ALG 12

This week we look at Cartesian graphs of straight lines again. The aim is to focus on the visual aspects of the graphs, in particular the slopes of various lines, and to find points on the lines by using notions of proportion and similarity, rather than by immediately expressing the lines as equations (though that would, of course, be a perfectly legitimate thing to do).
In many of the tasks we have deliberately chosen points with very different sets of coordinates [such as (0, 0) and (100, 101) in ALG 12A] so as to encourage students to visualise or sketch what's going on, rather than attempt to use accurate drawings.
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MONDAY: Many students should be able to find coordinates for some points on the red line, eg (50, 50.5) and (200, 202). Encourage students to seek out lots of points as some might provide insights that would help with parts i. and ii., eg (10, 10.1) and (1, 1.01).

It is likely that some students will suggest that the value of the y-coordinate in part i. is 102. How can we show (eg with the aid of a sketch) that a point with coordinates (101, 102) is not on the red line? How can we decide whether it is above or below?
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TUESDAY: We search for some more widely-spaced points.

A sketch might help; a table of values might help, to record moves towards the y-axis, say, in well-chosen steps (eg 10 to the left, 1 up).
PS: Here's a nice solution, posted on Twitter. Quite condensed but worth unpicking!

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WEDNESDAY: More points on a line, near and far...

Follow the light green straight line.... Not that different from the previous task, but no harm in consolidating earlier ideas. And you might like to think up further variations....
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THURSDAY: Two lines, by way of a change.....

Here we take a somewhat different approach by offering two methods for students to interpret. The first is quite grounded, the second is more abstract. You might, of course, prefer to ask students to come up with their own methods first, though the task itself is quite demanding.
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FRIDAY: By way of a further change, we home in on the microscopic - though you might want to zoom out again ....

Don't be put off by the fractions! The underlying algebraic relationships are not that difficult.... And there are also some nice geometric approaches that can be used to solve the task.
Commentary: A closer look at the coordinates reveals that for one of the line segments the coordinates add up to 1, for the other the y-coordinate is 3 times the x-coordinate. Which two numbers add up to 1, where one number is 3 times the other? We can get there fairly rapidly using trial and improvement. Or we could draw a 'bar' or 'rod' or number line (although this requires quite a difficult switch in thinking - the 'bar' is not a line segment in the Cartesian plane!):

We can also express the relations between the coordinates more formally, as the simultaneous equations x + y = 1 and y = 3x, leading to x + 3x = 1, so x = 1/4, y = 3/4.
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Here's a nice approach (by Matt L) that is more geometric. It involves constructing a new point on the line through the steeper line segment that is vertically below the left hand end-point of the other line segment. It also uses the fact that this line segment slopes at 45˚, so the two segments labelled 'a' are equal.

We can extend the two given line segments still further, until they touch a unit square. [In fact, this is what I started with when I designed the task!]

A couple of years ago there was a flurry of tasks on Twitter involving dissected squares of this sort, usually with one region shaded, the aim being to find 'What fraction of the square is shaded?' My favoured approach was to draw equally spaced parallel lines to produce equal intercepts, as here:

One could then argue along these lines: the diagonal of the square has been cut into 4 equal segments, so starting at the bottom left corner, say, of the square, point P is 1/4 units across and 3/4 units up. [Or the height of the triangle seated on the base of the square is 3/4 so its area is 3/8 of the unit square.]
A related and perhaps more direct approach (whether to locate P or to find areas) is to make use of the fact that this triangle on the base of the square is similar to the small 'upside down' triangle above it, with a base that is 3 times as long.
I remember someone coming up with an algebraic solution to one of these area tasks on Twitter. This took me by surprise and I was impressed by the power of such an approach. But it also seemed disappointing to 'abandon' the geometry! Of course, one might say the reverse about the ALG 12E task. Here an algebraic approach would seem an obvious choice and it is perhaps a bit perverse to go to the lengths of constructing a geometric approach instead!