Thursday, July 26, 2007

Making Connections: An Algebra Problem -- Or Is it Geometry? Or Both?


Lines L and M are perpendicular. Line L contains (0,0). Lines L and M intersect at A(5,2). Find x.

Comments:
(1) Would students in geometry be more likely to consider some or all of the following: similar triangles, altitudes on hypotenuse theorems, areas, Pythagorean approaches? If this were presented as a coordinate problem in Algebra 2, what would be the most likely approach? Would some students use the distance formula?
(2) As always, it is our role as educators to present these kinds of challenges and to encourage students to think more deeply. Making connections between algebra and geometry happens naturally for some, but certainly not for all! We must enable this dialogue via the classroomm environment we establish and the kinds of questions we ask. It is important to first decide the goal. What concepts are we trying to develop? Slope? Ratios of corresponding parts? How many students never make the connection between the two!
(3) The answer is 29/5 or 5.8. Think of at least two methods!

9 comments:

mathmom said...

The first method that came to me I think is an Algebra I method.

The slope of line L is 2/5. Since line M is perpendicular, its slope is -5/2. (Is that an AlgI concept?)

Using point-slope form (definitely Alg1 around here), the equation of line M is y - 2 = -5/2 (x-5)

Set y=0 and solve for x, reveals that x=29/5 as expected.


I don't see a clever geometric approach, but using that old standby Pythagoras:

|OA| = sqrt(29)
|AC| = sqrt(4+(x-5)^2)
x^2 = |OA|^2 + |AC|^2
= 29 + 4 + x^2 - 10x + 25
10x = 58 and again x=29/5 QED

Totally_clueless said...

If the students are not familiar with the relationship between the slopes of perpendicular lines, then I would expect them to use geometric approaches.

Approach 1.5 (from Mathmom's terminlogy) is to use the line equation of the form y=-5/2x+C, find C, and then find x

Approach 3 is to calculate twice the area of the triangle as the product of the two legs or the the base and the altitude, and equate them. It takes some algebra, and then you realize the quadratic you have has special properties, which makes the solution easier. It is far easier to make silly calculation mistakes as you make your methods more involved (unnecessarily, IMHO)

TC

Dave Marain said...

I had to post this problem quickly today to get the bad taste of those Roman Numerals out of my mouth!

Seriously, mathmom and tc, your solutions are great. Several of the stronger students used the equations of lines to solve it. Two students used the altitude on hypotenuse theorem (it always amazes me when students can retain this after a couple of years.)

Using O for the origin and dropping the altitude from point A to point B on the x-axis, we can apply the geometric mean idea in a nice way:
(AB)^2 = (OB)(BC) or
2^2 = (5)(BC) --> BC = 0.8.
Therefore, x = OC = 5.8. Voila!
Yes, students are more likely to recall the similar triangles approach or using areas but it's always nice to see how the geometric mean actually makes things easy!
Now, for the algebra...
Virtually none of the students used the definition of slope directly. Rather than worry about the equation of the line, proceed as follows:
1. slope of OA = 2/5
2. Therefore, slope of line AC = -5/2.
3. Therefore, (2-0)/(5-x) = -5/2 and x = 29/5. No need to worry about the point-slope or other forms.

mathmom said...

Therefore, (2-0)/(5-x) = -5/2 and x = 29/5. No need to worry about the point-slope or other forms.

Too simple. ;-)

mathmom said...

Approach 1.5 (from Mathmom's terminlogy) is to use the line equation of the form y=-5/2x+C, find C, and then find x


Wait, find C seems like a big handwave there. I think it's just as hard to find C (y-intercept) as it is to directly find x (x-intercept) unless I'm missing something.

mathmom said...

the altitude on hypotenuse theorem .... which is what, exactly?

Dave Marain said...

mathmom--
The altitude to the hypotenuse of a right triangle divides the triangle into 2 smaller right triangles, each of which is similar to the original. There are several proportions that result but the most famous one states that the altitude is the geometric mean between the two parts of the hypotenuse (hard to explain without a diagram). If the two segments on the hypotenuse have lengths a and b and the altitude has length h, then this can be set up as:
a/h = h/b or
h^2 = ab
or h = SQRT(ab), the definition of the geometric mean between a and b.
Hard for students to retain this fact but geometric means are fascinating stuff...

peter said...

hi mr. marain! this is peter from michael academy. haha, i spotted your blog after all.

Totally_clueless said...

Hi Mathmom,

You know the equation of the line is y=-5/2x+C and that is passes through (5,2). Thus, you can find C.

It is up to one's preference to wave or flail your hands when doing this :-)

TC