Title: Understanding Locus in Geometry – Concepts, Examples, and Practice Problems
What is Locus?
In mathematics, the term ‘locus’ (plural: loci) refers to a set of points that satisfy a certain condition or a set of conditions. In coordinate geometry, a locus is typically a curve or a geometric figure whose points fulfill specified algebraic or geometric rules. The concept of locus forms the foundational idea for analyzing geometric places under constraints, such as distance or angle, leading to the derivation of algebraic equations that represent curves like circles, parabolas, and ellipses.
Importance of Studying Locus
Studying locus is crucial as it enhances the understanding of various geometric shapes through algebraic expressions. Loci are central to fields like geometry, engineering design, robotics, and computer graphics. Knowing how to determine the path or position of a point based on mathematical conditions is an essential tool in spatial reasoning and advanced geometry.
Notable Mathematicians Studying Locus
Modern research in geometry involving loci has been expanded by mathematicians like Maryam Mirzakhani. Her work in hyperbolic geometry and geometric topology tangentially explored loci-related concepts. Scholars today in algebraic geometry continue to work on moduli spaces, parameterizing geometrical objects via their loci.
Join Skorminda to dive deeper and master locus-based geometry with expert instruction, live practice, and visual simulations that will sharpen your spatial reasoning and problem-solving skills.
Let’s dive into key locus concepts using specific keywords.
1. Locus of Points Equidistant from a Single Point
This locus is a circle; every point is at a constant distance (radius) from a fixed center point.
Let the fixed point be \((a, b)\) and the distance be \(r\). Then, the locus is:
$$
(x – a)^2 + (y – b)^2 = r^2
$$
✅ Easy Questions:
1.1 What is the locus of all points 5 units away from the origin?
Solution: The locus is a circle centered at \((0,0)\) with radius 5:
\[
x^2 + y^2 = 25
\]
1.2 Find the equation of the locus of points 3 units from point \((2, -1)\).
Solution:
\[
(x – 2)^2 + (y + 1)^2 = 9
\]
✅ Medium Questions:
1.3 Determine the locus of a point 4 units from the point \((-1, 2)\) and sketch it.
Solution:
\[
(x + 1)^2 + (y – 2)^2 = 16
\]
This represents a circle centered at \((-1,2)\) with radius 4.
1.4 A point moves such that its distance from \((3,4)\) is equal to 6. Write its locus.
Solution:
\[
(x – 3)^2 + (y – 4)^2 = 36
\]
✅ Hard Questions:
1.5 Show that the locus of points equidistant from the point \((1,2)\) and lying on the x-axis is a circle tangent to the x-axis.
Solution:
Let any point on the x-axis be \((x, 0)\), and let the distance to \((1,2)\) be equal to the radius. So:
\[
(x – 1)^2 + (0 – 2)^2 = r^2 \Rightarrow (x – 1)^2 + 4 = r^2
\]
This gives a set of points forming a circle tangent to the x-axis.
1.6 A moving point remains 7 units from \((0,0)\). Prove it forms a circular path and find its standard form.
Solution:
\[
x^2 + y^2 = 49
\]
2. Locus of Points Equidistant from Two Points
The locus is the perpendicular bisector of the segment joining the two points.
Let the points be \(A(x_1, y_1)\) and \(B(x_2, y_2)\). The locus is:
$$
\text{Set of points P such that } PA = PB
$$
✅ Easy Questions:
2.1 Find the locus of a point equidistant from \((0,0)\) and \((4,0)\).
Solution: Midpoint is \((2,0)\); perpendicular line to x-axis:
\[
x = 2
\]
2.2 A point is equidistant from \((0,2)\) and \((4,2)\). Find the locus.
Solution:
\[
x = 2
\]
✅ Medium Questions:
2.3 Find the equation of the locus of points equidistant from \(A(1,1)\) and \(B(5,5)\).
Solution: Equation of perpendicular bisector:
Midpoint is \((3,3)\); slope of AB = 1, so perpendicular slope = -1. Equation:
\[
y – 3 = -1(x – 3) \Rightarrow x + y = 6
\]
2.4 What is the perpendicular bisector of the segment joining \((2, -1)\) and \((4, 3)\)?
Solution:
Midpoint: \((3, 1)\), original slope: 2. Perpendicular slope: \(-\frac{1}{2}\). Equation:
\[
y – 1 = -\frac{1}{2}(x – 3)
\]
✅ Hard Questions:
2.5 Prove that the set of all points P equidistant from fixed points A and B lies on their perpendicular bisector.
Solution: Let \(PA^2 = PB^2\):
\[
(x – x_1)^2 + (y – y_1)^2 = (x – x_2)^2 + (y – y_2)^2
\]
Simplify and rearrange – get linear equation: the perpendicular bisector.
2.6 A point moves such that its distances from A(0,0) and B(2,6) are equal. Find its locus.
Solution: Let P \((x, y)\) satisfies
\[
x^2 + y^2 = (x – 2)^2 + (y – 6)^2
\Rightarrow x + 3y = 10
\]
3. Locus of Points at Constant Distance from a Line
This locus defines two lines parallel to the given line at a fixed perpendicular distance.
Let the line be:
\[
ax + by + c = 0
\]
Distance \(d\), shifted lines are:
\[
ax + by + c \pm d\sqrt{a^2 + b^2} = 0
\]
✅ Easy Questions:
3.1 Find the locus of points 3 units from the line \(y = 1\).
Solution: Parallel lines:
\[
y = 1 \pm 3 \Rightarrow y = 4 \text{ and } y = -2
\]
3.2 Write equations of lines 2 units from x-axis.
Solution:
\[
y = 2 \text{ and } y = -2
\]
✅ Medium Questions:
3.3 Find the lines 4 units from the line \(x + y = 0\).
Solution:
Lines are:
\[
x + y = 0 \pm \dfrac{4\sqrt{2}}{1.414}
\Rightarrow x + y = \pm 4\sqrt{2}
\]
3.4 A point moves 5 units away from the line \(2x – y + 4 = 0\). Find its locus.
Solution:
\[
2x – y + 4 \pm 5\sqrt{5} = 0
\]
✅ Hard Questions:
3.5 Prove that the locus of points at constant distance \(d\) from a line is two parallel lines.
Solution:
Using distance formula and solving:
\[
\frac{|ax + by + c|}{\sqrt{a^2 + b^2}} = d
\Rightarrow |ax + by + c| = d\sqrt{a^2 + b^2}
\]
3.6 The locus of a point remains 6 units from \(3x – 4y = 7\). Find equation.
Solution:
\[
3x – 4y = 7 \pm 6\sqrt{25} = 7 \pm 30
\Rightarrow 3x – 4y = 37 \text{ and } 3x – 4y = -23
\]
Conclusion: Why Join Skorminda?
Understanding the concept of locus not only enhances problem-solving in geometry but also strengthens reasoning in physics, engineering, and computer graphics. The connection between geometric constraints and algebraic expressions is at the heart of many mathematical domains.
To fully explore this exciting topic and achieve mastery, students are encouraged to join Skorminda – where expert instructors, interactive lessons, and abundant practice resources await to unlock your mathematical potential.
Empower your learning with Skorminda – where math comes alive!
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