By J. M. Aroca, R. Buchweitz, M. Giusti, M. Merle
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Additional info for Algebraic Geometry, la Rabida, Spain 1981: Proceedings
43. P, Q, , Find the condition that the point 45. joining (8,4) (-5, , should (x, y) lie on the straight line 2). ABCD is a rectangle, P any point in the plane. PA* Prove + PC* = PB2 + P/)a 47. are (2, 3), (10, 15). Show that the straight line PQ passes through the origin. 46. Find the coordinates of a point R such that RQ = %QP PR/RQ. 48. The points C, D divide the line AB harmonically. P; is the mid-point of AB, OA* = 00 (ii) if (iii) 2/C/)= l/BD+1/AD. 49. The coordinates of four points aro (1, 3), (3, 5), (4, 6), .
E. the locus 29 ^) satisfies the 0, on the axis of y it is therefore the point where the locus and Let this point be Q, represented by this equation cuts the axis of y. lies and : for convenience let k Q Clearly ^ = --. can be positive or negative ; if due regard the signs the following argument holds whether tity positive or negative. (b) we is paid to take this quan- Rectangular axes. y Suppose P (Xj Join and draw QA, PQ y) is any other point whose coordinates PA satisfy parallel to the coordinate axes, then The given equation can be written C A or "B 9 AP_ _4 hence QA~~ consequently for all positions of B P the AP ratio - 1 , is constant, hence (JA.
Find y = \/3# the length is to drawn through line, will illustrate the : we shall the point P('2, 3) parallel to meet the straight line 2x + iy = 27 in the PQ. r-axis straight line y of the equation straight line through (2, 3) parallel to it is The : hence the . , y) and the fixed point THE EQUATION OF THE FIRST DEGREE Now (2, 3). r-f ; r Hence 2 +4 2 r-f \ ^ J ^ / r+3 = hence - r is Ex. e. which 45 246, the required length of PQ. ii. To find the perpendicular distance Ax + By + C = the straight line of the point (a, I) from 0.
Algebraic Geometry, la Rabida, Spain 1981: Proceedings by J. M. Aroca, R. Buchweitz, M. Giusti, M. Merle