Loci (Locus and Its Constructions) — Important Questions

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

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

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 $4\,\text{cm}$ centred at $O$.

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

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

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

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

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

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

Steps of construction:

1. Draw $BC=6\,\text{cm}$. At $B$ construct $\angle B=60^{\circ}$ and cut off $BA=5\,\text{cm}$; join $AC$ to complete $\triangle ABC$.
2. The locus of points equidistant from lines $BA$ and $BC$ is the bisector of $\angle B$. Bisect $\angle ABC$ and draw the bisector.
3. The locus of points equidistant from $B$ and $C$ is the perpendicular bisector of $BC$. Construct it.
4. Their intersection is the required point $P$; it satisfies $PB=PC$ and lies equidistant from $BA$ and $BC$.

The point $T$ must satisfy two conditions.

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

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

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

Steps of construction:

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

Since $AB=7\,\text{cm}$, the perpendicular bisector passes at a distance $3.5\,\text{cm}$ from $A$, which is greater than the radius $3\,\text{cm}$. 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 $BC=7\,\text{cm}$.
2. At $B$ construct $\angle ABC=60^{\circ}$ using compasses (draw an arc, then step the radius once to mark $60^{\circ}$).
3. Along this arm cut off $BA=6\,\text{cm}$; join $AC$ to complete $\triangle ABC$.
4. Locus 1 — equidistant from $AB$ and $BC$: bisect $\angle ABC$; the bisector is the required locus.
5. Locus 2 — equidistant from $B$ and $C$: construct the perpendicular bisector of $BC$.
6. The two loci intersect at the required point $P$.
7. Join $PB$ and measure it; on an accurate figure $PB\approx 3.7\,\text{cm}$ (and $PB=PC$).

Steps of construction:

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

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

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

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

Take $PQ$ and $SR$ as the two long sides ($90\,\text{m}$ apart is $QR=60\,\text{m}$; the perpendicular distance between $PQ$ and $SR$ equals $QR=60\,\text{m}$).

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

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

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