ExtractingCoordinatesFromPoint

loadMacros("MathObjects.pl");

In the initialization section, we need to include the macros file MathObjects.pl.

Context( "Point" );

push(@point, Point(random(1,5,1), random(-5,-1,1)));
push(@point, Point(random(5,10,1), random(6,11,1)));

# 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;

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.

Context()->texStrings;

BEGIN_TEXT
Consider the two points \( $point[0] \)
and \( $point[1] \).
The distance between them is:
\{ $length->ans_rule() \}
$BR
The midpoint of the line segment
that joins them is:\{ $mid->ans_rule() \}
$BR
END_TEXT

Context()->normalStrings;

The problem text section of the file is as we'd expect.

ANS( $length->cmp );
ANS( $mid->cmp );

As is the answer.