Spatial Sense

Experiential Learning, Collaborative Learning, Talk for Learning and building on what others say.
Learning Experience: Learners in collaborative groups deduce the equation of a locus under given conditions.
Activity 1: Learners in groups collaborate and:
- discuss the meaning of locus and share their thoughts with the whole class.
- deduce that a set of points that satisfy a given condition is a circle.
- establish locus as a geometric path that can be written algebraically.
Activity 2: Learners in groups collaborate to construct loci under given conditions.
Condition 1: Locus equidistant from the centre Learners in mixed-ability groups:
- construct a locus whose distance is equidistant from the centre.
- establish that the locus of a fixed point is the circumference of a circle, as shown.

Condition 2: Locus equidistant from two fixed points Learners in mixed-ability groups brainstorm and discuss:
- a locus equidistant from two fixed points
- establish that a locus from two fixed points is a perpendicular bisector, as shown.

Condition 3: Locus from two fixed lines Learners in mixed-ability groups brainstorm and discuss:
- how to construct a locus equidistant from two fixed lines, say 𝐴 and 𝐵
- establish that a locus from two fixed lines is an angle bisector, and for any point lying on the angle bisector, the perpendicular distance is equidistant from the two lines.

Condition 4: Locus equidistant from a line Learners in mixed-ability groups brainstorm and discuss:
- how to construct a locus equidistant from a line, say 𝐴
- deduce that a locus from a line 𝐴 is a parallel line.

Condition 5: The equation of a locus from a fixed point Learners in groups deduce that since the locus of a point fixed point is the circumference of a circle, the equation of this locus is the equation of a circle. That is, the equation of a locus from a fixed point is 𝑟 ) = (𝑥 − ℎ) ) − (𝑦 − 𝑘) )
Example 1: Find the equation of the locus of a moving point 𝑃(𝑥, 𝑦), which is always at a distance of 3 units from a fixed point 𝑄(1,2). Solution: (𝑥 − 1) ) − (𝑦 − 2) ) = 3 𝑥 ) + 𝑦 ) − 2𝑥 + 4𝑦 − 12 = 0
Learners in mixed-ability groups: establish that since, a locus 𝑃 from two fixed points (𝐴) is a perpendicular bisector, then |𝐴| = |𝐵| apply their knowledge in finding the distance between two points to find the equation of this loci
Example 2: Find the equation of a locus of a point that is equidistant from the points 𝐴(−2,0) and 𝐵(3,2).
Solution w(𝑥 + 2) ) + (𝑦 − 0) ) = w(𝑥 − 3) ) + (𝑦 − 2) ) 10𝑥 + 4𝑦 − 9 = 0 Learners in mixed-ability apply their knowledge of dividing a line in a given ratio to find the equation of a locus involving a constant distance 𝑟 from two fixed points in the ratio 𝑚: 𝑛 Example 3: If 𝐴(2, 0) and 𝐵 (0, −2) are two fixed points and point 𝑃 moves with a ratio so that 𝐴 ∶ 𝐵 = 1 ∶ 3. Find the equation of the locus of point 𝑃 Solution: 𝐴 1 = 𝐵 3 (3𝐴) ) = 𝐵 ) ⟹ 9𝐴 ) = 𝐵 )
9[(𝑥 − 2) ) + (𝑦) ) ] = [(𝑥) ) + (𝑦 + 2) ) ] 8𝑥 ) + 8𝑦 ) − 36𝑥 − 4𝑦 + 32 = 0 Other conditions Learners in groups apply their knowledge in equations of perpendicular lines to find the equation of locus of a Point 𝑃(𝑥, 𝑦) such that 𝐴 is Perpendicular to 𝐵 where 𝐴 and 𝐵 are constants. Learners in mixed-ability groups brainstorm to solve an example.
Example 4: A point 𝑃(𝑥, 𝑦) moves so that 𝐴 and 𝐵 are perpendicular. Given that 𝐴(1,2) and 𝐵(2,4) find the equation of the locus 𝑃.
Solution: A*) A*, The product of slopes of perpendicular lines is -1, therefore i 0*! j i )*0 j = −1
The equation of the locus P is 𝑥 ) + 𝑦 ) − 3𝑥 − 6𝑦 + 10 = 0
Learners in groups collaborate to find the locus of a point 𝑃(𝑥, 𝑦) that is always at a constant distance of units from a given Line. Learners work in mixed-ability groups to solve examples.
Example 5: What is the locus of point 𝑃(𝑥, 𝑦) that is always 2 units from the line 𝑥 = 4? Solution 𝐴(4, 𝑦), 𝑃(𝑥, 𝑦) and the constant distance (𝑑) = 2 𝑃 = w(𝑥 − 4) ) + (𝑦 − 𝑦) ) = 2 𝑥 ) − 8𝑥 + 12 = 0 𝑥 = 2 𝑜 𝑥 = 6 Learners in their groups create questions involving other conditions to be solved.
Example 6: 𝐴(4,5) and 𝐵(−2,7) are given points. Find the equation of the line such that 2𝑃 = 𝑃.
Solution: 𝑃(𝑥, 𝑦), 𝐴(4,5) and 𝐵(−2,7) Given condition is 2𝑃 = 𝑃 ∴ 4𝑃 ) = 𝑃 ) 4[(𝑥 − 4) + (𝑦 − 5) ) ] = (𝑥 + 2) ) + (𝑦 − 7) ) ) The equation of the locus is 3𝑥 ) + 3𝑦 ) − 36𝑥 − 26𝑦 + 111 = 0
Teaching and Learning Resources:
- Worksheets
- Scientific Calculator
- Technological tools, apps, etc.
- SHS Additional Mathematics Curriculum
Assessment (2.2.1.AS.4). The document marks these depth-of-knowledge levels for this indicator: Level 4 Extended critical thinking and reasoning.