This problem shows how to extract the coordinates from a point.
Download file: ExtractingCoordinatesFromPoint.pg
DOCUMENT();
loadMacros('PGstandard.pl', 'PGML.pl', 'PGcourse.pl');
Preamble
These standard macros need to be loaded.Context('Point');
push(@point, Point(random(1, 5), random(-5, -1)));
push(@point, Point(random(5, 10), random(6, 11)));
# now we have two points, $point[0] = (x1,y1)
# and $point[1] = (x2,y2).
# the following makes $d1 = x1 - x2, $d2 = y1 - y2
($d1, $d2) = ($point[0] - $point[1])->value;
$length = Compute("sqrt( ($d1)^2+($d2)^2 )");
$mid = ($point[1] + $point[0]) / 2;
BEGIN_PGML
Consider the two points [`[$point[0]]`] and [`[$point[1]]`].
The distance between them is: [___]{$length}
The midpoint of the line segment that joins them is:
[___]{$mid}
END_PGML
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 (x) 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_SOLUTION Solution explanation goes here. END_PGML_SOLUTION ENDDOCUMENT();
Solution
A solution should be provided here.