Welcome to the World of Loci!
Hey everyone! Welcome to the topic of Loci. The word "locus" might sound a bit fancy, but the idea behind it is super simple and actually quite visual. Think of it like a treasure hunt. The instructions on a map ("stand 5 steps away from the big oak tree") describe a path or a set of possible locations. In math, that path is a locus!
In this chapter, you'll learn how to describe these paths both with words and pictures, and then, most importantly, with the language of algebra – equations! This skill is a fantastic bridge between geometry (shapes) and algebra (equations).
So, What Exactly is a Locus?
A locus is the set of all points that satisfy a given rule or condition. It's the complete path or region traced by a point as it moves according to that specific rule.
Imagine the tip of the second hand on a clock. As it moves, it's always the same distance from the center. The path it traces is a circle. That circle is the locus of the tip of the second hand.
Quick Review: The Distance Formula
Before we jump in, let's refresh a key tool we'll need: the distance formula. It helps us find the distance between any two points, say $$A(x_1, y_1)$$ and $$B(x_2, y_2)$$, on a coordinate plane.
The formula is:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$We'll be using this a LOT to turn geometric rules into equations. Don't worry, you'll get plenty of practice!
The "Big Five": Your Locus Toolkit
Let's explore the five most common types of loci. Understanding these is key to mastering the topic. For each one, we'll look at the condition, the shape it creates, and how to describe it.
Locus 1: A Fixed Distance from a Fixed Point
The Condition
A point moves so that it is always a fixed distance (let's call it r) from a fixed point (let's call it C).
The Result: A Circle
This is the classic definition of a circle!
Let's Visualize It
Imagine a dog tied to a pole with a leash. The pole is the fixed point and the length of the leash is the fixed distance. The path the dog can walk at the full extent of its leash is a perfect circle.
How to Describe It
The locus of the point is a circle with the fixed point as the centre and the fixed distance as the radius.
Locus 2: Equal Distance from Two Given Points
The Condition
A point moves so that it is always the same distance (equidistant) from two different fixed points (say, A and B).
The Result: The Perpendicular Bisector
The locus is the perpendicular bisector of the line segment joining the two points.
Let's Visualize It
Imagine two friends, Amy (point A) and Ben (point B), standing in a field. If you want to stand in a spot that is exactly the same distance from Amy as it is from Ben, you would have to stand somewhere on a straight line that cuts exactly between them at a right angle.
How to Describe It
The locus of the point is the perpendicular bisector of the line segment joining points A and B.
Locus 3: A Fixed Distance from a Line
The Condition
A point moves so that it is always a fixed distance from a fixed straight line.
The Result: A Pair of Parallel Lines
The locus is a pair of parallel lines, one on each side of the original line, at the specified fixed distance from it.
Let's Visualize It
Think of the painted lines that mark a swimming lane. The centre rope is the fixed line. The edges of the lane are always the same distance from the centre rope. They form two parallel lines.
How to Describe It
The locus of the point is a pair of straight lines parallel to the given line.
Locus 4: Equal Distance from Two Parallel Lines
The Condition
A point moves so that it is always the same distance from two parallel lines.
The Result: A Parallel Line in the Middle
The locus is a single straight line, parallel to the two given lines and lying exactly midway between them.
Let's Visualize It
Imagine a road with two parallel kerbs. The centre line painted on the road is always the same distance from both kerbs. That centre line is the locus.
How to Describe It
The locus of the point is a straight line parallel to and midway between the two given parallel lines.
Locus 5: Equal Distance from Two Intersecting Lines
The Condition
A point moves so that it is always the same distance from two intersecting lines.
The Result: A Pair of Angle Bisectors
The locus is the pair of angle bisectors of the angles formed by the two intersecting lines. These two bisectors will always be perpendicular to each other.
Let's Visualize It
Imagine two straight roads intersecting. If you walk in a way that you are always the same distance from the edge of both roads, you would be walking along the line that cuts the angle between the roads exactly in half. Since there are two pairs of angles, there are two such paths (which form a cross).
How to Describe It
The locus of the point is the pair of angle bisectors of the angles between the two given lines.
Key Takeaway for Sketching
When you read a locus problem, first identify which of these five conditions it matches. Visualizing the real-world analogy can really help you sketch the correct shape!
Speaking in Algebra: Finding the Locus Equation
Now for the really cool part: turning these geometric rules into algebraic equations. This is where you can prove what the shape is, without a doubt! Don't worry if this seems tricky at first, it's a process that gets easier with practice.
The Golden Rule for Finding a Locus Equation
Here’s the step-by-step method that works every time:
1. Define the Point: Let the moving point be $$P(x, y)$$. This is the point that will trace our locus.
2. Translate the Words into Math: Write down the given condition as a mathematical statement using coordinates. This usually involves the distance formula.
3. Simplify: Do the algebra to simplify the equation into its final form.
Example 1: The Perpendicular Bisector
Problem: Find the equation of the locus of a point P which is equidistant from A(1, 3) and B(5, 7).
Step-by-step Solution:
1. Define the Point: Let the moving point be $$P(x, y)$$.
2. Translate the Words into Math: The condition is "equidistant from A and B". This means the distance PA must equal the distance PB.
So, PA = PB.
Now, let's use the distance formula:
$$ \sqrt{(x - 1)^2 + (y - 3)^2} = \sqrt{(x - 5)^2 + (y - 7)^2} $$Memory Aid: To get rid of the annoying square roots, we can square both sides! It's much easier to work with $$PA^2 = PB^2$$.
$$ (x - 1)^2 + (y - 3)^2 = (x - 5)^2 + (y - 7)^2 $$3. Simplify: Now, expand the brackets. Be careful with your algebra!
$$ (x^2 - 2x + 1) + (y^2 - 6y + 9) = (x^2 - 10x + 25) + (y^2 - 14y + 49) $$The $$x^2$$ and $$y^2$$ terms appear on both sides, so they cancel out. Phew!
$$ -2x + 1 - 6y + 9 = -10x + 25 - 14y + 49 $$Let's collect all the terms together:
$$ -2x - 6y + 10 = -10x - 14y + 74 $$Now, move all the terms to one side to make the equation tidy. Let's move everything to the left side.
$$ (-2x + 10x) + (-6y + 14y) + (10 - 74) = 0 $$ $$ 8x + 8y - 64 = 0 $$We can simplify this by dividing everything by 8:
$$ x + y - 8 = 0 $$Answer: The equation of the locus is $$x + y - 8 = 0$$. This is the equation of a straight line, which confirms our geometric result (a perpendicular bisector).
Example 2: The Circle
Problem: Find the equation of the locus of a point P that is always 3 units away from the point C(2, -4).
Step-by-step Solution:
1. Define the Point: Let the moving point be $$P(x, y)$$.
2. Translate the Words into Math: The condition is "the distance from P to C is always 3".
So, PC = 3.
Using the distance formula:
$$ \sqrt{(x - 2)^2 + (y - (-4))^2} = 3 $$ $$ \sqrt{(x - 2)^2 + (y + 4)^2} = 3 $$3. Simplify: Again, let's square both sides to remove the square root.
$$ (x - 2)^2 + (y + 4)^2 = 3^2 $$ $$ (x - 2)^2 + (y + 4)^2 = 9 $$Answer: The equation of the locus is $$(x - 2)^2 + (y + 4)^2 = 9$$. This is the standard equation of a circle, just as we expected!
Example 3: The Parabola
Problem: Find the equation of the locus of a point P which is equidistant from the point F(0, 2) and the line L with the equation y = -2.
Step-by-step Solution:
1. Define the Point: Let the moving point be $$P(x, y)$$.
2. Translate the Words into Math: The condition is that the distance from P to F is equal to the distance from P to the line L.
The distance $$PF$$ is easy, using the distance formula:
$$ PF = \sqrt{(x - 0)^2 + (y - 2)^2} = \sqrt{x^2 + (y - 2)^2} $$The distance from a point $$(x,y)$$ to a horizontal line $$y=k$$ is simply $$|y-k|$$. So, the distance from $$P(x,y)$$ to the line $$y=-2$$ is $$|y - (-2)| = |y+2|$$.
Our condition is PF = (Distance from P to L).
$$ \sqrt{x^2 + (y - 2)^2} = |y + 2| $$3. Simplify: Let's square both sides.
$$ x^2 + (y - 2)^2 = (y + 2)^2 $$Expand the brackets:
$$ x^2 + (y^2 - 4y + 4) = (y^2 + 4y + 4) $$The $$y^2$$ and $$4$$ terms cancel out from both sides!
$$ x^2 - 4y = 4y $$Now, let's get y by itself.
$$ x^2 = 8y $$ $$ y = \frac{1}{8}x^2 $$Answer: The equation of the locus is $$y = \frac{1}{8}x^2$$. This is the equation of a parabola opening upwards, in the form $$y=ax^2+bx+c$$ (where b and c are zero here).
Watch Out! Common Mistakes & Exam Tips
Common Mistakes to Avoid
- Forgetting the "pair": For Locus 3 (fixed distance from a line) and Locus 5 (equidistant from intersecting lines), the answer is a PAIR of lines. Don't just draw one!
- Algebra Errors: Be very careful when expanding brackets, especially with negative signs. For example, $$(y-3)^2$$ is NOT $$y^2 - 9$$.
- Mixing up Conditions: Read the question carefully. Is it a "fixed distance from one point" (circle) or "equal distance from two points" (perpendicular bisector)? They sound similar but give very different results.
Top Exam Tips
- Sketch First!: Before you start any algebra, draw a quick, rough sketch of the points and/or lines. This will help you predict what the locus should look like. If your final equation is a circle, but your sketch suggested a straight line, you know you've made a mistake somewhere!
- Square Both Sides: To make your life easier, always use the $$(\text{distance})^2$$ form when dealing with "equidistant" conditions. It eliminates square roots right at the start.
- Show Your Steps: Clearly write down the condition (e.g., PA = PB) before you substitute in the formulas. This shows the examiner you know what you're doing, even if you make a small calculation error later.
Key Takeaways: Your Loci Cheat Sheet
Here is a quick summary of everything we've covered. Use this for your revision!
Condition 1: Fixed distance from a fixed point.
Locus: A Circle.
Condition 2: Equidistant from two fixed points.
Locus: The Perpendicular Bisector of the line segment joining them.
Condition 3: Fixed distance from a fixed line.
Locus: A Pair of Parallel Lines.
Condition 4: Equidistant from two parallel lines.
Locus: A single Parallel Line midway between them.
Condition 5: Equidistant from two intersecting lines.
Locus: The Pair of Angle Bisectors.
You've got this! Practice these concepts, and you'll find that Loci is one of the most logical and satisfying topics in geometry.