Testwiki:Reference desk/Archives/Mathematics/2009 November 17

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November 17

please explain

A problem that runs as follows:

 "If the points (2,-3), (4,3), and (5,k/2) are on the same straight line, then k equals "  

has been solved in the following manner:

 SOLN:  Interpolate and exterpolate: (2,-3), (3,0), (4,3), (5,6) From this k=12.

Please explain me in detail, what do the words "Interpolate" and "Exterpolate" mean. Also and how the four points (2,-3), (3,0), (4,3), (5,6) have been arrived at by those operations Interpolation and Exterpolation Kasiraoj (talk) 10:39, 17 November 2009 (UTC)

See interpolation and extrapolation. "Exterpolation" is only a typo; hopefully such a word does not exist . --pma (talk) 11:33, 17 November 2009 (UTC)
Maybe such a word should exist... The interplay between "internal" and "external" suggests "interpolation" and "exterpolation". To the OP: "Extrapolation" is correct - don't believe me in this case! ;) --PST 13:21, 17 November 2009 (UTC)
If (2, -3) and (4, 3) are points on a particular line, one notices that the slope of the line is given by
m=change in ychange in x=ΔyΔx
or numerically in this case, 3(3)42=62=3. Therefore, any change in x by one unit, will result in a change in y by three units. Since x=4 corresponds to y=3, x=5 corresponds to y=6, and consequently k=12. In this instance, extrapolation refers to "extending x by one unit (from x=4 to x=5)" to obtain the value of y when x=5, knowing the value of y when x=4. To do this of course, you must know the slope of the line (that is, how fast y changes relative to x), and thus the initial point, (2, -3), is required. Hope this helps. --PST 13:17, 17 November 2009 (UTC)
Another way to solve this is to graph it. Start by plotting the 2 points they gave you on graph paper:
  Y
 6↑               
 5|              
 4|              
 3|           ¤     
 2|               
 1|               
 0+--------------> X
-1|  1  2  3  4  5
-2|      
-3↓     ¤
Then draw a line with a ruler, extending it upwards, and find the other points:
  Y
 6↑              ¤
 5|             /
 4|            / 
 3|           ¤     
 2|          /    
 1|         /     
 0+--------¤-----> X
-1|  1  2 /3  4  5
-2|      /
-3↓     ¤
  
Note that the line will appear more vertical for you, since I spaced it out more in the X direction than the Y. Once you have the coords of the point (5,6), you can find K. StuRat (talk) 13:26, 18 November 2009 (UTC)