ExtractingCoordinatesFromPoint

Your title here: PG Code Snippet

This code snippet shows the essential PG code to evaluate antderivative and general antiderivative formulas. Note that these are insertions, not a complete PG file. This code will have to be incorporated into the problem file on which you are working.

This wiki page is under construction as of 6/13/08.

Problem Techniques Index

PG problem file	Explanation
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) ) ); ($d1, $d2) = ($point[0] - $point[1])->value; # $d1 = x1 - x2, $d2 = y1 - y2 $length = Compute("sqrt( ($d1)^2+($d2)^2 )"); $mid = ( $point[1] + $point[0] ) / 2;	In the problem setup section of the file, we don't use the Vector context and norm( $point[0] - $point[1] ) to determine length because we don't want to accept an answer like |<5,7>-<7,8>| You could use $length=norm( $point[0] - $point[1] ); with the Vector context if you didn't care about accepting 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 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.

Problem Techniques Index