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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.

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Let P be any point such that PA is equal to PB,

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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:

it is required to find the locus of a point equidistant from AB and CD.

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Let P be any point such that PM, the perpendicular on AB, is equal to PN, the perpendicular on CD,

join OP.

Then in the right-angled triangles OPM, OPN,

the side PM is equal to the side PN,

and the side OP is common to both,

therefore the angle POM is equal to the angle PON,

I. 20, Cor. 1.

therefore P is on one of the bisectors of the angles between AB and CD.

Also every point on these bisectors is equidistant from AB

and CD.

Let Q be such a point,

draw QR perpendicular to AB,

and QS perpendicular to CD.

Then in the triangles QRO, QSO,

the angle QRO is equal to the angle QSO, the angle QOR is equal to the angle QOS,

and the side QO is common to both, therefore QR is equal to QS.

I. 19.

Hence the bisectors of the angles between AB and CD are

the locus of a point equidistant from AB and CD.

EXERCISES.

103. Find the locus of a point at a given distance from the circumference of a given circle.

104. Find the locus of the vertex of an isosceles triangle on a given base.

105. Find the locus of the middle point of a straight line drawn from a given point to a given unlimited straight line which does not pass through the point.

106. Given the sum (or the difference) of the distances of a point from two intersecting and unlimited straight lines, find its locus.

INTERSECTION OF LOCI.

It follows from Def. 44 that if X is the locus of a point satisfying the condition A, and Y the locus of a point satisfying the condition B; then the intersections of X and Y, and these points only, satisfy both the conditions A and B.

i. To find a point equidistant from three given points not in the same straight line.

Let A, B, C be three given points not in the same straight line,

it is required to find a point equidistant from A, B and C.

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Draw the straight line DD' bisecting AB at right angles.

and EE' bisecting BC at right angles.

Prob. 4.

Prob. 4.

Then because every point equidistant from A and B lies on

DD', Locus iii. and every point equidistant from B and C lies on EE', Locus iii. therefore any point equidistant from A, B and C must lie on DD' and on EE'.

Now DD' and EE' intersect, since if they were parallel, AB and BC, which are perpendicular to them, would lie in one straight line;

let them intersect in O,

then a point O has been found equidistant from A, B and C.

Also, because DD' and EE' can intersect in one point only,
Ax. 2.

therefore O is the only point equidistant from A, B and C.

ii. To find a point equidistant from three given straight lines which intersect one another, but not in the same point.

Let AA', BB', CC' be three given straight lines which intersect so as to form a triangle DEF :

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