The scaffold came through several activities:
(1) The first activity of this lesson is to find a linear equation when given 2 points.
If A (2, -2) and B (-2, 4), find the equation of the line AB.
And hence, state the value of the x-intercpet and y-intercept.
Students were able to use the formula to find the gradient using the formula,
(y2-y1)/(x2-x1), which is - 3/2.
One question from the student was...
"when m could be written as 6/(-4) and subsequently (-3)/2.
So, does it matter where the negative sign is? Is it with numerator or the denominator?"
Hm... while the response to the student was, "It doesn't matter..."
Well, I think it's where the students always get confuse over... at any level.
So, would writing - 3/2 helps? Maybe... when the negative sign belongs to the entire fraction 3/2.
It was also noted that students were introduced to the fact that the y-intercept can be expressed as (0, c) that is, when the x coordinate is zero.
Well, this is an important concept that students must understand.
To find the y-intercept, it was quite obvious that students knew that they would substitute any of the coordinate to the "half-found" equation: y = -3/2 x + c
(2) Indeed, I enjoyed the next part of the activity :) an interesting one - getting students to learn the 'beauty' of expressing the equation in the form of x/a + y/b = 1
I could not recall my teacher was able to explain so clearly what's the significance of expressing a linear equation in the form x/a + y/b = 1
Hm... I think this was introduced in Secondary 3 Additional Mathematics then. The key deliverable at this point, I guess is leading the students to be able to how the x- and y-intercepts could be so well represented in an equation expressed in the form of x/a + y/b = 1
The process to elicit responses from students really made their minds spin. Hahah...
The best part is really asking the students "Why" the a is the intercept and why b is the y-intercept... and how they derive the explanation through application of the following concepts and solve algebraically:
- what happens to the equation when the line cuts the y axis (i.e. to find the y-intercept)? i.e. x coordinate = 0, hence y = b.
- similarly, to show why a is the x intercept. The same concept applies: when the line cuts x axis, the value of y is zero. Hence x/a + 0/b = 1; therefore, x = a.
It really required them to have solid understanding of the intercepts (i.e. the y-coordinate of the x-intercept is zero while x coordinate of the y-intercept is zero, too).
- It sounds easy, but definitely not something simple for a 13-year old mind who just started to touch-base with the Cartesian Coordinate Geometry.
- Indeed, I'm really curious to find out how many in the class would be able to explain...
In fact, I like it very much... it's like gaining enlightenment at this age - making new discovery in what I was familiar with!
Students observation, for students to the relationships of lines passing through the same point:
What kind of lines pass through the point (-2, 0)?
- What kind of gradient and y-intercepts do we get?
2 linear equations contributed by students:
- y = x + 2 [root = -2]
- y = x - 4 [root = 4]
- y = (x + 2)(x - 4) [roots = -2 and 4]
Why is the roots (x-intercepts) of the straight lines are the same as the roots of the parabola?
I guess that's where technology came in very handy that allows students to see how the lines coincide with each other.
Of course, the other part was when students were guided to check out the "signs" of the gradient and y-intercepts at the 4 different quadrants.
By the way, I wonder, after this first introduction, how many students were able to explain the process they went through in this activity.
All in all, I enjoyed this very energetic lesson :)
- The explanation was clear... though it's a bit quick in the pace.
- On the other hand, I think too much things were packed into this 1-hour lesson.
- Perhaps some breathing space for the students to attempt some tasks would help to check their understanding.