99. Construct a triangle having given the base, the opposite angle, and the difference of the remaining sides, 100. Construct a triangle having given the base, the difference of the base angles, and the difference of the remaining sides. 101. Construct a triangle having given the perimeter and the angles. 102. Construct a right-angled triangle, having given the hypote nuse and the perpendicular upon it from the right angle. SECTION V PLANE LOCI It may happen that the conditions given for the determination of a point do not suffice to fix its position absolutely, but are sufficient to limit it to some line, part of a line, or group of lines. In such cases the point is said to have a locus. DEF. 44. If any and every point on a line, part of a line, or group of lines (straight or curved), satisfies an assigned condition, and no other point does so, then that line, part of a line, or group of lines, is called the locus of the point satisfying that condition. Hence, in order that a line or group of lines X may be properly termed the locus of a point satisfying an assigned condition A, it is necessary and sufficient to demonstrate the following pair of associated theorems : If a point satisfies A, it is upon X, Instead of (1) we may prove that (1) (2) F If a point is not upon X, it does not satisfy A, and instead of (2) that If a point does not satisfy A, it is not upon X LOCUS i. To find the locus of a point at a given distance from a given point. Let A be the given point, B the given distance: it is required to find the locus of a point at distance B from A. B D P With centre A, and with radius equal to B, draw the circle CDE: the circumference CDE shall be the locus required. Take any point P on the circumference CDE, join AP, then AP is a radius of the circle CDE, and is therefore equal to B. Take any point Q not in the circumference CDE, Def. 41. join AQ, and let AQ, produced if necessary, meet the circumference in R, since Q and R do not coincide, AQ is not equal to AR, but AR is equal to B, therefore AQ is not equal to R. Ax. a. Def. 41. Hence the circumference CDE is the locus of a point at a distance from A equal to B. LOCUS ii. To find the locus of a point at a given distance from a given straight line. Let D be the given distance, and AB the given straight line: it is required to find the locus of a point at the distance D from AB. Let Q be a point at the distance D from AB. Draw AP perpendicular to AB on the same side of AB as Q. and equal to D. Draw a straight line through P and Q, and draw QM perpendicular to AB. Then QM is equal to D, and therefore to PA, and it is perpendicular to AB, and therefore parallel to PA. Hence PQ is parallel to AM, I. 22 Cor. I. 31, therefore Q lies upon the straight line drawn through P parallel to AB. AB. Also every point on this line is at the same distance from Take any point R upon the line PQ, and draw RS perpendicular to AB. Then RS is parallel to PA, and RSAP is a parallelogram, therefore RS is equal to PA, therefore RS is equal to D. I. 22, Cor. I. 29. Hence a straight line through P parallel to AB is the locus of a point, on the same side of AB as P, at a distance from AB equal to D. Similarly, by drawing AP' in a similar manner on the other side of AB, the line P'Q' through P' parallel to AB will be the locus of a point, on the other side of AB, at a distance from AB equal to D. Hence the locus of a point at a distance from AB equal to D is the pair of parallel straight lines PQ and P'Q'. LOCUS iii. To find the locus of a point equidistant from two given points. Let A, B be the two given points: it is required to find the locus of a point equidistant from A and B. M Let P be any point such that PA is equal to PB, and the side AP is equal to the side BP, therefore the angle AMP is equal to the angle BMP, therefore they are right angles, Hyp. I. 18. therefore P is on the straight line drawn through M perpendicular to AB. Also every point on this straight line is equidistant from A and B. Let Q be any point on MP, join QA and QB. Then in the triangles AMQ, BMQ the side AM is equal to the side BM, the side MQ is common to both, and the angle AMQ is equal to the angle BMQ, since they are both right angles, therefore QA is equal to QB. I. 5. Hence the straight line drawn through M perpendicular to AB is the locus of a point equidistant from A and B. LOCUS iv. To find the locus of a point equidistant from two intersecting straight lines. Let AB and CD be two straight lines intersecting at 0: |