Difference between revisions of "ExtractingCoordinatesFromPoint"
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| − | In the problem setup section of the file, we |
+ | In the problem setup section of the file, we put the value of the subtraction of two Points in two variables, <code>$d1</code>, the x coordinate, and <code>$d2</code>, the y coordinate. This is achieved by calling Point's <code>value</code> method, as shown. |
| − | + | </p> |
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| + | <p>Alternative method: If you want to get only one of the coordinates of a Point, you can use the <code>extract</code> method, for example: <code>$x = $point->extract(1);</code>. This gets the first coordinate of <code>$point</code> (x) and assigns it to the variable <code>$x</code>. |
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| + | We don't use <code>Context("Vector");</code> and <code>norm( $point[0] - $point[1] )</code> here |
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| + | to determine length because we don't want to accept an answer like <code>|<5,7>-<7,8>|</code>. |
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| + | </p> |
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| + | <p>Alternative method: You can use <code>$length=norm( $point[0] - $point[1] );</code> with <code>Context("Vector");</code> if you want to accept answers that are valid in the Vector context (such as the absolute value of a vector). |
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| − | We need to put parentheses around <code>$d1</code> and <code>$d2</code> because if <code>$d1 |
+ | We need to put parentheses around <code>$d1</code> and <code>$d2</code> because if <code>$d1 = -6</code>, then <code>-6^2 = -36</code>, not <code>36</code>, as desired. However, if the code is <code>($d1)^2</code> then that evaluates as <code>(-6)^2 = 36</code>, as desired. |
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Revision as of 21:07, 13 June 2008
Extracting coordinates from a Point: 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.
| PG problem file | Explanation |
|---|---|
loadMacros("MathObjects.pl");
|
In the initialization section, we need to include the macros file |
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 put the value of the subtraction of two Points in two variables, Alternative method: If you want to get only one of the coordinates of a Point, you can use the
We don't use Alternative method: You can use
We need to put parentheses around |
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. |
