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

= 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! ;) -- PS T  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
 * $$m={\mbox{change in } y \over \mbox{change in } x} = {\Delta y \over{\Delta x}}$$
 * or numerically in this case, $$\frac{3-(-3)}{4-2} = \frac{6}{2} = 3$$. 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. -- PS T  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)