(-1.5, 5) * A (1, -1)

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2 CS1-1 Consider the following trail that is the result of connecting a sequence of twelve trail-points.. Four such test-points are shown; they are labeled A, B, C, and D. The dashed lines indicate the nearest distance from each of these points over to the trail. (-6, 3) (-2.2, 4) (-3, 3.5) (-1.5, 5) * A (-1, 3) Y (7, 5.1) * C (-7, -1) B * (-3, -1) (1, -1) (6, -.5) X * D (7, -.5) (-7, -5) Fig. CS1-1 (-6.5, -7) (3, -6.5) (5, -7) (7, -5) We will begin by creating two text files. The first file will be named and will contain the sequence of 12 points as follows. (Be sure to enter parenthesis and commas.) (-6.5, -7) (-7, -5) (-7, -1) (-6, 3) (-3, 3.5) (-2.2, 4) (-1, 3) (1, -1) (3, -6.5) (5, -7) (7, -5) (6, -.5)

3 The second text file will be called and will contain data exactly as shown by the following (notice a space following the letters): A (-1.5, 5) B (-3, -1) C (7, 5.1) D (7, -.5) The final output of this program will be: Test point A distance to trail >>> Test point B distance to trail >>> Test point C distance to trail >>> Test point D distance to trail >>> ********************************************************************** CS1-2 This is a fairly long, sophisticated project and the secret to success is not to bite off too much at once. Do a small, fundamental part of the code, test it, and revise as necessary. Then do a little more, test, and get that part working too. This will be our approach here. The first thing we must do is create a project. Let s call our project and have it include a class called. To make things go a little faster, we will paste in the contents of the class developed in Lesson 27 and use that code to input the file. After creating two arrays, and (both dimensioned to a length of 12), strip off the parenthesis and commas and store the coordinates in. Similarly, store the coordinates in. Use the following temporary code for testing: for(int j = 0; j < 12; j++) { System.out.println(trailX[j] +, + traily[j] + ); } System.out.println( ); The output of this test should look like this: -6.5, , , , , , , , , , , , -0.5

4 The code for the project up to this point (and including the above test) can be found in the Blue Pelican Java Answer Book in the Case Study section titled. *************************************************************************** So far, so good. Next, we will bring in the file and store its parts in three different arrays. Remove the previous test code and add new code to that will create the following arrays with each dimensioned to a length of 4:. CS1-3 Now write code that will bring in this file and separate the parts of each line of text and store each in one of the new arrays just created. Use the following test code to verify that this section of the code is working: for(int j = 0; j < 4; j++) { System.out.println(testLetter[j] + + testx[j] +, + testy[j]); } System.out.println( ); The output of this test should appear as follows: A -1.5, 5.0 B -3.0, -1.0 C 7.0, 5.1 D 7.0, -0.5 The code for the project up to this point (including the test just above) can be found in the Blue Pelican Java Answer Book in the section titled. ************************************************************************** Now we come to the major part of the code for this project. This code will go in another class called. Briefly, the class can be described by the following list. At this point, do not try to implement any of this. Just scan the list and become somewhat familiar with the methods and state variables. Implementation will come later, step-by-step. 1. A constructor receives four parameters that represent the coordinates of the two. a. These four parameters are assigned to the state variables, seg,, and. b. Use the coordinates of the end points of the line segment to determine the equation of the line in Ax + By + C = 0 form. i. Be sure to handle the special case in which the line is vertical. ii. Store,, and in state variables of the same name. 2. Method

5 CS1-4 a. The parameters and represent the coordinates of a not necessarily on our line. b. Return the distance from the point to the line described by Ax + By + C = 0. This involves the use of a formula from Analytic Geometry that will be presented a little later. 3. Method a. This method tests to see if the perpendicular projection of (, tpy) onto line Ax + By + C = 0 falls on the segment defined by. b. Returns if on the segment. c. Returns if not on the segment. This analysis is somewhat complicated and will be explored later. 4. Create the following state variables: segx1, segy1, segx2, segy2, A, B, C Ok, time to get busy and start building the class. In your project, create the skeleton of the class. Create the state variables. Next, create part of the constructor and assign the parameters to the state variables,,, and. In the Answer Book this code is labeled as. *************************************************************************** Your next task is to finish the constructor by using the parameters to generate the equation of the line, thus producing,, and. Be very careful here. You should not immediately calculate the slope of the line because it may be infinite. Instead, find out first if it is infinite by testing the denominator of the slope formula ( m = (y2 - y1)/(x2 - x1) ). This test is: if( (x2 x1) = = 0 ) In fact, we can get in trouble if the difference between and is very, very small, but still nonzero. It is suggested that you use the following test instead: if( Math.abs(x2-x1) < ) Some IDE s like BlueJ will let your directly test your class without having to create test code in of the class. The table below shows the final values of,, and after passing the test parameters,,, and to the constructor. If your IDE does not permit such testing, you will need to hard code these tests into of the class.

6 Test Values Results x1 y1 x2 y2 A B C LineStuffPart 4 disttoline method disttoline (tpx, tpy) disttolinelinestuff disttoline Arguments Sent to Constructor Test-Point disttoline Point #1 Point #2 x y x y x y double Part 5

7 When projecting a test-point over to a line, there are two distinct cases. First, consider the scenario to the right in which a test-point projects over to a line and falls in the of the segment originally defining the line. In this case, the point of projection is the segment and the method should return a. (tpx, tpy) (segx1, segy1) CS1-6 (segx2, segy2) Point of projection (projx, projy) Fig. CS1-3 Projection the segment (segx1, segy1) (segx2, segy2) (tpx, tpy) Second, the test-point projects over to the line so as to fall the line segment originally defining the line. An example of this scenario is found in the drawing to the left. The method should return a. Point of projection (projx, projy) Fig. CS1-4 Projection on the line How can we distinguish between these two situations? This is accomplished by between the three points on the line. Let s examine these distances for the case when the point of projection falls the line segment. Fig CS1-5 shows three distances. The relative sizes of these distances will be used to verify that the point of projection does, indeed, fall on the line segment. (segx1, segy1) d1p (tpx, tpy) d12 d2p (projx, projy) (segx2, segy2) Fig. CS1-5 Point of projection falls in the of the line segment. From Fig. CS1-5 we see that the condition for the point of projection to fall in the of a line segment is for to be less than or equal to. This would result in the method returning a. You will need to calculate the distances,, and. Use the following distance formula to calculate, for example, the distance between (x 1, y 1 ) and (x 2, y 2 ). dist = (x 2 x 1 ) 2 + (y 2 y 1 ) 2

8 CS1-7 Now lets examine the case when the point of projection falls the line segment originally defining the line. (segx1, segy1) d12 d1p d2p (segx2, segy2) (tpx, tpy) (projx, projy) Fig CS1-6 Point of projection falls on the of the line segment. From Fig. CS1-6 we see that the condition for the point of projection to fall on the of the line segment is for to be greater than. This would result in the method returning a. This is all well and good; however, there is still one major obstacle. How do we find the point of projection? Very succinctly, here is how it s done. In Fig CS1-7 we note that the two lines labeled and are perpendicular (their slopes are negative reciprocals of each other). That will help us obtain the equation of. The equation for is already known; using the state variables,, and it is Ax + By + C = 0. line2 line1 (segx1, segy1) (tpx, tpy) (projx, projy) (segx2, segy2) Fig CS1-7 and are solved simultaneously to find the point of projection After finding the equation of, solve the two lines simultaneously as follows to find the desired intersection point : Assuming that the equation of is of the form (A1)x + (B1)y + C1 = 0, the solutions are: = A1(B) A(B1) x = [ -C1(B) + B1(C) ] / y = [ -C(A1) + A(C1) ] / The solution here,, is the desired intersection point. Following is a flow chart that should prove useful in putting all these ideas together.

9 onsegment onsegmentlinestuff onsegment Arguments Sent to Constructor Test-Point onsegment Point #1 Point #2 x y x y x y boolean onsegmentpart6 LineStuff Tester either or all onsegment disttolinedouble dist[ ]distarraycounter

10 4. Cycle through all twelve trail-points and determine the distance from the test-point to each. Store each of these distances in the array and increment each time. 5. Sort the array. 6. The first item in the array is the desired shortest distance. CS Produce output for each iteration of the loop described by item 1 above. The final output should appear as follows: Test point A distance to trail >>> Test point B distance to trail >>> Test point C distance to trail >>> Test point D distance to trail >>> The implementation of the code for the above seven steps is labeled in the Answer Book as. Following that is the complete code for the class. On the next page is a flow chart that is the equivalent of the above seven steps: ************************************************************************* Following is a practical application of a project such as this: In GIS (Geographical Information System) software there might be a trail of points representing a road or perhaps a pipeline. As a mouse pointer is moved across a map containing such a trail, we could repeatedly call a method implementing the ideas of this project to continuously show the distance from the mouse pointer to the trail.

11 CS1-10 Get next test-point Fill dist[ ] with large numbers Get next segment Get projection of test-point on seg. Proj on Segment? no yes Get dist of test-point to line Store distance in dist[ ] Set distcounter = 0 no Finished with all segs.? Increment distcounter yes Get next trail-pt Get dist of trailpoint to test-point Store in dist[ ] and increment distcounter no Finished with trail-points? yes Sort dist[ ] array Get distance in index 0 of array no Finished with test-points? Use it to produce printout yes end Fig. CS1-9 Flow chart for determining nearest point.

.(3, 2) Co-ordinate Geometry Co-ordinates. Every point has two co-ordinates. Plot the following points on the plane. A (4, 1) D (2, 5) G (6, 3)

.(3, 2) Co-ordinate Geometry Co-ordinates. Every point has two co-ordinates. Plot the following points on the plane. A (4, 1) D (2, 5) G (6, 3) Co-ordinate Geometry Co-ordinates Every point has two co-ordinates. (3, 2) x co-ordinate y co-ordinate Plot the following points on the plane..(3, 2) A (4, 1) D (2, 5) G (6, 3) B (3, 3) E ( 4, 4) H (6,