Perform calculations with Points.
Download file: CalculatingWithPoints.pg
DOCUMENT();
loadMacros('PGstandard.pl', 'PGML.pl', 'PGcourse.pl');
Preamble
These standard macros need to be loaded.Context("Point");
@points = (
Point(random(1, 5, 1), random(-5, -1, 1)),
Point(random(5, 10, 1), random(6, 11, 1))
);
# If $point[0] = (x1,y1) and $point[1] = (x2,y2),
# then the following makes $d1 = x1 - x2, $d2 = y1 - y2
($d1, $d2) = ($points[0] - $points[1])->value;
$length = Compute("sqrt( ($d1)^2+($d2)^2 )");
$mid = ($points[1] + $points[0]) / 2;
Setup
In the problem setup section of the file, we put the value of the subtraction of two Points in two variables, $d1, the x coordinate, and $d2, the y coordinate. This is achieved by calling Point’s value method, as shown.
Alternative method: If you want to get only one of the coordinates of a Point, you can use the extract method, for example: $x = $point->extract(1);. This gets the first coordinate of $point and assigns it to the variable $x.
We don’t use Context("Vector"); and norm( $point[0] - $point[1] ) here to determine length because we don’t want to accept an answer like |<5,7>-<7,8>|.
Alternative method: You can use $length=norm( $point[0] - $point[1] ); with Context("Vector"); if you want to accept answers that are valid in the Vector context (such as the absolute value of a vector).
We need to put parentheses around $d1 and $d2 in the Compute expression because if $d1 = -6, then -6^2 = -36, not 36, as desired. However, if the code is ($d1)^2 then that evaluates as (-6)^2 = 36, as desired.
BEGIN_PGML
Consider the two points [` [$points[0]] `]
and [` [$points[1]] `].
a. The distance between them is: [_______]{$length}
b. The midpoint of the line segment that joins them is:
[______]{$mid}
END_PGML
Statement
This is the problem statement in PGML.BEGIN_PGML_SOLUTION Solution explanation goes here. END_PGML_SOLUTION ENDDOCUMENT();
Solution
A solution should be provided here.