attempts the solution, and form a definite plan before he begins. He may very possibly not be able to carry out his original scheme, but his attempts to do so will probably suggest some method by which he may solve the problem; and he will, at any rate, avoid a practice very common to beginners, of working without any definite aim, and consequently introducing and combining equations and formulæ that only serve to embarrass him, without in any way aiding him in the solution."* 36. We would recommend the advanced student to investigate the principles on which the following general problems are solved :— (1) To find the angle between the staight lines represented by the equation Ax2+Bxy + Cy2 = 0, the axes being rectangular. (2) To find the equation to the straight lines which bisect the angles between the straight lines Ax2+Bxy+Cy2=0, the axes being rectangular. * G. H. 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