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Advanced Line Geometry Formulae - By Jenny Harrow

How to use this applet:

Click on either of the buttons on the start panel to see a further explanation of either the area of a triangle or the perpendicular distance from a point to a line. If the area of a triangle button is pressed you will see a new panel. To enter points into the applet, click on the graph. The reset button will reset the numbers and allow you to enter more and the back button will bring you back to the start panel again, allowing you to select another option. If the perpendicular distance from a point to a line button is pressed it will again bring up another screen. Enter numbers for a point and a line into the text fields and press the enter button. The answer will be displayed in the answer text field as a number. If the answer is a square root the actual number is calculated! Again the reset button clears the text fields and allows you to enter more numbers and the back button brings you back to the start panel.

Notes on the maths used in the applet:

Area of a triangle Formula: 1/2 |x1y2– x1y2| To find the area of a triangle we first need three points in the form (x, y). One of these points must be at (0, 0), the origin. If this is not the case take one of the points and transpose it to the origin. Lets take the third point and transpose it to the origin: (e, f) --> (0, 0) (a, b) --> (a-e, b-f) (c, d) --> (c-e, d-f) Now we have our three points (0, 0), (a’, b’) , (c’, d’) Re-label the last two point (x1, y2), (x1, y2) and substitute the numbers into the formula. The two vertical lines | | represent the modulus, meaning take the positive of the number generated. The reason for doing this is because, obviously, a triangle cannot have a negative area. Perpendicular distance from a point to a line Formula:
    |ax1=by1+c|    

    
 ____________
Öa2+b2
   
 
This formula represents the distance d from the point (x1, y1) to the line ax + by +c =0 The first thing to do is substitute x1 & y1 from the point into the equation of the line. The two vertical lines | | represent the modulus, meaning take the positive of the number generated. a is the coefficient of the x variable and b is the coefficient of the y variable. Square both numbers…and take the square of their sum.Substitute the numbers into the formula and viola, you have the perpendicular distance from a point to a line.