Spatial Sense

Learning Experience: Investigate how the formula for finding the gradient of a line can be applied to find the equation of a parallel and a perpendicular line and how to find the equation of a perpendicular bisector.
Activity 1: Use think pair and share to find the gradient of a parallel line.
Learners in pairs recollect the properties of parallel lines and share their findings.
Expected Response to build on: Parallel lines have the same gradient.
Building on the response, learners in pairs find the gradient of given coordinates and state their observations.
Example 1: Find the gradient of lines 𝐴 and 𝑃 if 𝐴(2, 3)𝐵(5, 6), 𝑃(−1, 4) 𝑄(5, 10)
/*" Solution: Gradient of line 𝐴 = .*) = 1
!(*, Gradient of line 𝑃 = .*(*!) = 1
Observation: Since the gradient of lines 𝐴 and 𝑃 are the same, they are parallel.
Learners, in pairs, solve examples and non-examples to reinforce the concept of parallel lines.
Activity 2: Think-pair-share the concept of the equation of a parallel line.
Learners in pairs apply their knowledge of parallel lines and the gradient of a line to find the equation of a line to find the equation of a parallel line.
Example: Find the equation of the line that is parallel to the line 𝑦 = - 2𝑥 + 6 and pass through the point 𝐴(1, 10).
Solution: The gradient of line 𝑦 = −2𝑥 + 6 is −2, if it passes through the point 𝐴(1,10) then 𝑦 − 10 = −2(𝑥 − 1) Therefore, the equation of the line is 𝑦 = −2𝑥 + 12
Activity 3: Learners in their mixed-ability groups find the gradient of a perpendicular line. Learners in mixed-ability groups recollect the properties of perpendicular lines and share their findings.
Expected Response to build on: Two lines are Perpendicular when they meet at a right angle;
- ! when the slope of one line is 𝑚, the slope of the perpendicular line is . OR If two lines are # perpendicular, then the product of their gradients is -1.
Building on the responses, learners in their mixed-ability groups find the gradient of lines and share their observations.
Example 1: Find the gradient of lines 𝐴 and 𝑃, If 𝐴( - 1, - 1), 𝐵 (0, 4)𝑃( - 4, 3), 𝑄(6, 1) and state your observation.
Solution: The gradient of line 𝐴 = ,*(*!) = 5. (*(*!)
!*"
- ! The gradient of line 𝑃 = /*(*,) = . .
- ! Observation: Since the gradient of lines 𝐴 = 5 and 𝑃 = . then 𝐴 is perpendicular to 𝑃.
Activity 4:
- Use Think-pair-share to deduce the equation of a perpendicular line.
- Learners in pairs apply their knowledge of perpendicular lines, the gradient of a line and finding the equation of a line to find the equation of a perpendicular line.
Example: Find the equation of the line that passes through the point (1, 3) and is perpendicular to the line whose equation is 𝑦 = 2𝑥 + 1.
Solution: The gradient of the line 𝑦 = 2𝑥 + 1 is 2. Therefore, the gradient of the perpendicular will
- !
- ! be . If the line passes through point (1,3), then 𝑦 − 3 = (𝑥 − 1) ) )
- 0 7 Therefore, the equation of the line will be 𝑦 = ) + ) .
Activity 5:
- In mixed-ability groups, learners discuss the equation of the perpendicular bisector of a line.
- Learners in their mixed-ability group discuss and extend their knowledge to find the equation of a perpendicular bisector of a line.
- Learners discuss keywords such as perpendicular and bisector. Learners brainstorm on how to find the equation of a perpendicular bisector.
Example 1: Find the equation of the perpendicular bisector of 𝐴, where 𝐴 and 𝐵 are the points (− 4, 8) and (0, − 2)
Solution: Gradient of line 𝐴 = (*(*,) = ) , then the gradient of the perpendicular bisector is .
- )*1
- . )
- ,$( 1$(*)) The midpoint of line 𝐴 = i , j, which is (−2,3) ) ) Therefore, the Equation of the perpendicular bisector of 𝐴 which passes through the midpoint (- 2, 3), is 2 𝑦 − 3 = (𝑥 + 2) 5 5𝑦 - 2𝑥 - 19 = 0
Teaching and Learning Resources:
- SHS Curriculum
- Mathematical set calculators.
Assessment (1.2.1.AS.5). The document marks these depth-of-knowledge levels for this indicator: Level 2 Skills of conceptual understanding.