advertisement

1 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 TASK 2.13.3: LOOK AGAIN! Solutions Suppose we are given two points in a plane, A = (a, b) and B = (c, d). 1. By changing the second coordinate in A from “b” to “m”, locate a new point C = (a, m) such that the slope of the line through the new point and B is double that of the line through A and B. What do you notice about the difference in the first coordinates of C and B and the difference in the first coordinates of A and B? First note that the slope of the line through A and B is C = (a, m) so the slope of the line through C and B is d!b . We want to find a point c!a d!m c! a . Note: We are looking for the value of m. Since we want to double the original slope, we get ( ) . This reduces d!m d !b =2 c! a c!a to d ! m = 2 ( d !b) and so m = 2b ! d and C = (a, 2b-d). Participants should notice that the difference between the x-values of points C and B is twice the difference between the xvalues of points A and B. We must now find the y-intercept of the lines AB and CB. Starting with the line AB, let I be the y-intercept. Then " d ! b% y=$ x+I # c ! a '& Substitute the coordinates of point B for x and y. " d ! b% d=$ (c) + I # c ! a '& " d ! b% I = d !$ (c) # c ! a '& Now for line CB, let I’ be the y-intercept. Then " d ! b% y = 2$ x+ I' # c ! a '& Substitute the coordinates of point B for x and y. " d ! b% d = 2$ (c) + I ' # c ! a '& " d ! b% I ' = d ! 2$ (c) # c ! a '& November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board. 2 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 Notice that in I’, we are subtracting from d twice the amount that is being subtracted from d in I. 2. By changing the second coordinate in A from “b” to “p”, locate a new point D = (a, p) such that the slope of the line through the new point and B is triple that of the line through A and B. What do you notice about the difference in the first coordinates of D and B and the difference in the first coordinates of A and B? This time we want to find the point D=(a, p) so that the slope of the line through D and B is " d ! b% d!p three times that of the line through A and B. So we want . This reduces to = 3$ c!a # c ! a '& d ! p = 3(d ! b) and p = 3b ! 2d . Therefore D=(a, 3b-2d). Participants should note that the difference in the x-values is three times that of the difference of the x-values of the points A and B. We must now find the y-intercept of the lines AB and DB. From before, we know that the yintercept of AB is d!b I=d! (c) c!a Now for line DB, let I’ be the y-intercept. Then " d ! b% y = 3$ x+ I' # c ! a '& Substitute the coordinates of point B for x and y. " d ! b% d = 3$ (c) + I ' # c ! a '& " d ! b% I ' = d ! 3$ (c) # c ! a '& Notice that in I’, we are subtracting from d three times that of the amount that is being subtracted from d in I. 3. By changing the second coordinate in A from “b” to “r”, locate a new point E = (a, r) such that the slope of the line through the new point and B is four times that of the line through A and B. What do you notice about the difference in the first coordinates of E and B and the difference in the first coordinates of A and B? November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board. 3 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 This time we want to find the point E=(a, r) so that the slope of the line through E and B is " d ! b% d!r four times that of the line through A and B. So we want . This reduces to = 4$ c!a # c ! a '& d ! r = 4(d ! b) and r = 4b ! 3d . Therefore D=(a, 4b-3d). Participants should note that the difference in the x-values is three times that of the difference of the x-values of the points A and B. We must now find the y-intercept of the lines AB and DB. From before, we know that the y-intercept of AB is " d ! b% I = d !$ (c) # c ! a '& Now for line DB, let I’ be the y-intercept. Then " d ! b% y = 4$ x+ I' # c ! a '& Substitute the coordinates of point B for x and y. " d ! b% d = 4$ (c) + I ' # c ! a '& " d ! b% I ' = d ! 4$ (c) # c ! a '& Notice that in I’, we are subtracting from d four times that of the amount that is being subtracted from d in I. 4. Let F be a point created by changing only the second coordinate of A. Make a conjecture about the location of a point F such that the slope of the line through F and B is n times the slope of the line through A and B where n is a natural number. Explain your reasons for choosing this point. Participants may conjecture that based on the previous exercises, the difference in the xvalues of the points F and B is n times that of the points A and B. Thus the difference will be n*(b – d). The point will be F = a, nb ! (n ! 1)d . To show that this is true, we can ( ) calculate the slope of the line between F and B. " d ! b% d ! (nb ! (n ! 1)d) !nb + nd n(d ! b) m= = = = n*$ c!a c!a c!a # c ! a '& Thus the slope of the line between F and B is n times that of the line between A and B. 5. Make a conjecture about the y-intercept of the line FB. Explain your reasons for choosing this value and then show that your choice was correct. November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board. 4 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 Participants should conjecture that in the y-intercept of FB, we are subtracting from d, n times the amount that is being subtracted from d in the y-intercept of AB. To show this, we must find the y-intercept of the lines AB and DB. From before, we know that the y-intercept of AB is " d ! b% I = d !$ (c) # c ! a '& Now for line FB, let I’ be the y-intercept. Then " d ! b% y = n$ x+ I' # c ! a '& Plugging in point B, we get " d ! b% d = n$ (c) + I ' # c ! a '& " d ! b% I ' = d ! n$ (c) # c ! a '& Notice that in I’, we are subtracting from d, n times the amount that is being subtracted from d in I. 6. What pattern did you notice in the y-intercepts as the slope was changed? (If you didn’t notice any pattern, go back and make sure that you were always consistent about which point you used to find the equation of the line. Try always using the point B.) Explain what created the pattern. If the point B is on the y-axis, what changes? " d ! b% The y-intercept of the original line is I = d ! $ (c) . # c ! a '& The y-intercept of the new line that was created by changing the slope by an integer multiple " d ! b% (say k) ends up being I = d ! (k) $ (c) . This is because of the fact that the y-intercept # c ! a '& is exactly y2 ! mx2 , where B=(x2, y2). This point is consistent throughout the process. All that is changing is the value of m which is being multiplied by k. If B is on the y-axis, then B is also the y-intercept of the new line. We can see that because B = (0, d) making I, the intercept of the new line, equal to d. November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board. 5 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 Teaching notes In this activity, it is essential that the participants always use the point B as the point used to determine the y-intercept. Failure to do so will make it much harder to observe the relationship that is sought. November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board. 6 Algebra I: Strand 2. Linear Functions; Topic 13. Scavenger Hunt; Task 2.13.3 TASK 2.13.3: LOOK AGAIN! Note: In this activity, we are specifically looking for the pattern that occurs naturally. The “Big Idea” is what’s important here. Suppose we are given two points in a plane, A = (a, b) and B = (c, d). 1. By changing the second coordinate in A from “b” A to “m”, locate a new point C = (a, m) such that the slope of the line through the new point and B is double that of the line through A and B. What do you notice about the difference in the first coordinates of C and B and the difference in the first coordinates of A and B? How does the y-intercept of the line AB relate to the y-intercept of the line CB? 2. By changing the second coordinate in A from “b” to “p”, locate a new point D = (a, p) such that the slope of the line through the new point and B is triple that of the line through A and B. What do you notice about the difference in the first coordinates of D and B and the difference in the first coordinates of A and B? How does the y-intercept of the line AB relate to the y-intercept of the line DB? 3. By changing the second coordinate in A from “b” to “r”, locate a new point E = (a, r) such that the slope of the line through the new point and B is four times that of the line through A and B. What do you notice about the difference in the first coordinates of E and B and the difference in the first coordinates of A and B? How does the y-intercept of the line AB relate to the y-intercept of the line EB? 4. Let F be a point created by changing only the second coordinate of A. Make a conjecture about the location of point F given that the slope of the line through F and B is n times the slope of the line through A and B where n is a natural number. Explain your reasons for choosing this point. 5. Make a conjecture about the y-intercept of the line FB. Explain your reasons for choosing this value and then show that your choice was correct. 6. What pattern did you notice in the y-intercepts as the slope was changed? (If you didn’t notice any pattern, go back and make sure that you were always consistent about which point you used to find the equation of the line. Try always using the point B.) Explain what created the pattern. If the point B is on the y-axis, what changes? November 22, 2004. Ensuring Teacher Quality: Algebra I, produced by the Charles A. Dana Center at The University of Texas at Austin for the Texas Higher Education Coordinating Board.