Loci (Locus and Its Constructions) — ICSE Class 10 Maths Important Questions

Answer: (b) The perpendicular bisector of $AB$AB.

Every point $P$P with $PA=PB$PA=PB lies on the perpendicular bisector of $AB$AB, and every point of that bisector is equidistant from $A$A and $B$B.

Answer: (b) The pair of bisectors of the angles between them.

Each point on an angle bisector is equidistant from both arms; the two bisectors are perpendicular to each other.

Answer: (b) A circle of radius $4\,\text{cm}$4\,cm centred at $O$O.

All points at a fixed distance $r$r from a fixed point lie on a circle of radius $r$r with that point as centre.

Answer: (c) A pair of lines parallel to $\ell$ at a distance of $3\,\text{cm}$3\,cm.

The centre stays $3\,\text{cm}$3\,cm from the line (the radius), so it lies on either of the two lines parallel to $\ell$ at a perpendicular distance of $3\,\text{cm}$3\,cm.

Answer: (a) Points equidistant from both arms trace the angle bisector, and this is precisely because every bisector point has equal perpendicular distances to the arms; so R correctly explains A.

The tip of the second hand stays at a fixed distance (its length) from the fixed centre, so its locus is a circle whose centre is the pivot and whose radius equals the length of the hand.

A point always $2\,\text{cm}$2\,cm from a fixed line $AB$AB traces a pair of straight lines parallel to $AB$AB, one on each side, at a perpendicular distance of $2\,\text{cm}$2\,cm.

If $\angle APB=90^{\circ}$APB=90^ then $P$P sees $AB$AB as a diameter, since the angle in a semicircle is a right angle.

Hence the locus of $P$P is the circle described on $AB$AB as diameter, i.e. a circle of radius $\dfrac{6}{2}=3\,\text{cm}$6/2=3\,cm with the midpoint of $AB$AB as centre (excluding the points $A$A and $B$B themselves).

(i) The centre is at the same distance (the radius) from $A$A and from $B$B, so it lies on the perpendicular bisector of $AB$AB.

(ii) The centre is at the same perpendicular distance (the radius) from both arms, so it lies on the bisector of $\angle XOY$XOY.

Steps of construction:

1. Draw $BC=6\,\text{cm}$BC=6\,cm. At $B$B construct $\angle ABC=90^{\circ}$ABC=90^ and cut off $BA=8\,\text{cm}$BA=8\,cm. Join $AC$AC (it measures $10\,\text{cm}$10\,cm).
2. Locus 1: points equidistant from $BA$BA and $BC$BC lie on the bisector of $\angle B$B. Construct it.
3. Locus 2: points equidistant from $CB$CB and $CA$CA lie on the bisector of $\angle C$C. Construct it.
4. The bisectors meet at $I$I. Being on both, $I$I is equidistant from all three sides (so it also lies on the bisector of $\angle A$A).
5. Draw $ID\perp BC$ID BC. With centre $I$I and radius $ID$ID, draw the circle. It touches all three sides.
6. Radius $\approx 2\,\text{cm}$2\,cm.

Check: area of $\triangle ABC=\tfrac{1}{2}\times 6\times 8=24\,\text{cm}^2$ABC=12× 6× 8=24\,cm^2 and the semi-perimeter $s=\dfrac{6+8+10}{2}=12\,\text{cm}$s=6+8+10/2=12\,cm, so $r=\dfrac{24}{12}=2\,\text{cm}$r=24/12=2\,cm.

The point $T$T must satisfy two conditions.

Condition 1 (equidistant from the two roads): the locus is the pair of bisectors of the angles between $AB$AB and $CD$CD. As the roads meet at right angles, these bisectors are the two lines through $O$O making $45^{\circ}$45^ with each road.

Condition 2 (a fixed distance $5\,\text{cm}$5\,cm from $O$O): the locus is a circle of radius $5\,\text{cm}$5\,cm centred at $O$O.

The treasure lies where these loci meet. The circle cuts each of the two bisectors in $2$2 points, giving $4$4 possible positions of $T$T, one in each of the four regions formed by the roads.

Steps of construction:

1. Draw $AB=7\,\text{cm}$AB=7\,cm.
2. The locus of a point $3\,\text{cm}$3\,cm from $A$A is a circle of radius $3\,\text{cm}$3\,cm centred at $A$A. With centre $A$A and radius $3\,\text{cm}$3\,cm draw this circle.
3. Construct the perpendicular bisector of $AB$AB (arcs of equal radius from $A$A and $B$B intersecting above and below, joined).

Since $AB=7\,\text{cm}$AB=7\,cm, the perpendicular bisector passes at a distance $3.5\,\text{cm}$3.5\,cm from $A$A, which is greater than the radius $3\,\text{cm}$3\,cm. Therefore the circle does not reach the perpendicular bisector, so the two loci do not intersect (no common point exists).

Steps of construction:

1. Draw $BC=7\,\text{cm}$BC=7\,cm.
2. At $B$B construct $\angle ABC=60^{\circ}$ABC=60^ using compasses (draw an arc, then step the radius once to mark $60^{\circ}$60^).
3. Along this arm cut off $BA=6\,\text{cm}$BA=6\,cm; join $AC$AC to complete $\triangle ABC$ABC.
4. Locus 1 — equidistant from $AB$AB and $BC$BC: bisect $\angle ABC$ABC; the bisector is the required locus.
5. Locus 2 — equidistant from $B$B and $C$C: construct the perpendicular bisector of $BC$BC.
6. The two loci intersect at the required point $P$P.
7. Join $PB$PB and measure it: $PB\approx 4.0\,\text{cm}$PB 4.0\,cm (and $PB=PC$PB=PC).

Check: $P$P is on the perpendicular bisector of $BC$BC, so it is $3.5\,\text{cm}$3.5\,cm along $BC$BC from $B$B, and on the bisector of $\angle B$B, which makes $30^{\circ}$30^ with $BC$BC. So $PB=\dfrac{3.5}{\cos 30^{\circ}}=\dfrac{7}{\sqrt{3}}\approx 4.04\,\text{cm}$PB=3.5 30^=7/√3 4.04\,cm.

Steps of construction:

1. Draw $AC=8\,\text{cm}$AC=8\,cm.
2. With centre $A$A and radius $5\,\text{cm}$5\,cm draw arcs above and below $AC$AC; with centre $C$C and radius $5\,\text{cm}$5\,cm cut them at $B$B (above) and $D$D (below).
3. Join $AB, BC, CD, DA$AB, BC, CD, DA to complete rhombus $ABCD$ABCD.

Locus (i): points equidistant from sides $AB$AB and $AD$AD lie on the bisector of $\angle A$A. In a rhombus the diagonal $AC$AC bisects $\angle A$A, so this locus is the diagonal $AC$AC itself.

Locus (ii): points equidistant from $A$A and $C$C lie on the perpendicular bisector of $AC$AC, which is the other diagonal $BD$BD (the diagonals of a rhombus bisect each other at right angles).

Proof of a common point: locus (i) is line $AC$AC and locus (ii) is line $BD$BD; the diagonals of a rhombus intersect at their common midpoint $O$O. Hence both loci pass through $O$O, the point of intersection of the diagonals, which is therefore equidistant from $AB$AB and $AD$AD and also equidistant from $A$A and $C$C.

$PQ$PQ and $SR$SR are the longer sides ($90\,\text{m}$90\,m each), and the perpendicular distance between them equals $QR=60\,\text{m}$QR=60\,m.

(i) The locus of points equidistant from the parallel sides $PQ$PQ and $SR$SR is the line midway between them, parallel to both, i.e. the horizontal centre-line of the park.

(ii) The locus of points equidistant from the corners $P$P and $Q$Q is the perpendicular bisector of $PQ$PQ, a line through the midpoint of $PQ$PQ perpendicular to it.

(iii) One locus runs parallel to $PQ$PQ and the other runs perpendicular to $PQ$PQ; two lines with different directions meet in exactly one point, so the two conditions fix a single position for $T$T (the centre of the park).

(iv) $T$T lies on the mid-line between $PQ$PQ and $SR$SR, so its distance from $PQ$PQ is half of $60\,\text{m}$60\,m, that is $\dfrac{60}{2}=30\,\text{m}$60/2=30\,m.