Difference between revisions of "ExtractingCoordinatesFromPoint"
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+ | <p style="font-size: 120%;font-weight:bold">This problem has been replaced with [https://openwebwork.github.io/pg-docs/sample-problems/problem-techniques/ExtractingCoordinatesFromPoint.html a newer version of this problem]</p> |
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<h2>Extracting coordinates from a Point: PG Code Snippet</h2> |
<h2>Extracting coordinates from a Point: PG Code Snippet</h2> |
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Context( "Point" ); |
Context( "Point" ); |
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− | push( |
+ | push(@point, Point(random(1,5,1), random(-5,-1,1))); |
− | push( |
+ | push(@point, Point(random(5,10,1), random(6,11,1))); |
− | ($d1, $d2) = ($point[0] - $point[1])->value; # $d1 = x1 - x2, $d2 = y1 - y2 |
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+ | # now we have two points, $point[0] = (x1,y1) |
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+ | # and $point[1] = (x2,y2). |
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+ | # the following makes $d1 = x1 - x2, $d2 = y1 - y2 |
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+ | ($d1, $d2) = ($point[0] - $point[1])->value; |
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</p> |
</p> |
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<p> |
<p> |
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− | We need to put parentheses around <code>$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>($d1)^2</code> then that evaluates as <code>(-6)^2 = 36</code>, as desired. |
+ | We need to put parentheses around <code>$d1</code> and <code>$d2</code> in the <code>Compute</code> expression because if <code>$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. |
</p> |
</p> |
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</td> |
</td> |
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BEGIN_TEXT |
BEGIN_TEXT |
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− | Consider the two points \( $point[0] \) |
+ | Consider the two points \( $point[0] \) |
− | + | and \( $point[1] \). |
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+ | The distance between them is: |
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+ | \{ $length->ans_rule() \} |
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$BR |
$BR |
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The midpoint of the line segment |
The midpoint of the line segment |
Latest revision as of 08:42, 28 June 2023
This problem has been replaced with a newer version of this problem
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 |
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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))); # 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, 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. |