Class 10 Mathematics Chapter 16 Loci (Locus And Its Constructions)

Key Concept: Locus of point equidistant from intersecting lines, Angle bisector theorem

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that any point $P$ equidistant from two intersecting lines $AB$ and $CD$ lies on their angle bisector. This is true based on Theorem 3, which proves that the locus of such points is indeed the angle bisector. The Reason correctly explains the Assertion, as it reiterates the theorem's conclusion that the angle bisector is the set of all points equidistant from the two lines.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Theorems Based on Symmetry

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion (A) states that the locus of a point equidistant from two intersecting lines is the perpendicular bisector, which is incorrect. According to Theorem 3, the correct locus is the angle bisector of the angles formed by the intersecting lines. The Reason (R) correctly explains the concept as per Theorem 3, stating that the angle bisector is the set of all points equidistant from both lines. Therefore, Assertion is false, but Reason is true.

Key Concept: Theorem on Locus of Points Equidistant from Two Lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion (A) states Theorem 3: The locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines, which is correct as proven in the theorem. The Reason (R) states that a point lying on the angle bisector is equidistant from both lines, which is also true and correctly explains why the assertion holds. Hence, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Locus of equidistant points, Perpendicular bisector properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Theorem 4, any point equidistant from two fixed points lies on the perpendicular bisector of the line segment joining them. Conversely, every point on the perpendicular bisector is equidistant from the two fixed points. Therefore, Assertion (A) is true as it directly follows from the property of perpendicular bisectors, and Reason (R) correctly explains why the Assertion holds true, as it states the fundamental theorem about the locus of points equidistant from two fixed points.

Key Concept: Locus of Equidistant Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $AB$. This is true because, by the theorem, any point $P$ on the perpendicular bisector will satisfy $PA = PB$. The reason correctly explains the assertion as it describes the defining property of the perpendicular bisector—that every point on it is equidistant from $A$ and $B$.
To verify:
1. Let $O$ be the midpoint of $AB$.
2. For any point $P$ on the perpendicular bisector, triangles $AOP$ and $BOP$ are congruent (by S.S.S. since $AO = BO$, $PO$ is common, and both have right angles at $O$).
3. Therefore, $PA = PB$ (corresponding parts).
Thus, both the Assertion and Reason are true, and the Reason explains the Assertion.

Key Concept: Perpendicular Bisector

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states a property of the perpendicular bisector: any point lying on it is equidistant from the endpoints of the segment. This is true as per Theorem 4. The reason provides the definition of the perpendicular bisector as the locus of points equidistant from two given points, which is also correct and explains why the assertion holds. Therefore, both are true, and the reason correctly explains the assertion.

Key Concept: Locus of a point, Angle bisector

d) Assertion is false, but Reason is true.
[Solution Description]
Step 1: The assertion states that the locus of a point equidistant from two intersecting lines is a circle. However, this is incorrect. According to the syllabus, the correct locus for such points is the angle bisector of the angle formed by the two lines.
Step 2: The reason correctly states that the locus of a point equidistant from two intersecting lines lies on the angle bisector, which aligns with the syllabus.
Conclusion: The assertion is false, but the reason is true because the locus is not a circle but rather an angle bisector.

Key Concept: Locus of a point equidistant from a fixed point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a circle is the locus of all points at a fixed distance from a given center point, which is correct based on the definition of a circle. The reason explains that the radius is the constant distance between the center and any point on the circumference, which is also true. Moreover, the reason correctly explains why the assertion is true, as the fixed distance (radius) defines the circle.
Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Basic Construction of a Circle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion (A) states the fundamental definition of a circle, which is indeed the locus of all points equidistant from a fixed center. The Reason (R) correctly describes that this fixed distance is called the radius. Therefore, both statements are true, and the Reason accurately explains the Assertion.

Key Concept: Locus of point equidistant from two intersecting lines and angle bisectors

d) Assertion is false, but Reason is true.
[Solution Description]
First, analyze the Assertion: The statement claims that for an equilateral triangle, the locus of points equidistant from its sides is a single straight line.
Next, analyze the Reason: It states that in an equilateral triangle, the angle bisectors coincide with the perpendicular bisectors of the sides.
The key concept here is the properties of an equilateral triangle. In such a triangle:
1. All three angle bisectors, medians, altitudes, and perpendicular bisectors coincide and meet at a single point (the centroid).
2. The locus of points equidistant from the sides of a triangle is the angle bisector(s). Since the triangle is equilateral, all three angle bisectors coincide and form a single point, not a straight line.
Thus:
- Assertion is false because the locus reduces to a single point, not a line.
- Reason is true as it correctly describes the property of an equilateral triangle.
Therefore, the correct evaluation is "Assertion is false, but Reason is true."

Key Concept: Locus of a Point Equidistant from Two Lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two parallel lines is a single line parallel to them and equidistant from both. From the syllabus, this is true as it matches the construction rule for such a locus. The reason correctly explains why the assertion is true, as it describes how all points on the constructed line satisfy the equidistant condition. Both are true, and the reason justifies the assertion.

Key Concept: Locus of a Point Equidistant from Two Fixed Points

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points is the angle bisector, which is incorrect. The correct statement is that it is the perpendicular bisector of the line segment joining the two points. The reason correctly states that the perpendicular bisector divides the line segment into two equal parts, but it does not explain the assertion since the assertion itself is false.
Thus, the assertion is false, but the reason is true.

Key Concept: Locus of Points, Equidistant Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of the center of a circle passing through two fixed points A and B is the perpendicular bisector of AB. This is true because the center must be equidistant from both A and B to form a circle passing through them. The reason states that all points on the perpendicular bisector of AB are equidistant from A and B, which is a property of perpendicular bisectors. Therefore, the reason correctly explains the assertion.

Key Concept: Locus of a Point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points A and B is the perpendicular bisector of AB. This is true because, by definition, the locus is the path satisfying this condition.
The reason states that every point on the perpendicular bisector of AB is equidistant from A and B. This is also true, as it follows directly from the properties of a perpendicular bisector.
Moreover, the reason correctly explains why the assertion is true: Since all points on the perpendicular bisector satisfy the condition of being equidistant from A and B, the perpendicular bisector is indeed the required locus.

Key Concept: Locus Definition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion (A) states that the locus of a point at a fixed distance from a given fixed point is a circle with the fixed point as its center. This is correct because, by definition, a circle is the set of all points in a plane that are at a fixed distance (radius) from a given point (center).
Reason (R) states that every point on the circumference of a circle is equidistant from its center. This is also correct and directly explains why Assertion (A) is true.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Key Concept: Locus of equidistant points, Locus in rhombus

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the locus of a point in rhombus ABCD equidistant from AB and BC is the angle bisector of $\angle ABC$. This is true because in a rhombus, all sides are equal, and the angle bisector of any angle will indeed be the set of points equidistant from the two adjacent sides.
The reason claims that in any quadrilateral, the set of points equidistant from two adjacent sides lies on the angle bisector of the angle formed by those sides. This is false because this property is specific to rhombuses (and certain other special quadrilaterals), not all quadrilaterals. For example, in a general trapezium, the angle bisector does not necessarily consist of points equidistant from the two adjacent sides.
Therefore, the Assertion is true, but the Reason is false.

Key Concept: Locus of Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point at a constant distance (4 cm) from a fixed point in space forms a sphere centered at that fixed point with radius 4 cm. This is correct because, by definition, a sphere is indeed the collection of all points in three-dimensional space that are equidistant from a central point. The reason provides the general definition of a sphere, which directly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Locus of points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- **Assertion (A):** This states that the locus of a point always 4 cm away from a fixed point in space is a sphere.
- **Reason (R):** This defines a sphere as the set of all points in space equidistant from a fixed center.
- **Analysis:** The Assertion correctly describes the locus, and the Reason provides the correct definition of a sphere. Hence, the Reason explains why the Assertion is true.

Key Concept: Angle Bisectors and Equidistant Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that point A, where the angle bisectors of $\angle L$ and $\angle N$ meet, is equidistant from all three sides of triangle LMN. This is true because the point of intersection of any two angle bisectors in a triangle is the incentre.
The Reason correctly explains that the incentre is equidistant from all sides of the triangle as it is the center of the incircle. Since both statements are true and the Reason correctly justifies the Assertion, the correct answer is option (a).

Key Concept: Angle Bisectors and Equidistant Points

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the point where the angle bisector of $\angle B$ and the perpendicular bisector of $AC$ meet is equidistant from all three vertices. This is false because the angle bisector does not necessarily pass through the circumcenter. The reason is also false because the circumcenter is the point of intersection of the perpendicular bisectors, not the angle bisector. Hence, the correct option is (d).

Key Concept: Bisectors and Equidistant Points

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the point of intersection of angle bisectors (incenter) is equidistant from all three sides of the triangle, which is true as per the properties of angle bisectors in a triangle. The Reason explains the basic definition of an angle bisector, but it does not explain why the incenter is equidistant from all three sides. Therefore, both statements are true, but the Reason is not the correct explanation for the Assertion.

Key Concept: Locus of points in a circle, Angle subtended by diameter

c) Assertion is true, but Reason is false.
[Solution Description]
According to Thales' theorem, if a triangle is inscribed in a semicircle such that one side coincides with the diameter, then the angle opposite to the diameter is always $90^\circ$. Thus, the assertion is correct as it describes the locus of points forming right angles with A and B. However, the reason states that the angle subtended by the diameter is $180^\circ$, which is false. The angle subtended by the diameter at the circumference is $90^\circ$, not $180^\circ$. Therefore, the assertion is true, but the reason is false.

Key Concept: Locus of Mid-Points of Parallel Chords in a Circle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of mid-points of all parallel chords in a circle is the diameter perpendicular to the given chords. This is true because any chord's mid-point lies on the diameter perpendicular to it due to the property of equal arcs subtended by equal chords. The reason claims that the mid-points of equal chords lie on a concentric circle, which is also true but does not explain why the locus is a diameter. Hence, both statements are correct, but the reason does not justify the assertion.

Key Concept: Locus of a point at a fixed distance from a given point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point moving in a plane at a constant distance (5 cm) from a fixed point forms a circle with radius 5 cm. This is correct because, by definition, a circle is indeed the collection of all points in a plane that are equidistant from a fixed point (the center). The reason correctly defines a circle as the set of all points in a plane at a given distance from a fixed point, which directly explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Angle between two lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine the angle between the lines, first find their slopes:
For the first line $3x + 4y = 5$, solve for y: $4y = -3x + 5$ $\Rightarrow$ $y = -\frac{3}{4}x + \frac{5}{4}$. Slope ($m_1$) is $-\frac{3}{4}$.
For the second line $4x - 3y = 7$, solve for y: $-3y = -4x + 7$ $\Rightarrow$ $y = \frac{4}{3}x - \frac{7}{3}$. Slope ($m_2$) is $\frac{4}{3}$.
Now, check if the product of the slopes is $-1$: $m_1 \times m_2 = -\frac{3}{4} \times \frac{4}{3} = -1$.
Since the product of the slopes is $-1$, the lines are perpendicular, meaning the angle between them is indeed $90^\circ$.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Angle between two lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if the angle between two lines is $90^\circ$, then the product of their slopes is $-1$. This is true because, for perpendicular lines, the product of their slopes is indeed $-1$ (assuming neither line is vertical or horizontal).
The reason states that two lines are perpendicular if and only if the product of their slopes is $-1$. This is also true, as it is a standard result in coordinate geometry. Moreover, the reason correctly explains the assertion since the condition given in the assertion arises from the perpendicularity condition stated in the reason.

Key Concept: Basic Angle Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is correct because two perpendicular lines always form a $90^\circ$ angle when they intersect.
The reason is also correct as two adjacent angles being equal implies each must be $90^\circ$, confirming perpendicularity.
Additionally, the reason logically explains why the assertion holds.

Key Concept: Locus of Points in Relation to a Fixed Angle

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that for any angle $\theta$ between $0^\circ$ and $180^\circ$, the locus of P such that $\angle APB = \theta$ is a circle. This is true because, by the Inscribed Angle Theorem, all points P subtending the same angle $\theta$ on chord AB lie on the circumference of a circle. However, the reason given is incomplete because it does not specify the condition that the angle must be constant ($\angle APB = \theta$). The correct explanation would involve the property that for a fixed chord AB and a fixed angle $\theta$, the locus of P is indeed a circular arc (or pair of arcs), but the Reason omits the critical detail about the angle being fixed. Thus, Assertion is true, but Reason is false.

Key Concept: Locus of Points in Relation to a Fixed Angle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of point $P$ is the circumference of the circle with $AB$ as diameter when angle $APB = 90^\circ$. This is true because, according to Thales' theorem, the angle subtended by the diameter of a circle at any point on its circumference is a right angle. Therefore, every point $P$ on the circumference satisfies angle $APB = 90^\circ$, making the locus the circumference of the circle. The reason correctly explains why the assertion is true by stating the property of angles subtended by the diameter.

Key Concept: Locus of Points in Relation to a Fixed Angle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of point P forming a right angle with two fixed points A and B is the circumference of the circle with AB as diameter. This is correct because, by the Thales' theorem, any angle inscribed in a semicircle is a right angle. The reason provided is also true and correctly explains why the assertion holds. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Incentre and Angle Bisector Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the incentre is the point of intersection of angle bisectors and is equidistant from all sides, which is true by definition. The reason states that an angle bisector divides the opposite side proportionally, which is the Angle Bisector Theorem and is also true. However, the reason does not explain why the incentre is equidistant from all sides, as it only explains a property of a single angle bisector, not their intersection. Hence, the assertion and reason are both true, but the reason does not explain the assertion.

Key Concept: Angle Bisector Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the point where the angle bisectors meet (incenter) is equidistant from all three sides of the triangle. This is a fundamental property of the incenter and is true. The reason states that the angle bisector divides the opposite side in the ratio of the adjacent sides, which is also a correct property (Angle Bisector Theorem). However, this theorem explains how the angle bisector divides the opposite side but does not directly explain why the incenter is equidistant from all sides. Hence, both are true, but the reason does not correctly explain the assertion.

Key Concept: Angle Bisector Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the point where the angle bisectors of a triangle meet is called the incentre, which is true by definition. The reason states that the incentre is equidistant from all the sides of the triangle, which is also true and correctly explains why it is called the incentre. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Locus of Points, Angle Bisector

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that the locus of points equidistant from two intersecting lines is a single straight line. However, according to the theorem, the locus consists of a pair of straight lines which are the angle bisectors of the angles formed by the intersecting lines. Hence, the Assertion is false.
The Reason correctly states that a pair of angle bisectors forms the locus of points equidistant from two intersecting lines. This is in accordance with the theorem. Therefore, the Reason is true.
Since the Assertion is false but the Reason is true, the correct option is (d).

Key Concept: Locus of Points at a Fixed Distance from Two Intersecting Lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the locus of points equidistant from two intersecting lines is the angle bisector, which is true as per the theorem discussed in the syllabus. The Reason correctly explains why this is the case because any point on the angle bisector will be equidistant from both arms of the angle due to congruent triangles formed by perpendicular distances (as shown in the proof). Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Locus of Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the theorem, the locus of a point equidistant from two intersecting lines is indeed the angle bisector of the angles formed by them. Additionally, every point on the angle bisector satisfies the condition of being equidistant from both lines. Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Parallel Lines, Transversals, Angle Properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that if two parallel lines are cut by a transversal, the alternate interior angles are equal. This is true as per the properties of parallel lines and transversals.
The reason states that the sum of the co-interior angles formed by a transversal with two parallel lines is always $180^\circ$. This is also true based on the properties of parallel lines and transversals.
However, the reason does not explain the assertion directly. While both statements are true, the reason provides a different property of parallel lines rather than explaining why alternate interior angles are equal.

Key Concept: Locus of Equidistant Points from Parallel Lines

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To find the locus of a point equidistant from two parallel lines:
1. The given lines are $3x + 4y = 12$ ($l_1$) and $3x + 4y = 24$ ($l_2$). These are parallel because their coefficients of $x$ and $y$ are proportional.
2. The distance between $l_1$ and $l_2$ is calculated as $\frac{|24 - 12|}{\sqrt{3^2 + 4^2}} = \frac{12}{5}$.
3. The locus is a line midway between $l_1$ and $l_2$, so its equation is obtained by taking the average of the constants: $\frac{12 + 24}{2} = 18$. Thus, the required line is $3x + 4y = 18$.
4. The Assertion is true as it correctly states the locus line.
5. The Reason calculates the distance between $l_1$ and $l_2$, which is correct but does not explain why the locus is $3x + 4y = 18$. Therefore, the Reason is true but not the correct explanation of the Assertion.

Key Concept: Conditions for Parallelism

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The slope of the first line $y = 3x + 5$ is $m_1 = 3$. The slope of the second line $y = 3x - 7$ is $m_2 = 3$. Since $m_1 = m_2$, the Assertion is true as they satisfy the condition for parallelism. The Reason states the correct condition for parallelism, which is the equality of slopes, and correctly explains why the lines are parallel.

Key Concept: Assertion-Reason

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the locus is a straight line when the condition given defines a circle. This is false because, by definition, the locus of a point at a constant distance from a fixed point is a circle. The reason correctly defines a circle as the set of points equidistant from a fixed center point, which is true. Therefore, the assertion is false, and the reason is true.

Key Concept: Locus of Points Equidistant from a Fixed Point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point at a fixed distance from a fixed point is a circle, which is true based on the definition of a circle. The reason correctly explains that a circle is indeed the collection of all points in a plane equidistant from a fixed center point. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Locus of Points Equidistant from a Fixed Point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point at a fixed distance of 5 cm from a fixed point $O$ is a circle with radius 5 cm. This is true because the definition of a circle (given in the reason) matches this condition. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Perpendicular Bisector, Locus of Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Theorem 4, the perpendicular bisector of AB is indeed the locus of points equidistant from A and B. This means if P lies on the perpendicular bisector, PA = PB (Assertion A is true). The Reason R correctly defines the property of the perpendicular bisector as the locus of equidistant points, which directly explains why Assertion A holds true.

Key Concept: Perpendicular Bisector Theorem

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that if a point $P$ is equidistant from points $A$ and $B$, it must lie on the perpendicular bisector of $AB$. This is true because, according to the Perpendicular Bisector Theorem, the set of all points equidistant from $A$ and $B$ forms the perpendicular bisector of $AB$. The Reason reiterates the theorem itself, which directly explains why the Assertion is true.
Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Perpendicular Bisector

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Assertion (A) states that the locus of a point equidistant from two given points is the perpendicular bisector of the line joining them, which is a fundamental property derived from congruent triangles (as proved in Theorem 4). Reason (R) describes an application of the perpendicular bisector property in triangles, where the circumcentre is equidistant from all three vertices. While both statements are true, Reason (R) does not explain Assertion (A); it provides additional information about another property of perpendicular bisectors in triangles. Therefore, the correct option is (b).

Key Concept: Locus of Points Equidistant from Two Fixed Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if two circles centered at $A$ and $B$ intersect at $P$, then the locus of all such points $P$ equidistant from $A$ and $B$ is the perpendicular bisector of $AB$. This is correct because, by definition, any point $P$ on the perpendicular bisector of $AB$ is equidistant from $A$ and $B$.
The reason states that the perpendicular bisector of $AB$ is the set of all points in the plane equidistant from $A$ and $B$. This is also true and directly explains why the assertion is correct.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Locus of Points Equidistant from Two Fixed Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points A and B is the perpendicular bisector of AB, which is true as per the definition given in the syllabus. The reason states that any point on the perpendicular bisector of AB is equidistant from A and B, which is also true and is the correct explanation for the assertion. This is because the perpendicular bisector is defined as the set of all points that are equidistant from the endpoints of the segment.

Key Concept: Locus of Points Equidistant from Two Fixed Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of points equidistant from two fixed points A and B is the perpendicular bisector of the line segment joining them. This is a fundamental theorem in geometry, as proven in Theorem 4. The reason correctly explains that every point on the perpendicular bisector satisfies the condition $PA = PB$ because the perpendicular bisector ensures congruency of triangles, leading to equal distances. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Locus of a point equidistant from two fixed points, Incentre and Circumcentre properties

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
- Assertion (A) states a fundamental property: The locus of a point equidistant from two fixed points A and B is indeed the perpendicular bisector of AB. This is a standard geometric result.
- Reason (R) describes another property: The circumcentre of a triangle is equidistant from all three vertices. This is also true as the circumcentre is the intersection point of the perpendicular bisectors of the sides, making it equidistant from the vertices.
- However, the Reason (R) does not explain the Assertion (A), as they are independent properties of different geometric configurations (locus vs. triangle centres). Both statements are true, but Reason does not justify Assertion.

Key Concept: Locus of a point equidistant from two intersecting lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, the locus of a point equidistant from two intersecting lines is indeed the bisector of the angles between the lines. This is because any point on the angle bisector satisfies the condition of being equidistant from both lines, as proved in the theorem mentioned in the syllabus. Additionally, any point not on the angle bisector will not be equidistant from both lines. Therefore, the assertion is true. The reason given is also true, as it correctly states that a point lying on the angle bisector will always be equidistant from both lines. Furthermore, the reason provides the correct explanation for why the assertion holds, as the property of the angle bisector directly implies the assertion's statement.

Key Concept: Geometric Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining them, which is a fundamental property of perpendicular bisectors in geometry. The reason correctly explains why this is true: any point on the perpendicular bisector of a segment will be equidistant from the endpoints of the segment due to congruent triangles formed by the perpendicular bisector and the segment.

Key Concept: Uniqueness of Locus, Geometric Uniqueness

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that the locus of points equidistant from two intersecting straight lines is a pair of angle bisectors. This is true because angle bisectors by definition are lines that divide the angle between two intersecting lines into two equal angles, and any point on an angle bisector is equidistant from both lines.
The Reason claims that angle bisectors are the only curves where every point is equidistant from both given lines. This is false because while angle bisectors satisfy this condition, they are not the only curves to do so. For example, in three-dimensional space, there are other surfaces (like hyperbolic paraboloids) that can also maintain equal distances from two lines under certain conditions.
Therefore, the Assertion is true but the Reason is false.

Key Concept: Uniqueness of Locus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points is always unique. According to the key properties of locus, this is true because the locus in this case is the perpendicular bisector of the line segment joining the two fixed points. The reason provided explains why the assertion is true: for any two distinct fixed points, there is only one perpendicular bisector, making the locus unique. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Key Properties of Locus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining them, which is correct based on the syllabus. The reason correctly explains why this is true: every point on the perpendicular bisector is indeed equidistant from both endpoints of the segment. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Properties of Loci: Locus of mid-points of parallel chords in a circle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the locus of the mid-points of all parallel chords in a given circle is a straight line, which is correct because the mid-points lie on a diameter perpendicular to the chords. The Reason correctly explains this by stating that the diameter perpendicular to the given parallel chords is indeed the locus of their mid-points. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Locus of Points Equidistant from Two Fixed Points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining them. This is a fundamental property of loci in geometry. The reason correctly explains that any point lying on the perpendicular bisector will indeed be equidistant from both fixed points, as it satisfies the condition of the locus. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Locus equidistant from two parallel lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the given syllabus, one of the key properties states: "The locus of a point, which remains equidistant from two given parallel lines, is a line parallel to the given lines and midway between them." This confirms that the Assertion is true. The Reason explains why the Assertion holds because the path traced by the point must satisfy the condition of being equidistant from both parallel lines, thus forming a new parallel line midway between them. Hence, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Definition of Locus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the locus of a point equidistant from two intersecting lines is the angle bisector. This is correct because, in geometry, the angle bisector is indeed the line that divides the angle into two equal parts and consists of all points equidistant from both sides of the angle. The Reason explains why this is true: the angle bisector is defined as the set of points that are equally distant from both sides of the angle. Therefore, the Reason correctly explains the Assertion.

Key Concept: Definition of Locus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
1. Let $A$ and $B$ be two fixed points. The perpendicular bisector of the line segment $AB$ is the set of all points that are equidistant from $A$ and $B$.
2. For any point $P$ on the perpendicular bisector, the distances $PA$ and $PB$ will be equal due to geometric properties of perpendicular bisectors.
3. Conversely, any point equidistant from $A$ and $B$ must lie on the perpendicular bisector.
4. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Definition of Locus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two parallel lines is a line parallel to them and midway between them. This is correct based on the definition of locus. The reason explains that every point on this line satisfies the condition of equidistance from both parallel lines, which is also true. Moreover, the reason correctly justifies the assertion because it describes why the locus must be a line midway between the parallel lines.

Key Concept: Locus of a point equidistant from two fixed points, Locus of mid-points of parallel chords in a circle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that the locus of a point equidistant from two fixed points A and B is the perpendicular bisector of AB. This is true by definition, as any point on the perpendicular bisector is equidistant from A and B.
The Reason states that all points on the perpendicular bisector are equidistant from A and B, which is correct. Additionally, it mentions that the midpoint of any chord of the circle with diameter AB lies on the perpendicular bisector. While the first part of the Reason correctly explains the Assertion, the second part introduces an unrelated concept about midpoints of chords, which does not contribute to explaining why the locus is the perpendicular bisector. Thus, the Reason is partially correct but includes unnecessary information.
Therefore, both Assertion and Reason are true, but the Reason is NOT the correct explanation of the Assertion.

Key Concept: Locus of a Point

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment joining them. This is a known property in geometry.
The reason states that the set of all points equidistant from two fixed points forms a circle. However, this is incorrect because the set of points equidistant from two fixed points forms the perpendicular bisector, not a circle. A circle is formed when considering points at a fixed distance from a single point (the center).
Therefore, the assertion is true, but the reason is false.

Key Concept: Locus of a point equidistant from two fixed points

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $AB$. This is correct because, by definition, the perpendicular bisector is the set of all points equidistant from $A$ and $B$. The reason explains why this is true: every point on the perpendicular bisector satisfies the condition of being equidistant from $A$ and $B$. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Perpendicular Bisector Theorem, Angle Bisector Theorem

b) $\sqrt{34}$ cm
[Solution Description]
Analyzing the configuration:
1. Since ABCD is a parallelogram (opposite sides equal), angle bisectors of A and C will meet along the diagonal AC.
2. The perpendicular bisectors of AB and CD in a parallelogram will meet at the center of symmetry.
3. For exact calculation: Place ABCD with A at origin, AB along x-axis.
Coordinates: A(0,0), B(10,0), C(10,6), D(0,6)
Midpoint of AB = (5,0), midpoint of CD = (5,6)
Perpendicular bisector of AB is vertical line x=5
Perpendicular bisector of CD is also x=5, so Q is at (5,y)
Since Q is equidistant from B and C: $\sqrt{(5-10)^2 + y^2} = \sqrt{(5-10)^2 + (y-6)^2}$
Solving gives $y = 3$, so Q is at (5,3)
Angle bisector of A: Slope 1 (45°), equation y=x
Angle bisector of C: Slope -1, equation y-6=-(x-10)
Intersection P: solving y=x and y=-x+16 gives P(8,8)
Distance PQ = $\sqrt{(8-5)^2 + (8-3)^2} = \sqrt{9 + 25} = \sqrt{34}$ cm

Key Concept: Locus of Points Equidistant from Lines, Congruent Triangles

b) $8$ cm
[Solution Description]
Given two lines intersecting at $60^\circ$ and point P equidistant (4 cm) from both:
1. By the angle bisector theorem, P must lie on the angle bisector.
2. The angle between OP and either line is $30^\circ$ (half of $60^\circ$).
3. The distance from P to a line forms a right angle, creating a $30^\circ-60^\circ-90^\circ$ triangle.
In such a triangle, the side opposite $30^\circ$ is half the hypotenuse. Here, the perpendicular distance (4 cm) is opposite $30^\circ$, so:
Hypotenuse OP = $\frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8$ cm

Key Concept: Angle Bisector Theorem, Perpendicular Bisector Theorem

c) $5\sqrt{2}$ cm
[Solution Description]
To find the distance of point P from A, we analyze its properties:
1. Since P lies on the angle bisector of $\angle B$, it is equidistant from BA and BC.
2. Since P lies on the perpendicular bisector of AC, it is equidistant from A and C.
Let the distance of P from BA be $d$. Then by angle bisector property, distance from BC is also $d$.
Using Pythagoras' theorem in triangles formed by dropping perpendiculars to BA and BC:
Distance of P from A: $\sqrt{(AB - d)^2 + d^2} = \sqrt{(6 - d)^2 + d^2}$
Distance of P from C: $\sqrt{(BC - d)^2 + d^2} = \sqrt{(8 - d)^2 + d^2}$
Since PA = PC:
$\sqrt{(6 - d)^2 + d^2} = \sqrt{(8 - d)^2 + d^2}$
Squaring both sides: $(6 - d)^2 = (8 - d)^2$
Solving gives $36 - 12d = 64 - 16d$ $\Rightarrow$ $4d = 28$ $\Rightarrow$ $d = 7$ cm
Substituting back: PA = $\sqrt{(6 - 7)^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$ cm

Key Concept: Congruency in equidistant points

a) The distance from $Q$ to $A$ is equal to the distance from $Q$ to $B$.
[Solution Description]
By the converse part of Theorem 4, any point lying on the perpendicular bisector of $AB$ is equidistant from $A$ and $B$. Therefore, triangles $AOQ$ and $BOQ$ are congruent, ensuring $AQ = BQ$.

Key Concept: Locus of a point equidistant from two given points

c) The point $P$ lies on the perpendicular bisector of $AB$.
[Solution Description]
According to Theorem 4, the locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment joining $A$ and $B$. Thus, the point $P$ must lie on this perpendicular bisector.

Key Concept: Locus of a point equidistant from two intersecting lines

a) The point $P$ lies on one of the angle bisectors of the angles formed by $AB$ and $CD$.
[Solution Description]
According to Theorem 3, the locus of a point equidistant from two intersecting lines is the angle bisector of the angles between them. Therefore, point $P$ must lie on the angle bisector of the angle formed by $AB$ and $CD$.

Key Concept: Verification of point lying on the angle bisector

c) It is equidistant from AB and CD
[Solution Description]
From Theorem 3 (ii), every point on the angle bisector of $\angle AOC$ is equidistant from AB and CD. Therefore, since Q lies on this bisector, it must be equidistant from AB and CD.

Key Concept: Locus of a point equidistant from two fixed points

b) The perpendicular bisector of AB
[Solution Description]
According to Theorem 4, the locus of a point equidistant from two fixed points is the perpendicular bisector of the line joining them. Since P is equidistant from A and B, its locus is the perpendicular bisector of AB.

Key Concept: Locus of a point equidistant from two intersecting lines

a) The angle bisector of $\angle AOC$
[Solution Description]
According to Theorem 3, the locus of a point equidistant from two intersecting lines is the angle bisector of the angles formed by them. Here, the point P is equidistant from AB and CD, so its locus must lie on the angle bisector of $\angle AOC$ or $\angle BOD$.

Key Concept: Construction, Perpendicular Bisector

c) Draw the perpendicular bisector of AB.
[Solution Description]
Steps to construct the locus of points equidistant from A and B:
1. Draw line segment AB = 8 cm.
2. Find the midpoint M of AB.
3. Draw a perpendicular line to AB passing through M.
This perpendicular line is the required locus.

Key Concept: Perpendicular Bisector, Angle Bisector

b) X lies on the angle bisector of $\angle ABC$.
[Solution Description]
To find the position of point X:
1. The locus of points equidistant from two intersecting lines (AB and BC) is the angle bisector of the angle formed by them ($\angle ABC$).
2. Hence, X must lie on the angle bisector of $\angle ABC$.

Key Concept: Perpendicular Bisector, Locus of Points

c) P lies on the perpendicular bisector of AB.
[Solution Description]
To determine where point P must lie if it is equidistant from A and B:
1. By definition, the locus of points equidistant from two fixed points A and B is the perpendicular bisector of the line segment AB.
2. Therefore, P must lie on the perpendicular bisector of AB.

Key Concept: Construction

c) Measuring the exact length of PQ.
[Solution Description]
To construct the perpendicular bisector of PQ, we need to draw arcs centered at P and Q with the same radius, intersecting above and below PQ, then join these intersections. Measuring the exact length of PQ is not required for the construction.

Key Concept: Equidistant Points

c) Every point on it is equidistant from A and B.
[Solution Description]
The locus of points equidistant from two fixed points A and B is the perpendicular bisector of the line joining them. Hence, every point on the perpendicular bisector of AB is equidistant from A and B.

Key Concept: Perpendicular Bisector Theorem

c) 5 cm
[Solution Description]
Since D lies on the perpendicular bisector of AB, by the Perpendicular Bisector Theorem, D is equidistant from A and B. Therefore, BD = AD = 5 cm.

Key Concept: Applications of Perpendicular Bisector

b) Point $D$ is equidistant from $B$ and $C$
[Solution Description]
The perpendicular bisector of $BC$ passes through all points equidistant from $B$ and $C$. Since $D$ lies on this bisector, it must satisfy $DB = DC$.

Key Concept: Construction of Perpendicular Bisector

c) Point $Q$ is equidistant from $A$ and $B$
[Solution Description]
Since $PQ$ is the perpendicular bisector of $AB$, by definition, point $Q$ lies on the perpendicular bisector. Therefore, $Q$ must be equidistant from $A$ and $B$, i.e., $QA = QB$.

Key Concept: Locus of Equidistant Points

c) The perpendicular bisector of $AB$
[Solution Description]
The locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment joining $A$ and $B$. This is because every point on the perpendicular bisector satisfies the condition $PA = PB$, where $P$ is any point on the bisector.

Key Concept: Equidistant Points, Construction

b) One
[Solution Description]
For three non-collinear points, there exists exactly one unique circle (circumcircle) that passes through all three points. This is because the perpendicular bisectors of the sides of triangle ABC will meet at a single point (circumcenter), which is equidistant from all three vertices.

Key Concept: Circle Construction, Locus

c) P can occupy exactly two positions
[Solution Description]
Point P must lie on the intersection of two loci:
1. A circle centered at M with radius 4 cm (since PM = 4 cm).
2. A circle centered at N with radius 3 cm (since PN = 3 cm).
However, since N itself moves on a circle around M (radius 5 cm), there will be two possible positions for P where both circles intersect.

Key Concept: Locus, Angle Bisector

c) The angle bisector of $\angle AOC$
[Solution Description]
The path of point P is the angle bisector of the angle formed by AB and CD, as per the theorem mentioned in the syllabus. Since AB and CD form a $60^\circ$ angle, the angle bisector will divide it into two angles of $30^\circ$ each.

Key Concept: Theorems Related to Locus

b) The angle bisectors of the angles formed by $AB$ and $CD$
[Solution Description]
The locus of points equidistant from two intersecting lines $AB$ and $CD$ is the pair of angle bisectors of the angles formed by the lines. This is because the angle bisectors divide the angles into two equal parts, ensuring that any point on the bisector is equidistant from both lines.

Key Concept: Construction Steps

b) Mark points at a distance $r$ from $O$ and connect them
[Solution Description]
The fundamental property of a circle is that all its points are equidistant from the center. Therefore, while constructing the circle, we must mark points at a constant distance $r$ from the center $O$. Drawing a freehand curve through these marked points will form the circle.

Key Concept: Locus of Points and Angle Bisector

c) $P$ is equidistant from $AB$ and $CD$
[Solution Description]
According to the theorem, a point equidistant from two intersecting lines lies on the angle bisector of the angle formed by the lines. Here, $P$ is given to be in the interior of $\angle AOC$. If $P$ is equidistant from both $AB$ and $CD$, then it must lie on the bisector of $\angle AOC$. Thus, the correct condition is that $P$ is equidistant from both $AB$ and $CD$.

Key Concept: Theorem Application

c) A circle with radius 5 cm
[Solution Description]
By the theorem, the locus of a point at a fixed distance from a fixed point is a circle. Here, the fixed point is $O$, and the fixed distance is 5 cm. Therefore, the locus of $P$ is a circle centred at $O$ with radius 5 cm.

Key Concept: Construction Steps

c) Compass
[Solution Description]
To draw a circle, we need to measure the radius accurately and rotate the compass 360° around the fixed centre. The primary tool used for this purpose is a compass.

Key Concept: Definition of a Circle

d) A circle
[Solution Description]
According to the definition, a circle is the set of all points in a plane equidistant from a fixed point called the centre. The constant distance is known as the radius. Hence, the correct answer is a circle.

Key Concept: Combined loci, Distance from parallel lines

b) y = $\frac{17}{3}$ and y = 11
[Solution Description]
The condition is distance from P(x,y) to y=3 is twice its distance to y=7.
1. Distance to y=3 is $|y - 3|$
2. Distance to y=7 is $|y - 7|$
3. Given condition: $|y - 3| = 2|y - 7|$
Solving:
Case 1: y ≥ 7
$y - 3 = 2(y - 7) ⇒ y - 3 = 2y - 14 ⇒ y = 11$
Case 2: 3 ≤ y < 7
$y - 3 = 2(7 - y) ⇒ y - 3 = 14 - 2y ⇒ 3y = 17 ⇒ y = \frac{17}{3}$
Case 3: y < 3
$3 - y = 2(7 - y) ⇒ 3 - y = 14 - 2y ⇒ y = 11$ (invalid as y <3)
Thus valid solutions are y=11 and y=$\frac{17}{3}$.

Key Concept: Locus from angle bisector, Angle between lines

c) A single line making 30° with L1
[Solution Description]
For two intersecting lines, the locus of points equidistant to both is the angle bisector(s).
1. Since the angle between L1 and L2 is 60°, there will be two angle bisectors: one for the acute angle and one for the obtuse angle.
2. The first angle bisector divides the 60° angle into two 30° angles.
3. The second angle bisector (for the 300° reflex angle) would make an angle of 150° with each line but we consider only the internal locus.
4. Therefore, within the 60° angle, the locus is the line making 30° with both L1 and L2.

Key Concept: Equidistant from two points, Perpendicular bisector

c) $x - 3y - 1 = 0$
[Solution Description]
The condition given is that point P is equidistant from points A(3, 4) and B(5, -2). This means P lies on the perpendicular bisector of AB.
1. Find the midpoint of AB:
Midpoint coordinates = $\left(\frac{3+5}{2}, \frac{4+(-2)}{2}\right) = (4, 1)$
2. Find slope of AB:
Slope of AB = $\frac{-2-4}{5-3} = \frac{-6}{2} = -3$
3. Slope of perpendicular bisector = negative reciprocal of AB's slope = $\frac{1}{3}$
4. Equation of line passing through (4, 1) with slope $\frac{1}{3}$:
$y - 1 = \frac{1}{3}(x - 4)$
Simplifying: $3y - 3 = x - 4$
Final equation: $x - 3y - 1 = 0$

Key Concept: Locus of a point equidistant from two intersecting lines

b) The angle bisectors of the angles formed by $m$ and $n$
[Solution Description]
The locus of a point equidistant from two intersecting lines $m$ and $n$ is the pair of angle bisectors of the angles formed by $m$ and $n$. Any point lying on either angle bisector will be equidistant from both lines.

Key Concept: Locus of a point at a constant distance from a fixed line

c) Two lines parallel to $l$ at 5 cm on either side
[Solution Description]
The locus of a point at a constant distance (5 cm) from a fixed line $l$ consists of two lines parallel to $l$, one on each side, both at a distance of 5 cm from $l$. This ensures that every point on these lines satisfies the condition.

Key Concept: Locus of a point equidistant from two fixed points

d) The perpendicular bisector of $AB$
[Solution Description]
The locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment joining $A$ and $B$. This is because any point on the perpendicular bisector will satisfy the condition $PA = PB$.

Key Concept: Locus of a point at a fixed distance from a given line

a) $x = -1$ and $x = 9$
[Solution Description]
The locus of a point at a fixed distance (5 units) from a vertical line $x = 4$ consists of two parallel lines: one to the left ($x = -1$) and one to the right ($x = 9$) of the given line.

Key Concept: Locus of a point equidistant from two intersecting lines

b) Two perpendicular lines bisecting the angles between $l_1$ and $l_2$
[Solution Description]
The locus of a point equidistant from two intersecting lines $l_1$ and $l_2$ is the pair of angle bisectors of the angles formed by these lines. Each angle bisector divides the angle into two equal parts, ensuring any point on it is equidistant from both lines.

Key Concept: Locus of a point equidistant from two fixed points

c) The perpendicular bisector of AB
[Solution Description]
The locus of a point equidistant from two fixed points A and B is the perpendicular bisector of the line segment joining them. This is because any point on the perpendicular bisector will be at the same distance from both A and B.

Key Concept: Locus of points equidistant from a fixed line

b) Two lines parallel to AB at 6 cm distance on either side
[Solution Description]
The locus of a point at a fixed distance ($6\text{ cm}$) from a given line AB is a pair of lines parallel to AB, each at a distance of $6\text{ cm}$ from AB on either side.

Key Concept: Locus of midpoints of equal chords in a circle

b) A circle with radius $4\sqrt{3}\text{ cm}$
[Solution Description]
For a chord of length $8\text{ cm}$ in a circle of radius $8\text{ cm}$, the perpendicular distance from the center to the chord can be calculated using the formula $d = \sqrt{r^2 - (\frac{L}{2})^2} = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}\text{ cm}$. Thus, the locus of midpoints is a concentric circle with radius $4\sqrt{3}\text{ cm}$.

Key Concept: Locus of points equidistant from two concentric circles

b) A circle with radius 7.5 cm
[Solution Description]
The locus of a point equidistant from two concentric circles is a circle concentric with them and midway between their circumferences. First, find the distance between the two circles: $10\text{ cm} - 5\text{ cm} = 5\text{ cm}$. The midpoint radius is $5 + \frac{5}{2} = 7.5\text{ cm}$ or $10 - \frac{5}{2} = 7.5\text{ cm}$.

Key Concept: Locus of mid-points of parallel chords in a circle

b) The diameter perpendicular to PQ
[Solution Description]
The locus of the mid-points of all chords parallel to a given chord PQ in a circle is the diameter of the circle that is perpendicular to PQ. This is because the perpendicular from the center to any chord bisects it, and all such mid-points lie on this diameter.

Key Concept: Locus of a point at a fixed distance from a given line

b) Two lines parallel to L at 4 cm distance on opposite sides
[Solution Description]
The locus of a point at a fixed distance (4 cm) from a given straight line L consists of two straight lines, each parallel to L and located 4 cm away from it on either side. This forms a pair of parallel lines.

Key Concept: Locus of a point equidistant from two fixed points

c) The perpendicular bisector of AB
[Solution Description]
The locus of a point equidistant from two fixed points A and B is the perpendicular bisector of the line segment joining A and B. This is because every point on the perpendicular bisector is equidistant from both A and B.

Key Concept: Locus of mid-points of equal chords in a circle

d) Another circle with the same center and radius equal to the distance of the equal chords from the center
[Solution Description]
The locus of the mid-points of all equal chords in a circle is the circumference of a concentric circle with a radius equal to the distance of these chords from the center. So, it forms another circle with the same center but smaller radius.

Key Concept: Locus of a point equidistant from two parallel lines

c) A line parallel to L1 and L2 and midway between them
[Solution Description]
The locus of a point equidistant from two parallel lines is a line parallel to both and midway between them. This line will be at an equal distance from both L1 and L2.

Key Concept: Locus of a point at a fixed distance from a fixed point

c) A circle with center P and radius 5 units
[Solution Description]
The locus of a point at a fixed distance from a fixed point is a circle with the fixed point as its center and the fixed distance as its radius. Here, the fixed point is P and the distance is 5 units. Therefore, the locus is a circle centered at P with radius 5.

Key Concept: Newton’s Law of Cooling, Exponential Decay

c) 45.48°C
[Solution Description]
Newton’s Law of Cooling: $T(t) = T_{\text{room}} + (T_0 - T_{\text{room}}) e^{-kt}$.
Given:
- $T(0) = 90^\circ C$,
- $T_{\text{room}} = 20^\circ C$,
- $T(5) = 70^\circ C$.
First, find $k$ using the 5-minute data:
$70 = 20 + (90 - 20) e^{-5k}$
Simplify:
$50 = 70 e^{-5k}$
Divide by 70:
$\frac{50}{70} = e^{-5k}$
Take the natural logarithm (ln):
$\ln\left(\frac{5}{7}\right) = -5k$
Solve for $k$:
$k = -\frac{1}{5} \ln\left(\frac{5}{7}\right) \approx 0.0673$
Now, find $T(15)$:
$T(15) = 20 + 70 e^{-0.0673 \times 15}$
Calculate exponent:
$e^{-1.0095} \approx 0.364$
Final calculation:
$T(15) = 20 + 70 \times 0.364 \approx 45.48^\circ C$

Key Concept: Compound Interest, Annual Compounding

c) 18.85 years
[Solution Description]
Compound interest formula: $A = P(1 + r)^t$, where:
- $A$ = Final amount (3 × 10,000 = \$30,000),
- $P$ = Principal (\$10,000),
- $r$ = Annual rate (0.06).
Plugging in the values:
$30,000 = 10,000(1.06)^t$
Divide both sides by 10,000:
$3 = (1.06)^t$
Take the natural logarithm (ln) of both sides:
$\ln(3) = t \ln(1.06)$
Solve for $t$:
$t = \frac{\ln(3)}{\ln(1.06)} \approx 18.85 \text{ years}$

Key Concept: Exponential Growth, Logarithmic Scaling

b) 10.14 years
[Solution Description]
The exponential growth formula is $P = P_0 e^{rt}$, where:
- $P$ = Final population (750,000),
- $P_0$ = Initial population (500,000),
- $r$ = Growth rate (0.04),
- $t$ = Time in years.
Rearrange the formula to solve for $t$:
$750,000 = 500,000 e^{0.04t}$
Divide both sides by 500,000:
$1.5 = e^{0.04t}$
Take the natural logarithm (ln) of both sides:
$\ln(1.5) = 0.04t$
Solve for $t$:
$t = \frac{\ln(1.5)}{0.04} \approx 10.14 \text{ years}$

Key Concept: Locus of points inside a circle equidistant from two fixed points

a) The entire perpendicular bisector of AB
[Solution Description]
Points equidistant from two given points A and B lie on the perpendicular bisector of the segment AB. However, since the points must also lie inside the circle, the locus is the portion of the perpendicular bisector of AB that lies within the circle.

Key Concept: Locus of door-handle

c) Arc of a circle
[Solution Description]
The door rotates about its hinges, which act as the fixed point or pivot. The door-handle moves in a circular arc centered at the hinge. Therefore, the locus of the door-handle is an arc of a circle with the hinge as the center.

Key Concept: Locus of a runner around a circular track

c) 101.5 m
[Solution Description]
The runner maintains a constant distance of 1.5 m from the inner edge of the circular track. The inner edge has a radius of 100 m. Therefore, the runner's path is a concentric circle with radius equal to the sum of the inner edge radius and the maintained distance. So, the radius of the runner's path is $100\, \text{m} + 1.5\, \text{m} = 101.5\, \text{m}$.

Key Concept: Locus inside a circle

c) The perpendicular bisector of the chord joining the two points (restricted inside the circle)
[Solution Description]
The locus of points inside a circle equidistant from two fixed points on its circumference is the perpendicular bisector of the chord joining those two points. This line passes through the center of the circle but includes only the portion inside the circle.

Key Concept: Locus of centers of circles

b) The perpendicular bisector of the line segment joining the two fixed points
[Solution Description]
The centers of all circles passing through two fixed points lie on the perpendicular bisector of the line segment joining the two points. This is because each center must be equidistant from both points.

Key Concept: Locus of a point

c) A sphere with radius 4 cm centered at the fixed point
[Solution Description]
The locus of a point at a constant distance from a fixed point is a circle in 2D or a sphere in 3D. Here, since it mentions "space," it refers to 3D, so the correct answer is a sphere.

Key Concept:

b) O is equidistant from AB and AD.
[Solution Description]
The angle bisector of $\angle A$ ensures that O is equidistant from AB and AD. Similarly, the angle bisector of $\angle D$ ensures that O is equidistant from DA and DC. However, since both angle bisectors meet at O, it does not guarantee that O is equidistant from any sides other than those adjacent to angles A and D. No information is provided to confirm its position relative to BC. Thus, O is only guaranteed to be equidistant from AB and AD.

Key Concept:

c) 2
[Solution Description]
To find point P, we draw two circles: one centered at A with radius 4 cm, and another centered at B with radius 3 cm. The distance between A and B is 5 cm. Since the sum of radii (4 + 3 = 7 cm) > AB (5 cm), and the difference of radii (4 - 3 = 1 cm) < AB, these two circles will intersect at exactly two distinct points. Hence, there are two possible positions for P.

Key Concept:

d) P is equidistant from all three sides of the triangle.
[Solution Description]
The angle bisector of $\angle B$ ensures that P is equidistant from BA and BC. The perpendicular bisector of AC ensures that P is equidistant from A and C. Therefore, P is equidistant from all three sides of the triangle (BA, BC, and AC) because it lies at the intersection of the angle bisector and the perpendicular bisector.

Key Concept: Angle Bisector and Perpendicular Bisector

c) Any point on it is equidistant from the endpoints of the segment.
[Solution Description]
The perpendicular bisector of a line segment $AB$ is the set of all points that are equidistant from $A$ and $B$. It intersects $AB$ at its midpoint and forms a right angle with it. Therefore, any point lying on the perpendicular bisector is equidistant from both $A$ and $B$.

Key Concept: Equidistant Points

a) (1, 0)
[Solution Description]
Since $P$ lies on the x-axis, its y-coordinate is 0. Let $P$ be $(x, 0)$.
The distance between $P$ and $A$ is $\sqrt{(x - 3)^2 + (0 - 4)^2}$, and the distance between $P$ and $B$ is $\sqrt{(x - 5)^2 + (0 + 2)^2}$.
Equating the distances:
$\sqrt{(x - 3)^2 + 16} = \sqrt{(x - 5)^2 + 4}.$
Squaring both sides:
$(x - 3)^2 + 16 = (x - 5)^2 + 4.$
Expanding and simplifying:
$x^2 - 6x + 9 + 16 = x^2 - 10x + 25 + 4.$
Solving for $x$:
$-6x + 25 = -10x + 29 \implies 4x = 4 \implies x = 1.$
Thus, $P$ is at $(1, 0)$.

Key Concept: Triangle Constructions

c) Obtuse-angled triangle
[Solution Description]
To determine the type of triangle, we can use the Pythagorean theorem to check if the sum of squares of two smaller sides equals the square of the longest side.
Step 1: Square all sides:
$5^2 = 25$,
$7^2 = 49$,
$9^2 = 81$.
Step 2: Check if $25 + 49 = 74$ equals $81$. Since $74 < 81$, the triangle is an obtuse-angled triangle.

Key Concept: Locus Constructions

c) The perpendicular bisector of AB
[Solution Description]
The locus of points equidistant from two fixed points A and B is the perpendicular bisector of the line segment joining A and B. This is because every point on the perpendicular bisector is at the same distance from both A and B.

Key Concept: Construction Using Ruler and Compasses

c) Ruler and compass
[Solution Description]
An angle bisector divides the angle into two equal parts. To construct it accurately, both a ruler (to draw straight lines) and a compass (to mark equal distances) are essential.

Key Concept: Bisectors and Equidistant Points

d) Incenter
[Solution Description]
The point where the angle bisectors of two angles in a triangle meet is known as the incenter. The incenter is equidistant from all three sides of the triangle.

Key Concept: Mid-Points of Parallel Chords, Circle Geometry

a) $y = x$
[Solution Description]
The locus of midpoints of parallel chords in a circle is the diameter perpendicular to those chords.
Step 1: The slope of the given line $x + y = 0$ is $m = -1$.
Step 2: The perpendicular slope is $m_{\perp} = 1$.
Step 3: The diameter passes through the center $O(0, 0)$ and has slope 1, so its equation is:
$y - 0 = 1(x - 0) \implies y = x$
Thus, the midpoints lie on the line $y = x$.

Key Concept: Fixed Distance from a Line, Parallel Lines

a) $2x - y + (1 - 3\sqrt{5}) = 0$ and $2x - y + (1 + 3\sqrt{5}) = 0$
[Solution Description]
The locus of points at a fixed distance $d$ from a line $Ax + By + C = 0$ are two parallel lines given by:
$\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} = d$
For the line $y = 2x + 1$, rewrite in standard form: $2x - y + 1 = 0$.
Step 1: Calculate the two parallel lines:
$\frac{|2x - y + 1|}{\sqrt{2^2 + (-1)^2}} = 3 \implies |2x - y + 1| = 3\sqrt{5}$
Step 2: Split into two equations:
$2x - y + 1 = 3\sqrt{5} \quad \text{and} \quad 2x - y + 1 = -3\sqrt{5}$
Rewriting:
$2x - y + (1 - 3\sqrt{5}) = 0 \quad \text{and} \quad 2x - y + (1 + 3\sqrt{5}) = 0$

Key Concept: Equidistant from Two Fixed Points, Perpendicular Bisector

a) $x + y - 11 = 0$
[Solution Description]
To find the locus of points equidistant from $A(3, 4)$ and $B(7, 8)$, we use the perpendicular bisector concept.
Step 1: Find the midpoint of $AB$. The midpoint $M$ is calculated as:
$M = \left( \frac{3 + 7}{2}, \frac{4 + 8}{2} \right) = (5, 6)$
Step 2: Find the slope of $AB$:
$m_{AB} = \frac{8 - 4}{7 - 3} = 1$
Step 3: The slope of the perpendicular bisector is the negative reciprocal of $m_{AB}$, which is $-1$.
Step 4: Using the point-slope form with $M(5, 6)$:
$y - 6 = -1(x - 5) \implies y = -x + 11$
Thus, the equation is $x + y - 11 = 0$.

Key Concept: Locus of vertices of isosceles triangles with common base

b) The perpendicular bisector of $AB$
[Solution Description]
For isosceles triangles with common base $AB$, the vertex must be equidistant from $A$ and $B$. Thus, the locus is the perpendicular bisector of the base $AB$.

Key Concept: Locus of a point in a rhombus equidistant from vertices

c) Diagonal $AC$
[Solution Description]
In a rhombus $ABCD$, the diagonals $AC$ and $BD$ are perpendicular bisectors of each other. Any point equidistant from $B$ and $D$ must lie on the perpendicular bisector of the line segment $BD$, which is the diagonal $AC$.

Key Concept: Locus of centers of circles passing through two fixed points

c) The perpendicular bisector of $AB$
[Solution Description]
The locus of the centers of all circles that pass through two fixed points $A$ and $B$ must be equidistant from both points. Therefore, these centers lie on the perpendicular bisector of the line segment joining $A$ and $B$.

Key Concept: Locus of a point equidistant from two parallel lines

c) A line parallel to l and m at a distance of 4 cm from each
[Solution Description] The locus of a point equidistant from two parallel lines l and m, which are 8 cm apart, is a line parallel to both l and m and exactly halfway (4 cm) between them.

Key Concept: Locus of a point at a fixed distance from a given point

d) A circle with center O and radius 5 cm
[Solution Description] The locus of a point P at a fixed distance of 5 cm from a fixed point O is a circle with center O and radius 5 cm.

Key Concept: Locus of a point equidistant from two fixed points

d) The perpendicular bisector of AB
[Solution Description] The locus of a point equidistant from two fixed points A and B is the perpendicular bisector of the line segment joining A and B.

Key Concept: Locus construction, distance constraints

d) 4 cm
[Solution Description]
Let's solve step-by-step:
1. Let distance from OY = d, then from OX = 2d.
2. Given 2d + d = 6 ⇒ d = 2 cm, so distance from OX = 4 cm.
3. Verify using angle formula: For θ=45°, ratio should be tanθ or tan(180°-θ).
4. tan45°=1, but our ratio is 2 (valid for modified condition).
5. Therefore possible solution exists with distance from OX as 4 cm.

Key Concept: Bisector theorem, congruence

d) 90°
[Solution Description]
1. Since AB = AC, triangle ABC is isosceles with base BC.
2. Given AD = AE, triangle ADE is also isosceles with base DE.
3. Angle A's bisector is also the median and altitude due to symmetry.
4. Therefore, angle AFE is 90° as the bisector is perpendicular to DE.

Key Concept: Angle bisector locus, distance from lines

b) 4
[Solution Description]
The two intersecting lines form four 60° angles. For any point to be at given distances from both lines, we need to consider all possible positions:
1. For angle bisectors: There are two bisectors (one for acute angle, one for obtuse).
2. Distance condition translates to $|d_1| = 4$ and $|d_2| = 4\sqrt{3}$.
3. Using the relation between distances and angle: $d_2/d_1 = \tanθ$ or $\tan(180°-θ)$.
Here, $4\sqrt{3}/4 = \sqrt{3} = \tan60°$, which matches the angle between lines.
4. Therefore, there will be 4 such points - two in adjacent angles and two in opposite angles.

Key Concept: Construction of angles

b) $60^\circ + 15^\circ$
[Solution Description]
An angle of $75^\circ$ can be constructed by combining a $60^\circ$ angle and a $15^\circ$ angle. First, construct an equilateral triangle to get $60^\circ$. Then, bisect a $30^\circ$ angle (which can be obtained by bisecting the $60^\circ$ angle) to get $15^\circ$. Adding these two angles gives $75^\circ$.

Key Concept: Locus of points equidistant from two intersecting lines

c) The angle bisector of $\angle BAC$
[Solution Description]
The locus of a point equidistant from two intersecting lines is the angle bisector of the angle formed by those two lines. Here, since lines AB and AC intersect at point A, forming a $60^\circ$ angle, the locus will be the angle bisectors of $\angle BAC$. Since every angle has two angle bisectors (one internal and one external), the locus consists of two straight lines that are angle bisectors of $\angle BAC$. However, standard geometrical problems generally consider the internal angle bisector unless specified otherwise.

Key Concept: Angle between two lines

c) $15^\circ$
[Solution Description]
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:
$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$
Given $m_1 = 1$, $m_2 = \sqrt{3}$. Substituting these values:
$\tan \theta = \left| \frac{\sqrt{3} - 1}{1 + 1 \cdot \sqrt{3}} \right| = \left| \frac{\sqrt{3} - 1}{1 + \sqrt{3}} \right|$
Rationalizing the denominator:
$\tan \theta = \left| \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} \right| = \left| \frac{\sqrt{3} - 3 - 1 + \sqrt{3}}{1 - 3} \right| = \left| \frac{2\sqrt{3} - 4}{-2} \right| = |2 - \sqrt{3}| = 2 - \sqrt{3}$
We know that $\tan 15^\circ = 2 - \sqrt{3}$. Thus, $\theta = 15^\circ$.

Key Concept: Locus of Points Equidistant from Two Lines

b) It forms two lines, each bisecting an angle between the given lines
[Solution Description]
The locus of points equidistant from two intersecting lines consists of the pair of angle bisectors of the angles formed by the two lines. This is derived from the Angle Bisector Theorem, ensuring all points on these bisectors are equidistant from both lines.

Key Concept: Construction of Equidistant Point

c) At the intersection of two parallel lines constructed as per the distances
[Solution Description]
To find point $P$:
1. Draw $\angle ABC = 120^\circ$.
2. Construct line $l_1$ parallel to $AB$ at a distance of 3 cm.
3. Construct line $l_2$ parallel to $BC$ at a distance of 2 cm.
4. The intersection of $l_1$ and $l_2$ gives the required point $P$.

Key Concept: Angle Bisector Theorem

c) The angle bisector of the angle formed by the two lines
[Solution Description]
According to the Angle Bisector Theorem, the locus of a point equidistant from two intersecting lines lies on the angle bisector of the angle formed by the two lines. The proof involves constructing perpendiculars from the point to both lines and showing that the triangles formed are congruent by the RHS (Right-Hypotenuse-Side) rule.

Key Concept: Locus of midpoints, Parallel chords

a) A diameter of length 10 cm perpendicular to PQ
[Solution Description]
The midpoint of PQ lies at distance $\sqrt{5^2 - (8/2)^2} = 3$ cm from center. All parallel chords will have midpoints lying on the diameter perpendicular to PQ (since it's the perpendicular bisector). This diameter has length 10 cm (radius × 2), but only a single straight line as locus.