# 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.

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 - \$point)->value; # \$d1 = x1 - x2, \$d2 = y1 - y2

\$length = Compute("sqrt( (\$d1)^2+(\$d2)^2 )");
\$mid = ( \$point + \$point ) / 2;
```

In the problem setup section of the file, we don't use the Vector context and `norm( \$point - \$point )` to determine length because we don't want to accept an answer like `|<5,7>-<7,8>|` You could use `\$length=norm( \$point - \$point );` 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 \( \$point \) and \( \$point \).
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 );
```