Locus problems (as required)

Imagine you have a tiny ant walking around. If you give the ant a rule, like "always stay exactly 5 cm away from this big rock," the path the ant walks would be a locus.

• Locus (pronounced 'LOW-cuss'): It's just a collection of all the points that fit a certain rule or condition. Think of it like a treasure map where 'X' marks every single spot that meets the treasure's description.

Let's break down some common types of paths (loci, which is the plural of locus) you might see:

• The path of points equidistant from a single point: "Equidistant" just means "the same distance." If our ant has to stay the same distance from a rock, what shape does it make? A circle! The rock is the center, and the distance is the radius.

• The path of points equidistant from two points: Imagine two rocks. Our ant has to stay exactly the same distance from Rock A as it is from Rock B. What path would it walk? A straight line that cuts exactly in the middle between the two rocks, and is perpendicular (at a perfect 90-degree angle) to the line connecting them.

• The path of points equidistant from a line: Now, what if our ant has to stay the same distance from a long, straight fence? It would walk a parallel line right next to the fence.

• The path of points equidistant from two intersecting lines: If our ant has to stay the same distance from two fences that cross each other, it would walk along a line that cuts the angle between them exactly in half. This is called an angle bisector.

Let's imagine you're at a funfair, and there's a ride where a little car spins around a central pole. The car is always the same distance from the pole. The path the car takes is a circle! This is a locus where all points are equidistant (the same distance) from a single point (the pole).

Now, let's say you're trying to place a new swing set in your garden. You want to make sure it's exactly the same distance from your house as it is from your big oak tree. If you drew all the possible spots where the swing set could go, you'd draw a straight line that cuts right between the house and the tree. This line is the locus of points equidistant from two fixed points (your house and your tree).

When you're asked to find the equation of a locus, you're basically translating a word problem into an algebraic equation.

1. Understand the Rule: Read the problem carefully to identify the condition or rule that the moving point must follow. Is it equidistant from a point? From a line? From two points?
2. Name Your Moving Point: Call the moving point P(x, y). This is the point whose path you are trying to describe.
3. Translate to Math: Use the distance formula or other geometric formulas to write down the condition from Step 1 using x and y.
4. Formulate the Equation: Set up an equation based on the condition. For example, if 'distance from A = distance from B', write that mathematically.
5. Simplify: Do the algebra to simplify the equation into a recognizable form (like y = mx + c for a line, or (x-h)^2 + (y-k)^2 = r^2 for a circle). This simplified equation is the equation of the locus.

Let's find the equation of the locus of a point P(x, y) that is always equidistant from the point A(2, 0) and the point B(6, 0).

1. The Rule: The distance from P to A must be equal to the distance from P to B.
2. Moving Point: P(x, y).
3. Translate to Math: We use the distance formula, which is like using Pythagoras' theorem to find the length of a slanted line: distance = √((x2-x1)² + (y2-y1)²).
   • Distance PA = √((x-2)² + (y-0)²) = √((x-2)² + y²)
   • Distance PB = √((x-6)² + (y-0)²) = √((x-6)² + y²)
4. Formulate the Equation: PA = PB, so √((x-2)² + y²) = √((x-6)² + y²).
5. Simplify: To get rid of the square roots, we square both sides:
   • (x-2)² + y² = (x-6)² + y²
   • x² - 4x + 4 + y² = x² - 12x + 36 + y²
   • Subtract x² and y² from both sides: -4x + 4 = -12x + 36
   • Add 12x to both sides: 8x + 4 = 36
   • Subtract 4 from both sides: 8x = 32
   • Divide by 8: x = 4

So, the equation of the locus is x = 4. This is a vertical line. It makes sense because a vertical line at x=4 is exactly in the middle of x=2 and x=6.

Here are some common traps students fall into and how to dodge them!

• Mistake 1: Forgetting to square both sides when dealing with square roots.

  • ❌ If you have √A = √B, don't just say A = B without squaring. You need to square both sides first.
  • ✅ Always square both sides of the equation when you have square roots on both sides (like from the distance formula) to simplify correctly. (√A)² = (√B)² becomes A = B.

• Mistake 2: Making algebraic errors when expanding brackets.

  • ❌ (x-a)² does not equal x² - a². Remember the middle term!
  • ✅ Remember the formula for expanding brackets: (a - b)² = a² - 2ab + b². So, (x-2)² = x² - 4x + 4. Take your time with these expansions.

• Mistake 3: Not simplifying the final equation into a recognizable form.

  • ❌ Leaving your answer as something messy like 2x - 8y + 10 = 0 when it could be x - 4y + 5 = 0.
  • ✅ Always simplify your equation as much as possible. Divide by common factors, rearrange into standard forms (like y = mx + c for a line or the general form for a circle). This makes it easier to check your answer and for the examiner to mark.